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Corollary 7.26 (Topological Invariance of \( {\pi }_{1} \) ). Homeomorphic spaces have isomorphic fundamental groups. Specifically, if \( \varphi : X \rightarrow Y \) is a homeomorphism, then \( {\varphi }_{ * } : {\pi }_{1}\left( {X, p}\right) \rightarrow {\pi }_{1}\left( {Y,\varphi \left( p\right) }\right) \) is an isomorphism.
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Proof. If \( \varphi \) is a homeomorphism, then \( {\left( {\varphi }^{-1}\right) }_{ * } \circ {\varphi }_{ * } = {\left( {\varphi }^{-1} \circ \varphi \right) }_{ * } = {\left( {\operatorname{Id}}_{X}\right) }_{ * } = \) \( {\operatorname{Id}}_{{\pi }_{1}\left( {X, p}\right) } \), and similarly \( {\varphi }_{ * } \circ {\left( {\varphi }^{-1}\right) }_{ * } \) is the identity on \( {\pi }_{1}\left( {Y,\varphi \left( p\right) }\right) \) .
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Yes
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Proposition 7.28. Suppose \( A \) is a retract of \( X \) . If \( r : X \rightarrow A \) is any retraction, then for any \( p \in A,{\left( {\iota }_{A}\right) }_{ * } : {\pi }_{1}\left( {A, p}\right) \rightarrow {\pi }_{1}\left( {X, p}\right) \) is injective and \( {r}_{ * } : {\pi }_{1}\left( {X, p}\right) \rightarrow \) \( {\pi }_{1}\left( {A, p}\right) \) is surjective.
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Proof. Since \( r \circ {\iota }_{A} = {\operatorname{Id}}_{A} \), the composition \( {r}_{ * } \circ {\left( {\iota }_{A}\right) }_{ * } \) is the identity on \( {\pi }_{1}\left( {A, p}\right) \) , from which it follows that \( {\left( {\iota }_{A}\right) }_{ * } \) is injective and \( {r}_{ * } \) is surjective.
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Yes
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Corollary 7.29. A retract of a simply connected space is simply connected.
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Proof. If \( A \) is a retract of \( X \), the previous proposition shows that \( {\left( {\iota }_{A}\right) }_{ * } : {\pi }_{1}\left( {A, p}\right) \rightarrow \) \( {\pi }_{1}\left( {X, p}\right) \) is injective. Thus if \( {\pi }_{1}\left( {X, p}\right) \) is trivial, so is \( {\pi }_{1}\left( {A, p}\right) \) .
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Yes
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For any \( n \geq 1 \), it is easy to check that the map \( r : {\mathbb{R}}^{n} \smallsetminus \{ 0\} \rightarrow {\mathbb{S}}^{n - 1} \) given by \( r\left( x\right) = x/\left| x\right| \) is a retraction.
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Because \( {\mathbb{S}}^{1} \) is not simply connected, it follows from Corollary 7.29 that \( {\mathbb{R}}^{2} \smallsetminus \{ 0\} \) is not simply connected. Thus \( {\mathbb{R}}^{2} \smallsetminus \{ 0\} \) is not homeomorphic to \( {\mathbb{R}}^{2} \) .
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No
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Proposition 7.34 (Fundamental Group of a Product). If \( {X}_{1},\ldots ,{X}_{n} \) are any topological spaces, the map \( P \) defined by (7.2) is an isomorphism.
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Proof. First we show that \( P \) is surjective. Let \( \left\lbrack {f}_{i}\right\rbrack \in {\pi }_{1}\left( {{X}_{i},{x}_{i}}\right) \) be arbitrary for \( i = 1,\ldots, n \) . Define a loop \( f \) in the product space by \( f\left( s\right) = \left( {{f}_{1}\left( s\right) ,\ldots ,{f}_{n}\left( s\right) }\right) \) . Since the component functions of \( f \) satisfy \( {f}_{i} = {p}_{i} \circ f \), we compute \( P\left\lbrack f\right\rbrack = \) \( \left( {{p}_{1 * }\left\lbrack f\right\rbrack ,\ldots ,{p}_{n * }\left\lbrack f\right\rbrack }\right) = \left( {\left\lbrack {{p}_{1} \circ f}\right\rbrack ,\ldots ,\left\lbrack {{p}_{n} \circ f}\right\rbrack }\right) = \left( {\left\lbrack {f}_{1}\right\rbrack ,\ldots ,\left\lbrack {f}_{n}\right\rbrack }\right) . \n\nTo show injectivity, suppose \( f \) is a loop in the product space such that \( P\left\lbrack f\right\rbrack \) is the identity element of \( {\pi }_{1}\left( {{X}_{1},{x}_{1}}\right) \times \cdots \times {\pi }_{1}\left( {{X}_{n},{x}_{n}}\right) \) . Writing \( f \) in terms of its component functions as \( f\left( s\right) = \left( {{f}_{1}\left( s\right) ,\ldots ,{f}_{n}\left( s\right) }\right) \), the hypothesis means that \( \left\lbrack {c}_{{x}_{i}}\right\rbrack = {p}_{{i}_{ * }}\left\lbrack f\right\rbrack = \left\lbrack {{p}_{i} \circ f}\right\rbrack = \left\lbrack {f}_{i}\right\rbrack \) for each \( i \) . If we choose homotopies \( {H}_{i} : {f}_{i} \sim {c}_{{x}_{i}} \) , it follows easily that the map \( H : I \times I \rightarrow {X}_{1} \times \cdots \times {X}_{n} \) given by\n\n\[ H\left( {s, t}\right) = \left( {{H}_{1}\left( {s, t}\right) ,\ldots ,{H}_{n}\left( {s, t}\right) }\right) \]\n\n is a homotopy from \( f \) to the constant loop \( {c}_{\left( {x}_{1},\ldots ,{x}_{n}\right) } \) .
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Yes
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Proposition 7.37. For any \( n \geq 1,{\mathbb{S}}^{n - 1} \) is a strong deformation retract of \( {\mathbb{R}}^{n} \smallsetminus \{ 0\} \) and of \( {\overline{\mathbb{B}}}^{n} \smallsetminus \{ 0\} \) .
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Proof. Define a homotopy \( H : \left( {{\mathbb{R}}^{n}\smallsetminus \{ 0\} }\right) \times I \rightarrow {\mathbb{R}}^{n} \smallsetminus \{ 0\} \) by\n\n\[ H\left( {x, t}\right) = \left( {1 - t}\right) x + t\frac{x}{\left| x\right| }.\]\n\nThis is just the straight-line homotopy from the identity map to the retraction onto the sphere (Fig. 7.13). The same formula works for \( {\overline{\mathbb{B}}}^{n} \smallsetminus \{ 0\} \) .
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Yes
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Let \( X \) be any space. If the identity map of \( X \) is homotopic to a constant map, we say that \( X \) is contractible. Other equivalent definitions are that any point of \( X \) is a deformation retract of \( X \), or \( X \) is homotopy equivalent to a one-point space (Exercise 7.42). Concretely, contractibility means that there exist a point \( p \in X \) and a continuous map \( H : X \times I \rightarrow X \) such that\n\n\[ H\left( {x,0}\right) = x\text{ for all }x \in X;\;H\left( {x,1}\right) = p\text{ for all }x \in X. \]
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In other words, the whole space \( X \) can be continuously shrunk to a point. Some simple examples of contractible spaces are convex subsets of \( {\mathbb{R}}^{n} \), and, more generally, any subset \( B \subseteq {\mathbb{R}}^{n} \) that is star-shaped, which means that there is some point \( {p}_{0} \in B \) such that for every \( p \in B \), the line segment from \( {p}_{0} \) to \( p \) is contained in \( B \) . Since a one-point space is simply connected, it follows that every contractible space is simply connected.
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Yes
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Lemma 7.45. Suppose \( \varphi ,\psi : X \rightarrow Y \) are continuous, and \( H : \varphi \simeq \psi \) is a homotopy. For any \( p \in X \), let \( h \) be the path in \( Y \) from \( \varphi \left( p\right) \) to \( \psi \left( p\right) \) defined by \( h\left( t\right) = H\left( {p, t}\right) \) , and let \( {\Phi }_{h} : {\pi }_{1}\left( {Y,\varphi \left( p\right) }\right) \rightarrow {\pi }_{1}\left( {Y,\psi \left( p\right) }\right) \) be the isomorphism defined in Theorem 7.13. Then the following diagram commutes:
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Proof. Let \( f \) be any loop in \( X \) based at \( p \) . What we need to show is\n\n\[ \n{\psi }_{ * }\left\lbrack f\right\rbrack = {\Phi }_{h}\left( {{\varphi }_{ * }\left\lbrack f\right\rbrack }\right) \n\]\n\n\[ \n\Leftrightarrow \psi \circ f \sim \bar{h} \cdot \left( {\varphi \circ f}\right) \cdot h \n\]\n\n\[ \n\Leftrightarrow h \cdot \left( {\psi \circ f}\right) \sim \left( {\varphi \circ f}\right) \cdot h. \n\]\n\nThis follows easily from the square lemma applied to the map \( F : I \times I \rightarrow Y \) defined by \( F\left( {s, t}\right) = H\left( {f\left( s\right), t}\right) \) (Fig. 7.15).
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Yes
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Proposition 7.46. With notation as above, if \( f \) is a homotopy equivalence, then \( \widetilde{Y} \) and \( \widetilde{X} \) are deformation retracts of \( {Z}_{f} \) . Thus two spaces are homotopy equivalent if and only if they are both homeomorphic to deformation retracts of a single space.
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Proof. For any \( \left( {x, s}\right) \in X \times I \), let \( \left\lbrack {x, s}\right\rbrack = q\left( {x, s}\right) \) denote its equivalence class in \( {Z}_{f} \) ; similarly, \( \left\lbrack y\right\rbrack = q\left( y\right) \) is the equivalence class of \( y \in Y \). First we show that \( \widetilde{Y} \) is a strong deformation retract of \( {Z}_{f} \), assuming only that \( f \) is continuous. We define a retraction \( A : {Z}_{f} \rightarrow {Z}_{f} \), which collapses \( {Z}_{f} \) down onto \( \widetilde{Y} \), by \[ A\left\lbrack {x, s}\right\rbrack = \left\lbrack {x,0}\right\rbrack \] \[ A\left\lbrack y\right\rbrack = \left\lbrack y\right\rbrack \] To be a bit more precise, we should define a map \( \widetilde{A} : Y \coprod \left( {X \times I}\right) \rightarrow {Z}_{f} \) by \( \widetilde{A}\left( {x, s}\right) = \left\lbrack {x,0}\right\rbrack \) and \( \widetilde{A}\left( y\right) = \left\lbrack y\right\rbrack \) . This map is evidently continuous because its restrictions to \( X \times I \) and \( Y \) are compositions of continuous maps. Because \( \widetilde{A}\left( {x,0}\right) = \left\lbrack {f\left( x\right) }\right\rbrack = \widetilde{A}\left( {f\left( x\right) }\right) ,\widetilde{A} \) respects the identifications made by \( q \), so it passes to the quotient to yield the continuous map \( A \) defined above. In this proof, we use this kind of standard argument repeatedly to show that a map from \( {Z}_{f} \) is continuous; we generally abbreviate it by saying something like \
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Yes
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Example 7.47 (Categories). Here are some familiar examples of categories, which we describe by specifying their objects and morphisms. In each case, the source and target of a morphism are its domain and codomain, respectively; the composition law is given by composition of maps; and the identity morphism is the identity map.
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In each case, the verification of the axioms of a category is straightforward. The main point is to show that a composition of the appropriate structure-preserving maps again preserves the structure. Associativity is automatic because it holds for composition of maps.
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No
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Theorem 7.51. For any categories \( \mathrm{C} \) and \( \mathrm{D} \), every (covariant or contravariant) functor from \( \mathrm{C} \) to \( \mathrm{D} \) takes isomorphisms in \( \mathrm{C} \) to isomorphisms in \( \mathrm{D} \) .
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Proof. The proof is exactly the same as the proof for the fundamental group functor (Corollary 7.26).
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No
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Theorem 7.54. If a product exists in any category, it is unique up to a unique isomorphism that respects the projections. More precisely, if \( \left( {P,\left( {\pi }_{\alpha }\right) }\right) \) and \( \left( {{P}^{\prime },\left( {\pi }_{\alpha }^{\prime }\right) }\right) \) are both products of the family \( \left( {X}_{\alpha }\right) \), there is a unique isomorphism \( f : P \rightarrow {P}^{\prime } \) satisfying \( {\pi }_{\alpha }^{\prime } \circ f = {\pi }_{\alpha } \) for each \( \alpha \) .
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Proof. Given \( \left( {P,\left( {\pi }_{\alpha }\right) }\right) \) and \( \left( {{P}^{\prime },\left( {\pi }_{\alpha }^{\prime }\right) }\right) \) as in the statement of the theorem, the defining property of products guarantees the existence of unique morphisms \( f : P \rightarrow {P}^{\prime } \) and \( {f}^{\prime } : {P}^{\prime } \rightarrow P \) satisfying \( {\pi }_{\alpha }^{\prime } \circ f = {\pi }_{\alpha } \) and \( {\pi }_{\alpha } \circ {f}^{\prime } = {\pi }_{\alpha }^{\prime } \) . If we take \( W = P \) and \( {f}_{\alpha } = {\pi }_{\alpha } \) in the diagram above, then the diagram commutes with either \( {f}^{\prime } \circ f \) or \( {\operatorname{Id}}_{P} \) in place of \( f \) . By the uniqueness part of the defining property of the product, it follows that \( {f}^{\prime } \circ f = {\operatorname{Id}}_{P} \) . A similar argument shows that \( f \circ {f}^{\prime } = {\operatorname{Id}}_{{P}^{\prime }} \) .
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Yes
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Proposition 8.1. Each point \( z \in {\mathbb{S}}^{1} \) has a neighborhood \( U \) with the following property (see Fig. 8.1):\n\n\( {\varepsilon }^{-1}\left( U\right) \) is a countable union of disjoint open intervals\n\n\( {\widetilde{U}}_{n} \) with the property that \( \varepsilon \) restricts to a homeomor-\n\n(8.1)\n\nphism from \( {\widetilde{U}}_{n} \) onto \( U \) .
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Proof. This is just a straightforward computation from the definition of \( \varepsilon \) . We can cover \( {\mathbb{S}}^{1} \) by the four open subsets\n\n\[ \n{X}_{ + } = \{ \left( {x + {iy}}\right) : x > 0\} ,\;{Y}_{ + } = \{ \left( {x + {iy}}\right) : y > 0\} , \n\]\n\n(8.2)\n\n\[ \n{X}_{ - } = \{ \left( {x + {iy}}\right) : x < 0\} ,\;{Y}_{ - } = \{ \left( {x + {iy}}\right) : y < 0\} . \n\]\n\nThe preimage of each of these sets is a countable union of disjoint open intervals of length \( \frac{1}{2} \), on each of which \( \varepsilon \) has a continuous local inverse. For example, \( {\varepsilon }^{-1}\left( {X}_{ + }\right) \) is the union of the intervals \( \left( {n - \frac{1}{4}, n + \frac{1}{4}}\right) \) for \( n \in \mathbb{Z} \), and for each of these intervals, a continuous local inverse \( {\varepsilon }^{-1} : {X}_{ + } \rightarrow \left( {n - \frac{1}{4}, n + \frac{1}{4}}\right) \) is given by\n\n\[ \n{\varepsilon }^{-1}\left( {x + {iy}}\right) = n + \frac{1}{2\pi }{\sin }^{-1}y \n\]\n\nOther local inverses are given by\n\n\[ \n{\varepsilon }^{-1}\left( {x + {iy}}\right) = n + \frac{1}{2} - \frac{1}{2\pi }{\sin }^{-1}y \n\]\n\n\[ \n{\varepsilon }^{-1}\left( {x + {iy}}\right) = n + \frac{1}{2\pi }{\cos }^{-1}x \n\]\n\non \( {Y}_{ + } \),\n\n\[ \n{\varepsilon }^{-1}\left( {x + {iy}}\right) = n - \frac{1}{2\pi }{\cos }^{-1}x \n\]\non \( {Y}_{ - } \).\n\nSince every point of \( {\mathbb{S}}^{1} \) is in at least one of the four sets listed in (8.2), this completes the proof.
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Yes
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Corollary 8.2 (Local Section Property of the Circle). Let \( U \subseteq {\mathbb{S}}^{1} \) be any evenly covered open subset. For any \( z \in U \) and any \( r \) in the fiber of \( \varepsilon \) over \( z \), there is a local section \( \sigma \) of \( \varepsilon \) over \( U \) such that \( \sigma \left( z\right) = r \) .
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Proof. Given \( z \in U \) and \( r \in {\varepsilon }^{-1}\left( z\right) \), let \( \widetilde{U} \subseteq \mathbb{R} \) be the component of \( {\varepsilon }^{-1}\left( U\right) \) containing \( r \) . By definition of an evenly covered open subset, \( \varepsilon : \widetilde{U} \rightarrow U \) is a homeomorphism. Thus \( \sigma = {\left( {\left. \varepsilon \right| }_{\widetilde{U}}\right) }^{-1} \) is the desired local section.
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Yes
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Corollary 8.5 (Path Lifting Property of the Circle). Suppose \( f : I \rightarrow {\mathbb{S}}^{1} \) is any path, and \( {r}_{0} \in \mathbb{R} \) is any point in the fiber of \( \varepsilon \) over \( f\left( 0\right) \) . Then there exists a unique lift \( \widetilde{f} : I \rightarrow \mathbb{R} \) of \( f \) such that \( \widetilde{f}\left( 0\right) = {r}_{0} \), and any other lift differs from \( \widetilde{f} \) by addition of an integer constant.
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Proof. A path \( f \) can be viewed as a homotopy between two maps from a one-point space \( \{ * \} \) into \( {\mathbb{S}}^{1} \), namely \( * \mapsto f\left( 0\right) \) and \( * \mapsto f\left( 1\right) \) . Thus the existence and uniqueness of \( \widetilde{f} \) follow from the homotopy lifting property. To prove the final statement of the corollary, suppose \( {\widetilde{f}}^{\prime } \) is any other lift of \( f \) . Then the fact that \( \varepsilon \left( {\widetilde{f}\left( s\right) }\right) = f\left( s\right) = \varepsilon \left( {{\widetilde{f}}^{\prime }\left( s\right) }\right) \) implies that \( \widetilde{f}\left( s\right) - {\widetilde{f}}^{\prime }\left( s\right) \) is an integer for each \( s \) . Because \( \widetilde{f} - {\widetilde{f}}^{\prime } \) is a continuous function from the connected space \( I \) into the discrete space \( \mathbb{Z} \), it must be constant.
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Yes
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Corollary 8.6 (Path Homotopy Criterion for the Circle). Suppose \( {f}_{0} \) and \( {f}_{1} \) are paths in \( {\mathbb{S}}^{1} \) with the same initial point and the same terminal point, and \( {\widetilde{f}}_{0},{\widetilde{f}}_{1} : I \rightarrow \) \( \mathbb{R} \) are lifts of \( {f}_{0} \) and \( {f}_{1} \) with the same initial point. Then \( {f}_{0} \sim {f}_{1} \) if and only if \( {\widetilde{f}}_{0} \) and \( {\widetilde{f}}_{1} \) have the same terminal point.
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Proof. If \( {\widetilde{f}}_{0} \) and \( {\widetilde{f}}_{1} \) have the same terminal point, then they are path-homotopic by Exercise 7.14, because \( \mathbb{R} \) is simply connected. It follows that \( {f}_{0} = \varepsilon \circ {\widetilde{f}}_{0} \) and \( {f}_{1} = \varepsilon \circ {\widetilde{f}}_{1} \) are also path-homotopic.\n\nConversely, suppose \( {f}_{0} \sim {f}_{1} \), and let \( H : I \times I \rightarrow {\mathbb{S}}^{1} \) be a path homotopy between them. Then the homotopy lifting property implies that \( H \) lifts to a homotopy \( \widetilde{H} : I \times I \rightarrow \mathbb{R} \) such that \( {\widetilde{H}}_{0} = {\widetilde{f}}_{0} \) . Because \( H \) is stationary on \( \{ 0,1\} \), so is \( \widetilde{H} \) , which means that it is a path homotopy. The path \( {\widetilde{H}}_{1} : I \rightarrow \mathbb{R} \) is a lift of \( {f}_{1} \) starting at \( {\widetilde{f}}_{0}\left( 1\right) \), so by uniqueness of lifts it must be equal to \( {\widetilde{f}}_{1} \) . Thus \( {\widetilde{f}}_{1} \) is path-homotopic to \( {\widetilde{f}}_{0} \), which implies in particular that they have the same terminal point.
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Yes
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Theorem 8.8 (Homotopy Classification of Loops in \( {\mathbb{S}}^{1} \) ). Two loops in \( {\mathbb{S}}^{1} \) based at the same point are path-homotopic if and only if they have the same winding number.
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Proof. Suppose \( {f}_{0} \) and \( {f}_{1} \) are loops in \( {\mathbb{S}}^{1} \) based at the same point. By the path lifting property (Corollary 8.5), they have lifts \( {\widetilde{f}}_{0},{\widetilde{f}}_{1} : I \rightarrow \mathbb{R} \) starting at the same point, and by the path homotopy criterion (Corollary 8.6) these lifts end at the same point if and only if \( {f}_{0} \sim {f}_{1} \) . By definition of the winding number, this means that \( {f}_{0} \) and \( {f}_{1} \) have the same winding number if and only if they are path-homotopic.
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Yes
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Theorem 8.9 (Fundamental Group of the Circle). The group \( {\pi }_{1}\left( {{\mathbb{S}}^{1},1}\right) \) is an infinite cyclic group generated by \( \left\lbrack \omega \right\rbrack \) .
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Proof. Define maps \( J : \mathbb{Z} \rightarrow {\pi }_{1}\left( {{\mathbb{S}}^{1},1}\right) \) and \( K : {\pi }_{1}\left( {{\mathbb{S}}^{1},1}\right) \rightarrow \mathbb{Z} \) by\n\n\[ J\left( n\right) = {\left\lbrack \omega \right\rbrack }^{n},\;K\left( \left\lbrack f\right\rbrack \right) = N\left( f\right) . \]\n\nBecause the winding number of a loop depends only on its path homotopy class, \( K \) is well defined. Because \( {\left\lbrack \omega \right\rbrack }^{n + m} = {\left\lbrack \omega \right\rbrack }^{n}{\left\lbrack \omega \right\rbrack }^{m}, J \) is a homomorphism (considering \( \mathbb{Z} \) as an additive group). To prove the theorem, it suffices to show that \( J \) is an isomorphism, which we do by showing that \( K \) is a two-sided inverse for it.\n\nFor this purpose, it is convenient to use a concrete representative of each path class \( {\left\lbrack \omega \right\rbrack }^{n} \), defined as follows. For any integer \( n \), let \( {\alpha }_{n} : I \rightarrow {\mathbb{S}}^{1} \) be the loop\n\n\[ {\alpha }_{n}\left( s\right) = {e}^{2\pi ins}. \]\n\nIt is easy to see that \( {\alpha }_{1} = \omega ,{\alpha }_{-1} = \bar{\omega } \) (the reverse path of \( \omega \) ), and \( {\alpha }_{n} \) is a reparam-etrization of \( {\alpha }_{n - 1} \cdot \omega \), so by induction \( \left\lbrack {\alpha }_{n}\right\rbrack = {\left\lbrack \omega \right\rbrack }^{n} \) for each \( n \) . By direct computation, the path \( {\widetilde{\alpha }}_{n} : I \rightarrow \mathbb{R} \) given by \( {\widetilde{\alpha }}_{n}\left( s\right) = {ns} \) is a lift of \( {\alpha }_{n} \), so the winding number of \( {\alpha }_{n} \) is \( {\widetilde{\alpha }}_{n}\left( 1\right) - {\widetilde{\alpha }}_{n}\left( 0\right) = n \).\n\nTo prove that \( K \circ J = {\operatorname{Id}}_{\mathbb{Z}} \), let \( n \in \mathbb{Z} \) be arbitrary, and note that \( K\left( {J\left( n\right) }\right) = \) \( K\left( {\left\lbrack \omega \right\rbrack }^{n}\right) = K\left( \left\lbrack {\alpha }_{n}\right\rbrack \right) = N\left( {\alpha }_{n}\right) = n \). To prove that \( J \circ K = {\operatorname{Id}}_{{\pi }_{1}\left( {{\mathbb{S}}^{1},1}\right) } \), suppose \( \left\lbrack f\right\rbrack \) is any element of \( {\pi }_{1}\left( {{\mathbb{S}}^{1},1}\right) \), and let \( n \) be the winding number of \( f \). Then \( f \) and \( {\alpha }_{n} \) are path-homotopic because they are loops based at 1 with the same winding number, so \( J\left( {K\left( \left\lbrack f\right\rbrack \right) }\right) = J\left( n\right) = {\left\lbrack \omega \right\rbrack }^{n} = \left\lbrack {\alpha }_{n}\right\rbrack = \left\lbrack f\right\rbrack \) .
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Yes
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Corollary 8.12 (Fundamental Groups of Tori). Let \( {\mathbb{T}}^{n} = {\mathbb{S}}^{1} \times \cdots \times {\mathbb{S}}^{1} \) be the \( n \) - dimensional torus with \( p = \left( {1,\ldots ,1}\right) \) as base point, and for each \( j = 1,\ldots, n \), let \( {\omega }_{j} \) denote the standard loop in the \( j \) th copy of \( {\mathbb{S}}^{1} \) :\n\n\[ \n{\omega }_{j}\left( s\right) = \left( {1,\ldots ,1,{e}^{2\pi is},1,\ldots ,1}\right) .\n\]\n\nThe map \( \varphi : {\mathbb{Z}}^{n} \rightarrow {\pi }_{1}\left( {{\mathbb{T}}^{n}, p}\right) \) given by \( \varphi \left( {{k}_{1},\ldots ,{k}_{n}}\right) = {\left\lbrack {\omega }_{1}\right\rbrack }^{{k}_{1}}\cdots {\left\lbrack {\omega }_{n}\right\rbrack }^{{k}_{n}} \) is an isomorphism.
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Proof. This is a direct consequence of Theorem 8.9 and Proposition 7.34.
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No
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Lemma 8.14 (Another Characterization of the Degree). If \( \varphi : {\mathbb{S}}^{1} \rightarrow {\mathbb{S}}^{1} \) is continuous, the degree of \( \varphi \) is equal to the degree of the following group endomorphism:\n\n\[ \n{\left( {\rho }_{\varphi } \circ \varphi \right) }_{ * } : {\pi }_{1}\left( {{\mathbb{S}}^{1},1}\right) \rightarrow {\pi }_{1}\left( {{\mathbb{S}}^{1},1}\right) .\n\]\n\nIn particular, if \( \varphi \left( 1\right) = 1 \), then \( \deg \varphi = \deg {\varphi }_{ * } \) .
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Proof. Let \( \varphi \) be as in the statement of the lemma, and let \( n \) be the degree of \( \varphi \), which is the winding number of the loop \( \varphi \circ \omega \) . By Exercise 8.7, the winding number of \( {\rho }_{\varphi } \circ \varphi \circ \omega \) is also \( n \) . By Theorem 8.8, this implies that \( {\rho }_{\varphi } \circ \varphi \circ \omega \sim {\alpha }_{n} \), where \( {\alpha }_{n} \) is the representative of \( {\left\lbrack \omega \right\rbrack }^{n} \) given by (8.3). Therefore,\n\n\[ \n{\left( {\rho }_{\varphi } \circ \varphi \right) }_{ * }\left\lbrack \omega \right\rbrack = \left\lbrack {{\rho }_{\varphi } \circ \varphi \circ \omega }\right\rbrack = \left\lbrack {\alpha }_{n}\right\rbrack = {\left\lbrack \omega \right\rbrack }^{n},\n\]\nwhich is exactly the statement that the degree of \( {\left( {\rho }_{\varphi } \circ \varphi \right) }_{ * } \) is also \( n \) .
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Yes
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Theorem 8.17 (Homotopy Classification of Maps of the Circle). Two continuous maps from \( {\mathbb{S}}^{1} \) to itself are homotopic if and only if they have the same degree.
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Proof. One direction was proved in Proposition 8.15. To prove the converse, suppose \( \varphi \) and \( \psi \) have the same degree. First consider the special case in which \( \varphi \left( 1\right) = \psi \left( 1\right) = 1 \) . Then the hypothesis means that \( \varphi \circ \omega \) and \( \psi \circ \omega \) are loops based at 1 with the same winding number, and therefore they are path-homotopic by Theorem 8.8. Let \( H : I \times I \rightarrow {\mathbb{S}}^{1} \) be a path homotopy from \( \varphi \circ \omega \) to \( \psi \circ \omega \) . Note that \( \omega \times \operatorname{Id} : I \times I \rightarrow {\mathbb{S}}^{1} \times I \) is a quotient map by the closed map lemma. Since \( H \) respects the identifications made by this map, it descends to a continuous map \( \widetilde{H} : {\mathbb{S}}^{1} \times I \rightarrow {\mathbb{S}}^{1} \), which is easily seen to be a homotopy between \( \varphi \) and \( \psi \) .\n\nTo handle the general case, let \( {\rho }_{\varphi } \) and \( {\rho }_{\psi } \) be the rotations taking \( \varphi \left( 1\right) \) to 1 and \( \psi \left( 1\right) \) to 1, respectively, so that \( \deg \left( {{\rho }_{\varphi } \circ \varphi }\right) = \deg \left( {{\rho }_{\psi } \circ \psi }\right) \) . Since both maps take 1 to 1, it follows from the argument in the preceding paragraph that \( {\rho }_{\varphi } \circ \varphi \simeq {\rho }_{\psi } \circ \psi \) , and since every rotation is homotopic to the identity map, we conclude finally that \( \varphi \simeq \psi \)
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Yes
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Theorem 8.18. Let \( \varphi : {\mathbb{S}}^{1} \rightarrow {\mathbb{S}}^{1} \) be continuous. If \( \deg \varphi \neq 0 \), then \( \varphi \) is surjective.
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Proof. We prove the contrapositive. If \( \varphi \) is not surjective, then it actually maps into the subset \( {\mathbb{S}}^{1} \smallsetminus \{ c\} \) for some \( c \in {\mathbb{S}}^{1} \) . But \( {\mathbb{S}}^{1} \smallsetminus \{ c\} \) is homeomorphic to \( \mathbb{R} \) (by the 1-dimensional version of stereographic projection, for example) and is therefore contractible. It follows that \( \varphi \) is homotopic to a constant map and thus has degree zero.
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Yes
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Theorem 8.19. Let \( \varphi : {\mathbb{S}}^{1} \rightarrow {\mathbb{S}}^{1} \) be continuous. If \( \deg \varphi \neq 1 \), then \( \varphi \) has a fixed point.
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Proof. Again we prove the contrapositive. Assuming \( \varphi \) has no fixed point, it follows that for every \( z \in {\mathbb{S}}^{1} \), the line segment in \( \mathbb{C} \) from \( \varphi \left( z\right) \) to \( - z \) does not pass through the origin. Thus we can define a homotopy from \( \varphi \) to the antipodal map by\n\n\[ H\left( {z, t}\right) = \frac{\left( {1 - t}\right) \varphi \left( z\right) - {tz}}{\left| \left( 1 - t\right) \varphi \left( z\right) - tz\right| }.\]\n\nBecause the antipodal map has degree 1, so does \( \varphi \) .
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Yes
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Proposition 9.1. Given an indexed family of groups \( {\left( {G}_{\alpha }\right) }_{\alpha \in A} \), their free product is a group under the multiplication operation induced by multiplication of words.
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Proof. First we need to check that multiplication of words respects the equivalence relation. If \( {V}^{\prime } \) is obtained from \( V \) by an elementary reduction, then it is easy to see that \( {V}^{\prime }W \) is similarly obtained from \( {VW} \), as is \( W{V}^{\prime } \) from \( {WV} \) . If \( V \sim {V}^{\prime } \) and \( W \sim {W}^{\prime } \), it follows by induction on the number of elementary reductions that \( {VW} \sim {V}^{\prime }{W}^{\prime } \) . Thus multiplication is well defined on equivalence classes.\n\nThe equivalence class of the empty word ( ) is an identity element for multiplication of equivalence classes, and multiplication is associative on equivalence classes because it already is on words. Finally, for any word \( \left( {{g}_{1},\ldots ,{g}_{m}}\right) \), it is easy to check that\n\n\[ \left( {{g}_{1},\ldots ,{g}_{m}}\right) \left( {{g}_{m}^{-1},\ldots ,{g}_{1}^{-1}}\right) \sim \left( \right) \sim \left( {{g}_{m}^{-1},\ldots ,{g}_{1}^{-1}}\right) \left( {{g}_{1},\ldots ,{g}_{m}}\right) ,\]\n\nso the equivalence class of the word \( \left( {{g}_{m}^{-1},\ldots ,{g}_{1}^{-1}}\right) \) is an inverse for that of \( \left( {{g}_{1},\ldots ,{g}_{m}}\right) \) .\n\nHenceforth, we denote the identity element of the free product (the equivalence class of the empty word) by 1 .
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Yes
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Example 9.3. Let \( \mathbb{Z}/2 \) denote the group of integers modulo 2 . The free product \( \mathbb{Z}/2 * \mathbb{Z}/2 \) can be described as follows. If we let \( \beta \) and \( \gamma \) denote the nontrivial elements of the first and second copies of \( \mathbb{Z}/2 \), respectively, each element of \( \mathbb{Z}/2 * \mathbb{Z}/2 \) other than the identity has a unique representation as a string of alternating \( \beta \) ’s and \( \gamma \) ’s. Multiplication is performed by concatenating the strings and deleting all consecutive pairs of \( \beta \) ’s or \( \gamma \) ’s.
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For example,\n\n\[ \left( {\beta \gamma \beta \gamma \beta }\right) \left( {\gamma \beta \gamma \beta }\right) = {\beta \gamma \beta \gamma \beta \gamma \beta \gamma \beta } \]\n\n\[ \left( {\gamma \beta \gamma \beta }\right) \left( {\beta \gamma \beta \gamma \beta }\right) = \beta \text{.} \]\n\nBecause these two products are not equal, this group is not abelian.
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Yes
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Theorem 9.5 (Characteristic Property of the Free Product). Let \( {\left( {G}_{\alpha }\right) }_{\alpha \in A} \) be an indexed family of groups. For any group \( H \) and any collection of homomorphisms \( {\varphi }_{\alpha } : {G}_{\alpha } \rightarrow H \), there exists a unique homomorphism \( \Phi : { * }_{\alpha \in A}{G}_{\alpha } \rightarrow H \) such that for each \( \alpha \) the following diagram commutes:
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Proof. Suppose we are given a collection of homomorphisms \( {\varphi }_{\alpha } : {G}_{\alpha } \rightarrow H \) . The requirement that \( \Phi \circ {\iota }_{\alpha } = {\varphi }_{\alpha } \) implies that the desired homomorphism \( \Phi \) must satisfy\n\n\[ \Phi \left( g\right) = {\varphi }_{\alpha }\left( g\right) \;\text{ if }g \in {G}_{\alpha }, \]\n\n(9.4)\n\nwhere, as usual, we identify \( {G}_{\alpha } \) with its image under \( {\iota }_{\alpha } \) . Since \( \Phi \) is supposed to be a homomorphism, it must satisfy\n\n\[ \Phi \left( {{g}_{1}\cdots {g}_{m}}\right) = \Phi \left( {g}_{1}\right) \cdots \Phi \left( {g}_{m}\right) . \]\n\n(9.5)\n\nTherefore, if \( \Phi \) and \( \widetilde{\Phi } \) both satisfy the conclusion, they must be equal because both must satisfy (9.4) and (9.5). This proves that \( \Phi \) is unique if it exists.\n\nTo prove existence of \( \Phi \), we use (9.4) and (9.5) to define it. This is clearly a homomorphism that satisfies the required properties, provided that it is well defined. To verify that it is well defined, we need to check that it gives the same result when applied to equivalent words. As usual, we need only check elementary reductions. If \( {g}_{i},{g}_{i + 1} \in {G}_{\alpha } \), we have Free Groups\n\n\[ \Phi \left( {{g}_{i}{g}_{i + 1}}\right) = {\varphi }_{\alpha }\left( {{g}_{i}{g}_{i + 1}}\right) = {\varphi }_{\alpha }\left( {g}_{i}\right) {\varphi }_{\alpha }\left( {g}_{i + 1}\right) = \Phi \left( {g}_{i}\right) \Phi \left( {g}_{i + 1}\right) ,\]\n\nfrom which it follows that the definition of \( \Phi \) is unchanged by multiplying together successive elements of the same group. Similarly, \( \Phi \left( {1}_{\alpha }\right) = {\varphi }_{\alpha }\left( {1}_{\alpha }\right) = 1 \in H \), which shows that \( \Phi \) is unchanged by deleting an identity element. This completes the proof.
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Yes
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Theorem 9.5 (Characteristic Property of the Free Product). Let \( {\left( {G}_{\alpha }\right) }_{\alpha \in A} \) be an indexed family of groups. For any group \( H \) and any collection of homomorphisms \( {\varphi }_{\alpha } : {G}_{\alpha } \rightarrow H \), there exists a unique homomorphism \( \Phi : { * }_{\alpha \in A}{G}_{\alpha } \rightarrow H \) such that for each \( \alpha \) the following diagram commutes:
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Proof. Suppose we are given a collection of homomorphisms \( {\varphi }_{\alpha } : {G}_{\alpha } \rightarrow H \) . The requirement that \( \Phi \circ {\iota }_{\alpha } = {\varphi }_{\alpha } \) implies that the desired homomorphism \( \Phi \) must satisfy\n\n\[ \Phi \left( g\right) = {\varphi }_{\alpha }\left( g\right) \;\text{ if }g \in {G}_{\alpha }, \]\n\n(9.4)\n\nwhere, as usual, we identify \( {G}_{\alpha } \) with its image under \( {\iota }_{\alpha } \) . Since \( \Phi \) is supposed to be a homomorphism, it must satisfy\n\n\[ \Phi \left( {{g}_{1}\cdots {g}_{m}}\right) = \Phi \left( {g}_{1}\right) \cdots \Phi \left( {g}_{m}\right) . \]\n\n(9.5)\n\nTherefore, if \( \Phi \) and \( \widetilde{\Phi } \) both satisfy the conclusion, they must be equal because both must satisfy (9.4) and (9.5). This proves that \( \Phi \) is unique if it exists.\n\nTo prove existence of \( \Phi \), we use (9.4) and (9.5) to define it. This is clearly a homomorphism that satisfies the required properties, provided that it is well defined. To verify that it is well defined, we need to check that it gives the same result when applied to equivalent words. As usual, we need only check elementary reductions. If \( {g}_{i},{g}_{i + 1} \in {G}_{\alpha } \), we have Free Groups\n\n\[ \Phi \left( {{g}_{i}{g}_{i + 1}}\right) = {\varphi }_{\alpha }\left( {{g}_{i}{g}_{i + 1}}\right) = {\varphi }_{\alpha }\left( {g}_{i}\right) {\varphi }_{\alpha }\left( {g}_{i + 1}\right) = \Phi \left( {g}_{i}\right) \Phi \left( {g}_{i + 1}\right) ,\]\n\nfrom which it follows that the definition of \( \Phi \) is unchanged by multiplying together successive elements of the same group. Similarly, \( \Phi \left( {1}_{\alpha }\right) = {\varphi }_{\alpha }\left( {1}_{\alpha }\right) = 1 \in H \), which shows that \( \Phi \) is unchanged by deleting an identity element. This completes the proof.
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Yes
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Corollary 9.6. The free product is the coproduct in the category of groups.
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Proof. The characteristic property is exactly the defining property of the coproduct in a category.
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No
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The free product is the unique group (up to isomorphism) satisfying the characteristic property.
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Proof. Theorem 7.57 shows that coproducts in any category are unique up to isomorphism.
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No
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Theorem 9.9 (Characteristic Property of the Free Group). Let \( S \) be a set. For any group \( H \) and any map \( \varphi : S \rightarrow H \), there exists a unique homomorphism \( \Phi : F\left( S\right) \rightarrow H \) extending \( \varphi \) :
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Proof. This can be proved directly as in the proof of Theorem 9.5. Alternatively, recalling that the free group is defined as a free product, we can proceed as follows. There is a one-to-one correspondence between set functions \( \varphi : S \rightarrow H \) and collections of homomorphisms \( {\varphi }_{\sigma } : F\left( \sigma \right) \rightarrow H \) for all \( \sigma \in S \), by the equation\n\n\[{\varphi }_{\sigma }\left( {\sigma }^{n}\right) = \varphi {\left( \sigma \right) }^{n}\]\n\nTranslating the characteristic property of the free product to this special case and using this correspondence yields the result. The details are left as an exercise.
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No
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Theorem 9.9 (Characteristic Property of the Free Group). Let \( S \) be a set. For any group \( H \) and any map \( \varphi : S \rightarrow H \), there exists a unique homomorphism \( \Phi : F\left( S\right) \rightarrow H \) extending \( \varphi \).
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Proof. This can be proved directly as in the proof of Theorem 9.5. Alternatively, recalling that the free group is defined as a free product, we can proceed as follows. There is a one-to-one correspondence between set functions \( \varphi : S \rightarrow H \) and collections of homomorphisms \( {\varphi }_{\sigma } : F\left( \sigma \right) \rightarrow H \) for all \( \sigma \in S \), by the equation\n\n\[{\varphi }_{\sigma }\left( {\sigma }^{n}\right) = \varphi {\left( \sigma \right) }^{n}\]\n\nTranslating the characteristic property of the free product to this special case and using this correspondence yields the result. The details are left as an exercise.
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No
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Proposition 9.12. A group \( G \) is free if and only if it has a generating set \( S \subseteq G \) such that every element \( g \in G \) other than the identity has a unique expression as a product of the form\n\n\[ g = {\sigma }_{1}^{{n}_{1}}\cdots {\sigma }_{k}^{{n}_{k}} \]\n\nwhere \( {\sigma }_{i} \in S,{n}_{i} \) are nonzero integers, and \( {\sigma }_{i} \neq {\sigma }_{i + 1} \) for each \( i = 1,\ldots, k - 1 \) .
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Proof. See Problem 9-3.
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No
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Lemma 9.18. If an abelian group \( G \) has a finite basis, then every finite basis has the same number of elements.
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Proof. Suppose \( G \) has a basis with \( n \) elements. Then \( G \cong {\mathbb{Z}}^{n} \) by Proposition 9.14(b), and the quotient group \( G/{2G} \) is easily seen to be isomorphic to \( {\left( \mathbb{Z}/2\right) }^{n} \) , which has exactly \( {2}^{n} \) elements. Since the order of \( G/{2G} \) is independent of the choice of basis, every finite basis must have \( n \) elements.
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Yes
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Proposition 9.19. Suppose \( G \) is a free abelian group of finite rank. Every subgroup of \( G \) is free abelian of rank less than or equal to that of \( G \) .
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Proof. We may assume without loss of generality that \( G = {\mathbb{Z}}^{n} \) . We prove the proposition by induction on \( n \) . For \( n = 1 \), it follows from the fact that every subgroup of a cyclic group is cyclic.\n\nSuppose the result is true for subgroups of \( {\mathbb{Z}}^{n - 1} \), and let \( H \) be any subgroup of \( {\mathbb{Z}}^{n} \) . Identifying \( {\mathbb{Z}}^{n - 1} \) with the subgroup \( \left\{ \left( {{k}_{1},\ldots ,{k}_{n - 1},0}\right) \right\} \) of \( {\mathbb{Z}}^{n} \), the inductive hypothesis guarantees that \( H \cap {\mathbb{Z}}^{n - 1} \) is free abelian of rank \( m - 1 \leq n - 1 \), so has a basis \( \left\{ {{h}_{1},\ldots ,{h}_{m - 1}}\right\} \) . If \( H \subseteq {\mathbb{Z}}^{n - 1} \), we are done. Otherwise, the image of \( H \) under the projection \( {\pi }_{n} : {\mathbb{Z}}^{n} \rightarrow \mathbb{Z} \) onto the \( n \) th factor is a nontrivial cyclic subgroup of \( \mathbb{Z} \) . Let \( c \in \mathbb{Z} \) be a generator of this subgroup, and let \( {h}_{m} \) be an element of \( H \) such that \( {\pi }_{n}\left( {h}_{m}\right) = c \) . The proof will be complete once we show that \( \left\{ {{h}_{1},\ldots ,{h}_{m}}\right\} \) is a basis for \( H \) .\n\nSuppose \( {a}_{1}{h}_{1} + \cdots + {a}_{m}{h}_{m} = 0 \) . Applying \( {\pi }_{n} \) to this equation yields \( {a}_{m}c = 0 \), so \( {a}_{m} = 0 \) . Then \( {a}_{1} = \cdots = {a}_{m - 1} = 0 \) because of the independence of \( \left\{ {{h}_{1},\ldots ,{h}_{m - 1}}\right\} \) , so \( \left\{ {{h}_{1},\ldots ,{h}_{m}}\right\} \) is linearly independent. Now suppose \( h \in H \) is arbitrary. Then \( {\pi }_{n}\left( h\right) = {ac} \) for some integer \( a \), so \( h - a{h}_{m} \in H \cap {\mathbb{Z}}^{n - 1} \) . This element can be written as a linear combination of \( \left\{ {{h}_{1},\ldots ,{h}_{m - 1}}\right\} \), which shows that \( H \) is generated by \( \left\{ {{h}_{1},\ldots ,{h}_{m}}\right\} \) .
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Yes
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Proposition 9.21. Any abelian group that is finitely generated and torsion-free is free abelian of finite rank.
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Proof. Suppose \( G \) is such a group. If \( S \subseteq G \) is a linearly independent subset, then the subgroup \( \langle S\rangle \subseteq G \) generated by \( S \) is easily seen to be free abelian with \( S \) as a basis.\n\nThe crux of the proof is the following claim: there exists a nonzero integer \( n \) and a finite linearly independent set \( S \subseteq G \) such that \( {nG} \subseteq \langle S\rangle \) . Assuming this, the rest of the proof goes as follows. Let \( \varphi : G \rightarrow G \) be the homomorphism \( \varphi \left( g\right) = {ng} \) . It is injective because \( G \) is torsion-free, and the claim implies that \( \varphi \left( G\right) \subseteq \langle S\rangle \) . Thus \( G \) is isomorphic to the subgroup \( \varphi \left( G\right) \) of the free abelian group \( \langle S\rangle \), so by Proposition \( {9.19}, G \) is free abelian of finite rank.\n\nWe prove the claim by induction on the number of elements in a generating set for \( G \) . If \( G \) is generated by one element \( g \), the claim is true with \( n = 1 \) and \( S = \{ g\} \) , because the fact that \( G \) is torsion-free implies that \( \{ g\} \) is a linearly independent set.\n\nNow assume that the claim is true for every torsion-free abelian group generated by \( m - 1 \) elements, and suppose \( G \) is generated by a set \( T = \left\{ {{g}_{1},\ldots ,{g}_{m}}\right\} \) with \( m \) elements. If \( T \) is linearly independent, we just take \( S = T \) . If not, there is a relation of the form \( {a}_{1}{g}_{1} + \cdots + {a}_{m}{g}_{m} = 0 \) with at least one of the coefficients, say \( {a}_{m} \), not equal to zero. Letting \( {G}^{\prime } \) denote the subgroup of \( G \) generated by \( \left\{ {{g}_{1},\ldots ,{g}_{m - 1}}\right\} \) , this means that \( {a}_{m}{g}_{m} \in {G}^{\prime } \) . Since \( {G}^{\prime } \) is generated by \( m - 1 \) elements, by induction there exist a nonzero integer \( {n}^{\prime } \) and a finite linearly independent set \( S \subseteq {G}^{\prime } \) such that \( {n}^{\prime }{G}^{\prime } \subseteq \langle S\rangle \) . Let \( n = {a}_{m}{n}^{\prime } \) . Since \( G \) is generated by \( T \), for any \( g \in G \) we have\n\n\[ \n{ng} = {a}_{m}{n}^{\prime }\left( {{b}_{1}{g}_{1} + \cdots + {b}_{m}{g}_{m}}\right) \n\]\n\n\[ \n= {n}^{\prime }\left( {{a}_{m}{b}_{1}{g}_{1} + \cdots + {a}_{m}{b}_{m - 1}{g}_{m - 1}}\right) + {n}^{\prime }{b}_{m}\left( {{a}_{m}{g}_{m}}\right) . \n\]\n\nBoth terms above are in \( {n}^{\prime }{G}^{\prime } \subseteq \langle S\rangle \) . It follows that \( {nG} \subseteq \langle S\rangle \), which completes the proof.
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Yes
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Proposition 9.23. Suppose \( G \) and \( H \) are abelian groups and \( f : G \rightarrow H \) is a homomorphism. Then \( G \) is finitely generated if and only if both \( \operatorname{Im}f \) and \( \operatorname{Ker}f \) are finitely generated, in which case \( \operatorname{rank}G = \operatorname{rank}\left( {\operatorname{Im}f}\right) + \operatorname{rank}\left( {\operatorname{Ker}f}\right) \) .
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Proof. Replacing \( H \) by the image of \( f \), we may as well assume that \( f : G \rightarrow H \) is surjective. Write \( K = \operatorname{Ker}f \subseteq G \). If \( G \) is finitely generated, then so is \( K \) by Corollary 9.20; and since \( f \) is surjective, it takes a set of generators for \( G \) to a set of generators for \( H \), so \( H \) is also finitely generated. Conversely, suppose that \( K \) and \( H \) are finitely generated. We can choose generating sets \( \left\{ {{k}_{1},\ldots ,{k}_{p}}\right\} \) for \( K \) and \( \left\{ {{h}_{1},\ldots ,{h}_{q}}\right\} \) for \( H \), and the fact that \( f \) is surjective means that there are elements \( {g}_{j} \in G \) such that \( f\left( {g}_{j}\right) = {h}_{j} \). We show that \( \left\{ {{k}_{1},\ldots ,{k}_{p},{g}_{1},\ldots ,{g}_{q}}\right\} \) is a generating set for \( G \). If \( g \in G \) is arbitrary, we can write \( f\left( g\right) = \mathop{\sum }\limits_{i}{n}_{i}{h}_{i} \) for some integers \( {n}_{1},\ldots ,{n}_{q} \). It follows that \( g - \mathop{\sum }\limits_{i}{n}_{i}{g}_{i} \in K \), so it can be written as \( \mathop{\sum }\limits_{j}{m}_{j}{k}_{j} \). Thus \( g = \mathop{\sum }\limits_{i}{n}_{i}{g}_{i} + \mathop{\sum }\limits_{j}{m}_{j}{k}_{j} \) as required. Now assume that \( G \) (and hence also \( K \) and \( H \) ) is finitely generated. Assume also for the moment that \( G \) and \( H \) are free abelian; then \( K \) is too by Proposition 9.19. Choose bases \( \left\{ {{k}_{1},\ldots ,{k}_{p}}\right\} \) for \( K \) and \( \left\{ {{h}_{1},\ldots ,{h}_{q}}\right\} \) for \( H \). By surjectivity, there are elements \( {g}_{j} \in G \) such that \( f\left( {g}_{j}\right) = {h}_{j} \). The set \( \left\{ {{k}_{i},{g}_{j}}\right\} \) is linearly independent, because a relation of the form \( g = \mathop{\sum }\limits_{i}{m}_{i}{k}_{i} + \mathop{\sum }\limits_{j}{n}_{j}{g}_{j} = 0 \) implies 0 = f\left( g\right) = \mathop{\sum }\limits_{j}{n}_{j}f\left( {g}_{j}\right) = \mathop{\sum }\limits_{j}{n}_{j}{h}_{j}, which implies \( {n}_{j} = 0 \) for each \( j \). Therefore, \( g = \mathop{\sum }\limits_{i}{m}_{i}{k}_{i} = 0 \), so \( {m}_{i} = 0 \). On the other hand, if \( g \in G \) is arbitrary, then we can write \( f\left( g\right) = \mathop{\sum }\limits_{j}{n}_{j}{h}_{j} \), which implies that \( g - \mathop{\sum }\limits_{j}{n}_{j}{g}_{j} \) is in \( K \) and thus can be written in the form \( \mathop{\sum }\limits_{i}{m}_{i}{k}_{i} \), so \( g = \mathop{\sum }\limits_{j}{n}_{j}{g}_{j} + \mathop{\sum }\limits_{i}{m}_{i}{k}_{i} \). It follows that \( \left\{ {{k}_{i},{g}_{j}}\right\} \) is a basis for \( G \), which therefore has rank equal to \( p + q = \operatorname{rank}K + \operatorname{rank}H \).
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Yes
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Theorem 10.3 (Presentation of an Amalgamated Free Product). Let \( {f}_{1} : H \rightarrow \) \( {G}_{1} \) and \( {f}_{2} : H \rightarrow {G}_{2} \) be group homomorphisms. Suppose \( {G}_{1},{G}_{2} \), and \( H \) have the following finite presentations:\n\n\[ \n{G}_{1} \cong \left\langle {{\alpha }_{1},\ldots ,{\alpha }_{m} \mid {\rho }_{1},\ldots ,{\rho }_{r}}\right\rangle \n\]\n\n\[ \n{G}_{2} \cong \left\langle {{\beta }_{1},\ldots ,{\beta }_{n} \mid {\sigma }_{1},\ldots ,{\sigma }_{s}}\right\rangle \n\]\n\n\[ \nH \cong \left\langle {{\gamma }_{1},\ldots ,{\gamma }_{p} \mid {\tau }_{1},\ldots ,{\tau }_{t}}\right\rangle \n\]\n\nThen the amalgamated free product has the presentation\n\n\[ \n{G}_{1}{ * }_{H}{G}_{2} \cong \left\langle {{\alpha }_{1},\ldots ,{\alpha }_{m},{\beta }_{1},\ldots ,{\beta }_{n} \mid }\right.\n\]\n\n(10.4)\n\n\[ \n\left. {{\rho }_{1},\ldots ,{\rho }_{r},{\sigma }_{1},\ldots ,{\sigma }_{s},{u}_{1} = {v}_{1},\ldots ,{u}_{p} = {v}_{p}}\right\rangle ,\n\]\n\nwhere \( {u}_{a} \) is an expression for \( {f}_{1}\left( {\gamma }_{a}\right) \in {G}_{1} \) in terms of the generators \( \left\{ {\alpha }_{i}\right\} \), and \( {v}_{a} \) similarly expresses \( {f}_{2}\left( {\gamma }_{a}\right) \in {G}_{2} \) in terms of \( \left\{ {\beta }_{j}\right\} \) .
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Proof. This is an immediate consequence of Problems 9-4(b) and 9-5.
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No
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Corollary 10.4 (First Special Case: Simply Connected Intersection). Assume the hypotheses of the Seifert-Van Kampen theorem, and suppose in addition that \( U \cap V \) is simply connected. Then \( \Phi \) is an isomorphism between \( {\pi }_{1}\left( {U, p}\right) * {\pi }_{1}\left( {V, p}\right) \) and \( {\pi }_{1}\left( {X, p}\right) \) . If the fundamental groups of \( U \) and \( V \) have presentations
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\[ {\pi }_{1}\left( {U, p}\right) \cong \left\langle {{\alpha }_{1},\ldots ,{\alpha }_{m} \mid {\rho }_{1},\ldots ,{\rho }_{r}}\right\rangle ,\] \[ {\pi }_{1}\left( {V, p}\right) \cong \left\langle {{\beta }_{1},\ldots ,{\beta }_{n} \mid {\sigma }_{1},\ldots ,{\sigma }_{s}}\right\rangle \] then \( {\pi }_{1}\left( {X, p}\right) \) has the presentation \[ {\pi }_{1}\left( {X, p}\right) \cong \left\langle {{\alpha }_{1},\ldots ,{\alpha }_{m},{\beta }_{1},\ldots ,{\beta }_{n} \mid {\rho }_{1},\ldots ,{\rho }_{r},{\sigma }_{1},\ldots ,{\sigma }_{s}}\right\rangle ,\] where the generators \( {\alpha }_{a},{\beta }_{a} \) are represented by the same loops as in the original presentations, but now considered as loops in \( X \) instead of \( U \) or \( V \) .
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Yes
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Lemma 10.6. Suppose \( {p}_{i} \in {X}_{i} \) is a nondegenerate base point for \( i = 1,\ldots, n \) . Then \( * \) is a nondegenerate base point in \( {X}_{1} \vee \cdots \vee {X}_{n} \) .
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Proof. For each \( i \), choose a neighborhood \( {W}_{i} \) of \( {p}_{i} \) that admits a strong deformation retraction \( {r}_{i} : {W}_{i} \rightarrow \left\{ {p}_{i}\right\} \), and let \( {H}_{i} : {W}_{i} \times I \rightarrow {W}_{i} \) be the associated homotopy from \( {\operatorname{Id}}_{{W}_{i}} \) to \( {\iota }_{\left\{ {p}_{i}\right\} } \circ {r}_{i} \) . Define a map \( H : \left( {\mathop{\coprod }\limits_{i}{W}_{i}}\right) \times I \rightarrow \mathop{\coprod }\limits_{i}{W}_{i} \) by letting \( H = {H}_{i} \) on \( {W}_{i} \times I \) . The restriction of the quotient map \( q \) to the saturated open subset \( \mathop{\coprod }\limits_{i}{W}_{i} \) is a quotient map to a neighborhood \( W \) of \( * \), and thus \( q \times {\operatorname{Id}}_{I} : \left( {\mathop{\coprod }\limits_{i}{W}_{i}}\right) \times I \rightarrow W \times I \) is a quotient map by Lemma 4.72. Since \( q \circ H \) respects the identifications made by \( q \times {\operatorname{Id}}_{I} \), it descends to the quotient and yields a strong deformation retraction of \( W \) onto \( \{ * \} \) .
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Yes
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Theorem 10.7. Let \( {X}_{1},\ldots ,{X}_{n} \) be spaces with nondegenerate base points \( {p}_{j} \in {X}_{j} \) . The map\n\n\[ \Phi : {\pi }_{1}\left( {{X}_{1},{p}_{1}}\right) * \cdots * {\pi }_{1}\left( {{X}_{n},{p}_{n}}\right) \rightarrow {\pi }_{1}\left( {{X}_{1} \vee \cdots \vee {X}_{n}, * }\right) \]\n\ninduced by \( {\iota }_{{j}_{ * }} : {\pi }_{1}\left( {{X}_{j},{p}_{j}}\right) \rightarrow {\pi }_{1}\left( {{X}_{1} \vee \cdots \vee {X}_{n}, * }\right) \) is an isomorphism.
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Proof. First consider the wedge sum of two spaces \( {X}_{1} \vee {X}_{2} \) . We would like to use Corollary 10.4 to the Seifert-Van Kampen theorem with \( U = {X}_{1}, V = {X}_{2} \) (considered as subspaces of the wedge sum), and \( U \cap V = \{ * \} \) . The trouble is that these spaces are not open in \( {X}_{1} \vee {X}_{2} \), so the corollary does not apply directly. To remedy this, we replace them by slightly \
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No
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The preceding proposition shows that the bouquet \( {\mathbb{S}}^{1} \vee \cdots \vee {\mathbb{S}}^{1} \) of \( n \) circles has a fundamental group isomorphic to \( \mathbb{Z} * \cdots * \mathbb{Z} \), which is a free group on \( n \) generators.
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In fact, it shows more: since the isomorphism is induced by inclusion of each copy of \( {\mathbb{S}}^{1} \) into the bouquet, we can write explicit generators of this free group. If \( {\omega }_{i} \) denotes the standard loop in the \( i \) th copy of \( {\mathbb{S}}^{1} \), then the fundamental group of the bouquet is just the free group \( F\left( {\left\lbrack {\omega }_{1}\right\rbrack ,\ldots ,\left\lbrack {\omega }_{n}\right\rbrack }\right) \) .
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Yes
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Proposition 10.11. Every finite connected graph contains a spanning tree.
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Proof. Let \( \Gamma \) be a finite connected graph. If \( \Gamma = \varnothing \), then the empty subgraph is a spanning tree. Otherwise, we begin by showing that \( \Gamma \) contains a maximal tree, meaning a subgraph that is a tree and is not properly contained in any larger tree in \( \Gamma \) . To prove this, start with any nonempty tree \( {T}_{0} \subseteq \Gamma \) (e.g., a single vertex). If it is not maximal, then it is contained in a strictly larger tree \( {T}_{1} \) . Continuing in this way by induction, we obtain a sequence of trees \( {T}_{0} \subseteq {T}_{1} \subseteq \cdots \), each properly contained in the next. Because \( \Gamma \) is finite, the process cannot go on forever, so eventually we obtain a tree \( T \subseteq \Gamma \) that is not contained in any strictly larger tree.\n\nTo show that \( T \) is a spanning tree, suppose for the sake of contradiction that there is a vertex \( v \in \Gamma \) that is not contained in \( T \) . Because \( \Gamma \) is connected, there is an edge path from a vertex \( {v}_{0} \in T \) to \( v \), say \( \left( {{v}_{0},{e}_{1},\ldots ,{e}_{k},{v}_{k} = v}\right) \) . Let \( {v}_{i} \) be the last vertex in the edge path that is contained in \( T \) . Then the edge \( {e}_{i + 1} \) is not contained in \( T \) , because if it were, \( {v}_{i + 1} \) would also be in \( T \) because \( T \) is a subgraph. The subgraph \( {T}^{\prime } = T \cup \overline{{e}_{i + 1}} \) properly contains \( T \), so it is not a tree, and therefore contains a cycle. This cycle must include \( {e}_{i + 1} \) or \( {v}_{i + 1} \), because otherwise it would be a cycle in \( T \) . However, since \( {e}_{i + 1} \) is the only edge of \( {T}^{\prime } \) that is incident with \( {v}_{i + 1} \), and \( {v}_{i + 1} \) is the only vertex of \( {T}^{\prime } \) incident with \( {e}_{i + 1} \), there can be no such cycle.
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Yes
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Proposition 10.13 (Attaching a Disk). Let \( X \) be a path-connected topological space, and let \( \widetilde{X} \) be the space obtained by attaching a closed 2-cell \( D \) to \( X \) along an attaching map \( \varphi : \partial D \rightarrow X \) . Let \( v \in \partial D,\widetilde{v} = \varphi \left( v\right) \in X \), and \( \gamma = {\varphi }_{ * }\left( \alpha \right) \in {\pi }_{1}\left( {X,\widetilde{v}}\right) \) , where \( \alpha \) is a generator of the infinite cyclic group \( {\pi }_{1}\left( {\partial D, v}\right) \) . Then the homomorphism \( {\pi }_{1}\left( {X,\widetilde{v}}\right) \rightarrow {\pi }_{1}\left( {\widetilde{X},\widetilde{v}}\right) \) induced by inclusion \( X \hookrightarrow \widetilde{X} \) is surjective, and its kernel is the smallest normal subgroup containing \( \gamma \) . If \( {\pi }_{1}\left( {X,\widetilde{v}}\right) \) has a finite presentation\n\n\[ \n{\pi }_{1}\left( {X,\widetilde{v}}\right) \cong \left\langle {{\beta }_{1},\ldots ,{\beta }_{n} \mid {\sigma }_{1},\ldots ,{\sigma }_{s}}\right\rangle ,\n\]\n\nthen \( {\pi }_{1}\left( {\widetilde{X},\widetilde{v}}\right) \) has the presentation\n\n\[ \n{\pi }_{1}\left( {\widetilde{X},\widetilde{v}}\right) \cong \left\langle {{\beta }_{1},\ldots ,{\beta }_{n} \mid {\sigma }_{1},\ldots ,{\sigma }_{s},\tau }\right\rangle \n\]\n\nwhere \( \tau \) is an expression for \( \gamma \in {\pi }_{1}\left( {X,\widetilde{v}}\right) \) in terms of \( \left\{ {{\beta }_{1},\ldots ,{\beta }_{n}}\right\} \) .
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Proof. Let \( q : X \coprod D \rightarrow \widetilde{X} \) be the quotient map. As usual, we identify \( X \) with its image under \( q \), so we can consider \( X \) as a subspace of \( \widetilde{X} \) . First we set up some notation (see Fig. 10.7). Choose a point \( z \in \operatorname{Int}D \), set \( U = \operatorname{Int}D \) and \( V = X \coprod (D \smallsetminus \) \( \{ z\} ) \), and let \( \widetilde{U} = q\left( U\right) ,\widetilde{V} = q\left( V\right) \subseteq \widetilde{X} \) . Since \( U \) and \( V \) are saturated open subsets, the restrictions of \( q \) to \( U \) and \( V \) are quotient maps, and their images \( \widetilde{U},\widetilde{V} \) are open in \( \widetilde{X} \) . Moreover, \( \widetilde{U} \) and \( \widetilde{U} \cap \widetilde{V} \) are path-connected because they are continuous images of path-connected sets, and \( \widetilde{V} \) is path-connected because it is the union of the path-connected sets \( X \) and \( \widetilde{U} \cap \widetilde{V} \) that have the point \( \widetilde{v} \) in common.\n\nIn order to apply the Seifert-Van Kampen theorem in this situation, we need to work with a base point in \( \widetilde{U} \cap \widetilde{V} \) . Choose \( p \in \operatorname{Int}D \smallsetminus \{ z\} \approx {\mathbb{B}}^{2} \smallsetminus \{ 0\} \), and let \( c : I \rightarrow \) Int \( D \smallsetminus \{ z\} \) be a loop based at \( p \) whose path class generates \( {\pi }_{1}\left( {\operatorname{Int}D\smallsetminus \{ z\}, p}\right) \) . Then let \( \widetilde{p} = q\left( p\right) \in \widetilde{U} \cap \widetilde{V} \), and \( \widetilde{c} = q \circ c \) . In general, we use symbols without tildes to denote sets, points, or paths in \( D \), and the same symbols with tildes to denote their images in \( \widetilde{X} \) .\n\nThe restriction of \( q \) to \( U \) is a one-to-one quotient map and therefore a homeomorphism onto its image. Since \( U \) is simply connected, so is \( \widetilde{U} \) . On the other hand, \( \widetilde{U} \cap \widetilde{V} \) is the image under \( q \) of the saturated open subset Int \( D \smallsetminus \{ z\} \), so \( q : \) Int \( D \smallsetminus \{ z\} \rightarrow \widetilde{U} \cap \widetilde{V} \) is an injective quotient map and thus a homeomorphism. It follows that \( {\pi }_{1}\left( {\widetilde{U} \cap \widetilde{V},\widetilde{p}}\right) \) is the infinite cyclic group generated by \( \left\lbrack \widetilde{c}\right\rbrack \) . Now Corollary 10.5 implies that inclusion
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Yes
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Proposition 10.14 (Attaching an \( n \) -cell). Let \( X \) be a path-connected topological space, and let \( \widetilde{X} \) be a space obtained by attaching an \( n \) -cell to \( X \), with \( n \geq 3 \) . Then inclusion \( X \hookrightarrow \widetilde{X} \) induces an isomorphism of fundamental groups.
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Proof. We define open subsets \( \widetilde{U},\widetilde{V} \subseteq \widetilde{X} \) just as in the preceding proof. In this case, \( \widetilde{U} \cap \widetilde{V} \) is simply connected, because it is homeomorphic to \( {\mathbb{B}}^{n} \smallsetminus \{ 0\} \), and the result follows.
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No
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Theorem 10.15 (Fundamental Group of a Finite CW Complex). Suppose \( X \) is a connected finite \( {CW} \) complex, and \( v \) is a point in the 1-skeleton of \( X \) that is contained in the closure of every 2-cell. Let \( {\beta }_{1},\ldots ,{\beta }_{n} \) be generators for the free group \( {\pi }_{1}\left( {{X}_{1}, v}\right) \), and let \( {e}_{1},\ldots ,{e}_{k} \) be the 2-cells of \( X \) . For each \( i = 1,\ldots, k \) , let \( {\Phi }_{i} : {D}_{i} \rightarrow X \) be a characteristic map for \( {e}_{i} \) that takes \( {v}_{i} \in \partial {D}_{i} \) to \( v \), let \( {\varphi }_{i} = {\left. {\Phi }_{i}\right| }_{\partial {D}_{i}} : \partial {D}_{i} \rightarrow {X}_{1} \) be the corresponding attaching map, let \( {\alpha }_{i} \) be a generator of \( {\pi }_{1}\left( {\partial {D}_{i},{v}_{i}}\right) \), and let \( {\sigma }_{i} \) be an expression for \( {\left( {\varphi }_{i}\right) }_{ * }\left( {\alpha }_{i}\right) \in {\pi }_{1}\left( {{X}_{1}, v}\right) \) in terms of the generators \( \left\{ {\beta }_{i}\right\} \) . Then \( {\pi }_{1}\left( {X, v}\right) \) has the following presentation:\n\n\[ \n{\pi }_{1}\left( {X, v}\right) \cong \left\langle {{\beta }_{1},\ldots ,{\beta }_{n} \mid {\sigma }_{1},\ldots ,{\sigma }_{k}}\right\rangle .\n\]
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Proof. This follows immediately by induction from the two preceding propositions, using the result of Exercise 5.19.
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No
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Theorem 10.16 (Fundamental Groups and Polygonal Presentations). Let \( M\;{be} \) a topological space with a polygonal presentation \( \left\langle {{a}_{1},\ldots ,{a}_{n} \mid W}\right\rangle \) with one face, in which all vertices are identified to a single point. Then \( {\pi }_{1}\left( M\right) \) has the presentation \( \left\langle {{a}_{1},\ldots ,{a}_{n} \mid W}\right\rangle \)
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Proof. As we observed in Chapter 6, a polygonal presentation determines a CW decomposition of \( M \) in a natural way. Under the assumption that all the vertices are identified to a single point, the 1 -skeleton \( {M}_{1} \) is a wedge sum of circles, one for each symbol in the presentation, and thus its fundamental group has the presentation \( \left\langle {{a}_{1},\ldots ,{a}_{n} \mid \varnothing }\right\rangle \) . The attaching map of the single 2-cell maps the boundary of the polygon onto the loop in \( {M}_{1} \) obtained by following the generators in the order specified by the word \( \bar{W} \) (see Fig. 10.8). The result follows immediately from Proposition 10.13.
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Yes
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Corollary 10.17 (Fundamental Groups of Compact Surfaces). The fundamental groups of compact connected surfaces have the following presentations:\n\n(a) \( {\pi }_{1}\left( {\mathbb{S}}^{2}\right) \cong \langle \varnothing \mid \varnothing \rangle \) (the trivial group).\n\n(b) \( {\pi }_{1}\left( {{\mathbb{T}}^{2}\# \cdots \# {\mathbb{T}}^{2}}\right) \cong \left\langle {{\beta }_{1},{\gamma }_{1},\ldots ,{\beta }_{n},{\gamma }_{n} \mid {\beta }_{1}{\gamma }_{1}{\beta }_{1}^{-1}{\gamma }_{1}^{-1}\cdots {\beta }_{n}{\gamma }_{n}{\beta }_{n}^{-1}{\gamma }_{n}^{-1} = 1}\right\rangle \) .\n\n(c) \( {\pi }_{1}\left( {{\mathbb{P}}^{2}\# \cdots \# {\mathbb{P}}^{2}}\right) \cong \left\langle {{\beta }_{1},\ldots ,{\beta }_{n} \mid {\beta }_{1}^{2}\cdots {\beta }_{n}^{2} = 1}\right\rangle \) .
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Proof. For \( {\mathbb{S}}^{2} \), this follows from Theorem 7.20. For all of the other surfaces, it follows from Theorem 10.16, using the standard presentations of Example 6.13, and noting that for each surface other than the sphere, the standard presentation identifies all of the vertices to one point, as you can easily check.\n\nIn particular, for the torus this gives \( {\pi }_{1}\left( {\mathbb{T}}^{2}\right) \cong \langle \beta ,\gamma \mid {\beta \gamma } = {\gamma \beta }\rangle \), which agrees with the result we derived earlier. In the case of the projective plane, this gives \( {\pi }_{1}\left( {\mathbb{P}}^{2}\right) \cong \left\langle {\beta \mid {\beta }^{2} = 1}\right\rangle \cong \mathbb{Z}/2. \)
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Yes
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Theorem 10.22 (Classification of Compact Surfaces, Part II). Every nonempty, compact, connected 2-manifold is homeomorphic to exactly one of the surfaces \( {\mathbb{S}}^{2} \) , \( {\mathbb{T}}^{2}\widetilde{\# }\cdots \# {\mathbb{T}}^{2} \), or \( {\mathbb{P}}^{2}\# \cdots \# {\mathbb{P}}^{2} \) .
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Proof. Theorem 6.15 showed that every nonempty, compact, connected surface is homeomorphic to one of the surfaces on the list, so we need only show that no two surfaces on the list are homeomorphic to each other. First note that the sphere cannot be homeomorphic to a connected sum of tori or projective planes, because one has a trivial fundamental group and the other does not. Next, if \( M \) is a connected sum of projective planes, then \( \operatorname{Ab}\left( {{\pi }_{1}\left( M\right) }\right) \) contains a nontrivial torsion element, whereas the abelianized fundamental groups of connected sums of tori are torsion-free. Therefore, no connected sum of projective planes can be homeomorphic to a connected sum of tori. If \( M \) is a connected sum of \( n \) tori, then its abelianized fundamental group has rank \( {2n} \) . Thus the genus (i.e., the number of tori in the connected sum) can be recovered from the fundamental group, so the genus of an orientable surface is a topological invariant. Similarly, a connected sum of \( n \) projective planes has abelianized fundamental group of rank \( n - 1 \), so once again the genus is a topological invariant.
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Yes
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Corollary 10.23. A connected sum of projective planes is not orientable.
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Proof. By the argument in Chapter 6, if a manifold admits an oriented presentation, then it is homeomorphic to a sphere or a connected sum of tori. The preceding corollary showed that a connected sum of projective planes is not homeomorphic to any of these surfaces.
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No
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Corollary 10.24. Orientability of a compact surface is a topological invariant.
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Proof. Combining the results of Proposition 6.20 and Corollary 10.23, we can conclude that no surface that has an oriented presentation is homeomorphic to one that does not.
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Yes
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Corollary 10.25. The Euler characteristic of a surface presentation is a topological invariant.
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Proof. Suppose \( \mathcal{P} \) and \( \mathcal{Q} \) are polygonal surface presentations such that \( \left| \mathcal{P}\right| \approx \left| \mathcal{Q}\right| \) . Each of these presentations can be transformed into one of the standard ones by elementary transformations, and since the surfaces represented by different standard presentations are not homeomorphic, both presentations must reduce to the same standard one. Since the Euler characteristic of a presentation is unchanged by elementary transformations, the two presentations must have had the same Euler characteristic to begin with.
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Yes
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The exponential quotient map \( \varepsilon : \mathbb{R} \rightarrow {\mathbb{S}}^{1} \) given by \( \varepsilon \left( x\right) = {e}^{2\pi ix} \) is a covering map;
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this is the content of Proposition 8.1.
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No
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The \( n \) th power map \( {p}_{n} : {\mathbb{S}}^{1} \rightarrow {\mathbb{S}}^{1} \) given by \( {p}_{n}\left( z\right) = {z}^{n} \) is also a covering map.
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For each \( {z}_{0} \in {\mathbb{S}}^{1} \), the set \( U = {\mathbb{S}}^{1} \smallsetminus \left\{ {-{z}_{0}}\right\} \) has preimage equal to \( \left\{ {z \in {\mathbb{S}}^{1} : {z}^{n} \neq - {z}_{0}}\right\} \), which has \( n \) components, each of which is an open arc mapped homeomorphically by \( {p}_{n} \) onto \( U \) .
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Yes
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Lemma 11.10 (Existence of Local Sections). Let \( q : E \rightarrow X \) be a covering map. Given any evenly covered open subset \( U \subseteq X \), any \( x \in U \), and any \( {e}_{0} \) in the fiber over \( x \), there exists a local section \( \sigma : U \rightarrow E \) such that \( \sigma \left( x\right) = {e}_{0} \) .
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Proof. Let \( {\widetilde{U}}_{0} \) be the sheet of \( {q}^{-1}\left( U\right) \) containing \( {e}_{0} \) . Since the restriction of \( q \) to \( {\widetilde{U}}_{0} \) is a homeomorphism, we can just take \( \sigma = {\left( {\left. q\right| }_{{\widetilde{U}}_{0}}\right) }^{-1} \) .
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Yes
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Proposition 11.11. For every covering map \( q : E \rightarrow X \), the cardinality of the fibers \( {q}^{-1}\left( x\right) \) is the same for all fibers.
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Proof. Define an equivalence relation on \( X \) by saying that \( x \sim {x}^{\prime } \) if and only if \( {q}^{-1}\left( x\right) \) and \( {q}^{-1}\left( {x}^{\prime }\right) \) have the same cardinality. Suppose \( x \in X \), and let \( U \) be an evenly covered neighborhood of \( x \) . Then each sheet of \( {q}^{-1}\left( U\right) \) contains exactly one point of each fiber, so for any \( {x}^{\prime } \in U \), there are one-to-one correspondences\n\n\[ \n{q}^{-1}\left( x\right) \leftrightarrow \left\{ {\text{ sheets of }{q}^{-1}\left( U\right) }\right\} \leftrightarrow {q}^{-1}\left( {x}^{\prime }\right) ,\n\]\n\nwhich shows that \( {x}^{\prime } \sim x \) . It follows that \( U \) is contained in the equivalence class \( \left\lbrack x\right\rbrack \), so each equivalence class is open. Thus, by Exercise 4.3, there is only one equivalence class.
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Yes
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Theorem 11.15 (Monodromy Theorem). Let \( q : E \rightarrow X \) be a covering map. Suppose \( f \) and \( g \) are paths in \( X \) with the same initial point and the same terminal point, and \( {\widetilde{f}}_{e},{\widetilde{g}}_{e} \) are their lifts with the same initial point \( e \in E \) .\n\n(a) \( {\widetilde{f}}_{e} \sim {\widetilde{g}}_{e} \) if and only if \( f \sim g \) .\n\n(b) If \( f \sim g \), then \( {\widetilde{f}}_{e}\left( 1\right) = {\widetilde{g}}_{e}\left( 1\right) \) .
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Proof. If \( {\widetilde{f}}_{e} \sim {\widetilde{g}}_{e} \), then \( f \sim g \) because composition with \( q \) preserves path homotopy. Conversely, suppose \( f \sim g \), and let \( H : I \times I \rightarrow X \) be a path homotopy between them. Then the homotopy lifting property implies that \( H \) lifts to a homotopy \( \widetilde{H} \) : \( I \times \) \( I \rightarrow E \) between \( {\widetilde{f}}_{e} \) and some lift of \( g \) starting at \( e \), which must be equal to \( {\widetilde{g}}_{e} \) by the unique lifting property. This proves (a). To prove (b), just note that \( f \sim g \) implies that \( {\widetilde{f}}_{e} \) and \( {\widetilde{g}}_{e} \) are path-homotopic by (a), so they have the same terminal point.
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Yes
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Theorem 11.16 (Injectivity Theorem). Let \( q : E \rightarrow X \) be a covering map. For any point \( e \in E \), the induced homomorphism \( {q}_{ * } : {\pi }_{1}\left( {E, e}\right) \rightarrow {\pi }_{1}\left( {X, q\left( e\right) }\right) \) is injective.
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Proof. Suppose \( \left\lbrack f\right\rbrack \in {\pi }_{1}\left( {E, e}\right) \) is in the kernel of \( {q}_{ * } \) . This means that \( {q}_{ * }\left\lbrack f\right\rbrack = \left\lbrack {c}_{x}\right\rbrack \) , where \( x = q\left( e\right) \) . In other words, \( q \circ f \sim {c}_{x} \) in \( X \) . By the monodromy theorem, therefore, any lifts of \( q \circ f \) and \( {c}_{x} \) that start at the same point must be path-homotopic in \( E \) . Now, \( f \) is a lift of \( q \circ f \) starting at \( e \), and the constant loop \( {c}_{e} \) is a lift of \( {c}_{x} \) starting at the same point; therefore, \( f \sim {c}_{e} \) in \( E \), which means that \( \left\lbrack f\right\rbrack = 1 \) .
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Yes
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Theorem 11.18 (Lifting Criterion). Suppose \( q : E \rightarrow X \) is a covering map. Let \( Y \) be a connected and locally path-connected space, and let \( \varphi : Y \rightarrow X \) be a continuous map. Given any points \( {y}_{0} \in Y \) and \( {e}_{0} \in E \) such that \( q\left( {e}_{0}\right) = \varphi \left( {y}_{0}\right) \), the map \( \varphi \) has a lift \( \widetilde{\varphi } : Y \rightarrow E \) satisfying \( \widetilde{\varphi }\left( {y}_{0}\right) = {e}_{0} \) if and only if the subgroup \( {\varphi }_{ * }{\pi }_{1}\left( {Y,{y}_{0}}\right) \) of \( {\pi }_{1}\left( {X,\varphi \left( {y}_{0}\right) }\right) \) is contained in \( {q}_{ * }{\pi }_{1}\left( {E,{e}_{0}}\right) \) .
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Proof. The necessity of the algebraic condition is easy to prove (and, in fact, does not require any connectivity assumptions about \( Y \) ). If \( \widetilde{\varphi } \) satisfies the conditions in the statement of the theorem, the following diagram commutes:\n\n\n\nTherefore, \( {\varphi }_{ * }{\pi }_{1}\left( {Y,{y}_{0}}\right) = {q}_{ * }{\widetilde{\varphi }}_{ * }{\pi }_{1}\left( {Y,{y}_{0}}\right) \subseteq {q}_{ * }{\pi }_{1}\left( {E,{e}_{0}}\right) .\n\nTo prove the converse, we \
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No
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Theorem 11.22 (The Monodromy Action). Suppose \( q : E \rightarrow X \) is a covering map and \( x \in X \) . There is a transitive right action of \( {\pi }_{1}\left( {X, x}\right) \) on the fiber \( {q}^{-1}\left( x\right) \), called the monodromy action, given by \( e \cdot \left\lbrack f\right\rbrack = {\widetilde{f}}_{e}\left( 1\right) \) for \( e \in {q}^{-1}\left( x\right) \) and \( \left\lbrack f\right\rbrack \in {\pi }_{1}\left( {X, x}\right) \) .
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Proof. If \( e \) is any point in \( {q}^{-1}\left( x\right) \), the path lifting property shows that every loop \( f \) based at \( x \) has a unique lift to a path \( {\widetilde{f}}_{e} \) starting at \( e \) . The fact that \( f \) is a loop guarantees that \( {\widetilde{f}}_{e}\left( 1\right) \in {q}^{-1}\left( x\right) \), and the monodromy theorem guarantees that \( {\widetilde{f}}_{e}\left( 1\right) \) depends only on the path class of \( f \) ; therefore, \( e \cdot \left\lbrack f\right\rbrack \) is well defined.\n\nTo see that this is a group action, we need to check two things:\n\n(i) \( e \cdot \left\lbrack {c}_{x}\right\rbrack = e \) .\n\n(ii) \( \left( {e \cdot \left\lbrack f\right\rbrack }\right) \cdot \left\lbrack g\right\rbrack = e \cdot \left( {\left\lbrack f\right\rbrack \cdot \left\lbrack g\right\rbrack }\right) \) .\n\nFor (i), just observe that the constant path \( {c}_{e} \) is the unique lift of \( {c}_{x} \) starting at \( e \) , and therefore \( e \cdot \left\lbrack {c}_{x}\right\rbrack = {c}_{e}\left( 1\right) = e \) . To prove the composition property (ii), suppose \( f \) and \( g \) are two loops based at \( x \), and let \( z = e \cdot \left\lbrack f\right\rbrack = {\widetilde{f}}_{e}\left( 1\right) \) . Then by definition, \( \left( {e \cdot \left\lbrack f\right\rbrack }\right) \cdot \left\lbrack g\right\rbrack = {\widetilde{g}}_{z}\left( 1\right) \) (Fig. 11.7). On the other hand, \( {\widetilde{f}}_{e} \cdot {\widetilde{g}}_{z} \) is the lift of \( f \cdot g \) starting at \( e \), which means that\n\n\[ e \cdot \left( {\left\lbrack f\right\rbrack \cdot \left\lbrack g\right\rbrack }\right) = e \cdot \left\lbrack {f \cdot g}\right\rbrack = \left( {{\widetilde{f}}_{e} \cdot {\widetilde{g}}_{z}}\right) \left( 1\right) = {\widetilde{g}}_{z}\left( 1\right) = \left( {e \cdot \left\lbrack f\right\rbrack }\right) \cdot \left\lbrack g\right\rbrack . \]\n\nNow we need to show that the action is transitive. Because \( E \) is path-connected, any two points \( e,{e}^{\prime } \) in the fiber over \( x \) are joined by a path \( h \) in \( E \) . Setting \( f = q \circ h \) , we see immediately that \( h \) is the lift of \( f \) starting at \( e \), and therefore \( e \cdot \left\lbrack f\right\rbrack = {e}^{\prime } \) .
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Yes
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Proposition 11.23 (Isotropy Groups of Transitive \( G \) -sets). Suppose \( G \) is a group and \( S \) is a transitive right \( G \) -set.\n\n(a) For each \( s \in S \) and \( g \in G \), \n\n\[ \n{G}_{s \cdot g} = {g}^{-1}{G}_{s}g \n\] \n\n(b) The set \( \\left\\{ {{G}_{s} : s \in S}\\right\\} \) of all isotropy groups is exactly one conjugacy class of subgroups of \( G \). This conjugacy class is called the isotropy type of \( S \).
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Proof. The proof of (a) is just a computation: for \( s \in S \) and \( g \in G \), \n\n\[ \n{G}_{s \cdot g} = \\left\\{ {{g}^{\\prime } \in G : \\left( {s \cdot g}\\right) \cdot {g}^{\\prime } = s \cdot g}\\right\\} \n\] \n\n\[ \n= \\left\\{ {{g}^{\\prime } \in G : s \cdot \\left( {g{g}^{\\prime }{g}^{-1}}\\right) = s}\\right\\} \n\] \n\n\[ \n= \\left\\{ {{g}^{\\prime } \in G : g{g}^{\\prime }{g}^{-1} \in {G}_{s}}\\right\\} \n\] \n\n\[ \n= {g}^{-1}{G}_{s}g.\\text{.} \n\] \n\nThen (b) follows from (a): if \( s \) and \( {s}^{\\prime } = s \cdot g \) are any two elements of \( S \), their isotropy groups are conjugate by (11.1); and conversely, if \( {G}_{s} \) is the isotropy group of some element \( s \in S \) and \( H = g{G}_{s}{g}^{-1} \) is any subgroup conjugate to \( {G}_{s} \), then \( H \) is the isotropy group of \( s \cdot {g}^{-1} \) .
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Yes
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Proposition 11.24 (Properties of \( G \) -Equivariant Maps). Suppose \( G \) is a group, and \( {S}_{1},{S}_{2} \) are transitive right \( G \) -sets.\n\n(a) Any two \( G \) -equivariant maps from \( {S}_{1} \) to \( {S}_{2} \) that agree on one element of \( {S}_{1} \) are identical.
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Proof. Suppose \( \varphi ,{\varphi }^{\prime } : {S}_{1} \rightarrow {S}_{2} \) are \( G \) -equivariant and \( \varphi \left( {s}_{1}\right) = {\varphi }^{\prime }\left( {s}_{1}\right) \) for some \( {s}_{1} \in \) \( {S}_{1} \) . Any \( s \in {S}_{1} \) can be written \( s = {s}_{1} \cdot g \) for some \( g \in G \) (because \( G \) acts transitively), and then it follows from equivariance that\n\n\[ \varphi \left( s\right) = \varphi \left( {{s}_{1} \cdot g}\right) = \varphi \left( {s}_{1}\right) \cdot g = {\varphi }^{\prime }\left( {s}_{1}\right) \cdot g = {\varphi }^{\prime }\left( {{s}_{1} \cdot g}\right) = {\varphi }^{\prime }\left( s\right) . \]\n\nThis proves (a).
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Yes
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Proposition 11.27 (Orbit Criterion for \( G \) -Automorphisms). Suppose \( S \) is a transitive right \( G \) -set. For any \( {s}_{1},{s}_{2} \in S \), there exists a (necessarily unique) \( \varphi \in \) \( {\operatorname{Aut}}_{G}\left( S\right) \) such that \( \varphi \left( {s}_{1}\right) = {s}_{2} \) if and only if the isotropy groups \( {G}_{{s}_{1}} \) and \( {G}_{{s}_{2}} \) are equal.
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Proof. This is an immediate consequence of Proposition 11.26(a).
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No
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Theorem 11.28 (Algebraic Characterization of \( G \) -Automorphism Groups). Let \( S \) be a transitive right \( G \) -set, and let \( {s}_{0} \) be any element of \( S \) . For each \( \gamma \in {N}_{G}\left( {G}_{{s}_{0}}\right) \) , there is a unique \( G \) -automorphism \( {\varphi }_{\gamma } \in {\operatorname{Aut}}_{G}\left( S\right) \) such that \( {\varphi }_{\gamma }\left( {s}_{0}\right) = {s}_{0} \cdot \gamma \) . The map \( \gamma \mapsto {\varphi }_{\gamma } \) is a surjective group homomorphism from \( {N}_{G}\left( {G}_{{s}_{0}}\right) \) to \( {\operatorname{Aut}}_{G}\left( S\right) \) whose kernel is \( {G}_{{s}_{0}} \), and thus descends to an isomorphism\n\n\[ \n{N}_{G}\left( {G}_{{s}_{0}}\right) /{G}_{{s}_{0}} \cong {\operatorname{Aut}}_{G}\left( S\right) .\n\]
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Proof. Suppose \( \gamma \in {N}_{G}\left( {G}_{{s}_{0}}\right) \) . Then \( {\gamma }^{-1} \in {N}_{G}\left( {G}_{{s}_{0}}\right) \) as well. Together with (11.1), this implies \( {G}_{{s}_{0}} = {\gamma }^{-1}{G}_{{s}_{0}}\gamma = {G}_{{s}_{0}} \cdot \gamma \) . Then Proposition 11.27 shows that there is a unique \( G \) -automorphism \( {\varphi }_{\gamma } \) taking \( {s}_{0} \) to \( {s}_{0} \cdot \gamma \) .\n\nTo show that the map \( \gamma \mapsto {\varphi }_{\gamma } \) is a homomorphism, let \( {\gamma }_{1},{\gamma }_{2} \in {N}_{G}\left( {G}_{{s}_{0}}\right) \) be arbitrary. Then\n\n\[ \n{\varphi }_{{\gamma }_{1}} \circ {\varphi }_{{\gamma }_{2}}\left( {s}_{0}\right) = {\varphi }_{{\gamma }_{1}}\left( {{s}_{0} \cdot {\gamma }_{2}}\right) = {\varphi }_{{\gamma }_{1}}\left( {s}_{0}\right) \cdot {\gamma }_{2} = \left( {{s}_{0} \cdot {\gamma }_{1}}\right) \cdot {\gamma }_{2} = {s}_{0} \cdot {\gamma }_{1}{\gamma }_{2} = {\varphi }_{{\gamma }_{1}{\gamma }_{2}}\left( {s}_{0}\right) .\n\]\n\nSince two \( G \) -automorphisms that agree on one element are equal, this shows that \( {\varphi }_{{\gamma }_{1}} \circ {\varphi }_{{\gamma }_{2}} = {\varphi }_{{\gamma }_{1}{\gamma }_{2}}. \)\n\nTo prove surjectivity, let \( \varphi \in {\operatorname{Aut}}_{G}\left( S\right) \) be arbitrary. By transitivity, there is some \( \gamma \in G \) such that \( {s}_{0} \cdot \gamma = \varphi \left( {s}_{0}\right) \) . By Proposition 11.27, this implies that \( {G}_{{s}_{0}} = {G}_{{s}_{0} \cdot \gamma } = {\gamma }^{-1}{G}_{{s}_{0}}\gamma \), so \( {\gamma }^{-1} \in {N}_{G}\left( {G}_{{s}_{0}}\right) \), which implies that \( \gamma \) is too. It follows that there is a unique \( G \) -automorphism \( {\varphi }_{\gamma } \) such that \( {\varphi }_{\gamma }\left( {s}_{0}\right) = {s}_{0} \cdot \gamma \), and since \( \varphi \) is such an automorphism, we must have \( {\varphi }_{\gamma } = \varphi \).\n\nFinally,\n\n\[ \n{\varphi }_{\gamma } = {\operatorname{Id}}_{S}\; \Leftrightarrow \;{\varphi }_{\gamma }\left( {s}_{0}\right) = {s}_{0}\; \Leftrightarrow \;{s}_{0} \cdot \gamma = {s}_{0}\; \Leftrightarrow \;\gamma \in {G}_{{s}_{0}},\n\]\n\nwhich shows that the kernel of the map \( \gamma \mapsto {\varphi }_{\gamma } \) is exactly \( {G}_{{s}_{0}} \) .
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Yes
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Theorem 11.29 (Isotropy Groups of the Monodromy Action). Suppose \( q \) : \( E \rightarrow \) \( X \) is a covering map and \( x \in X \) . For each \( e \in {q}^{-1}\left( x\right) \), the isotropy group of \( e \) under the monodromy action is \( {q}_{ * }{\pi }_{1}\left( {E, e}\right) \subseteq {\pi }_{1}\left( {X, x}\right) \) .
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Proof. Let \( e \in {q}^{-1}\left( x\right) \) be arbitrary, and suppose first that \( \left\lbrack f\right\rbrack \) is in the isotropy group of \( e \) . This means \( {\widetilde{f}}_{e}\left( 1\right) = e \cdot \left\lbrack f\right\rbrack = e \), which is to say that \( {\widetilde{f}}_{e} \) is a loop and thus represents an element of \( {\pi }_{1}\left( {E, e}\right) \) . It is easy to check that \( {q}_{ * }\left\lbrack {\widetilde{f}}_{e}\right\rbrack = \left\lbrack f\right\rbrack \), so \( \left\lbrack f\right\rbrack \in {q}_{ * }{\pi }_{1}\left( {E, e}\right) \) . Conversely, if \( \left\lbrack f\right\rbrack \in {q}_{ * }{\pi }_{1}\left( {E, e}\right) \), then there is a loop \( g : I \rightarrow E \) based at \( e \) such that \( {q}_{ * }\left\lbrack g\right\rbrack = \left\lbrack f\right\rbrack \), which means that \( q \circ g \sim f \) . If we let \( {f}^{\prime } = q \circ g \) , then \( g = {\widetilde{f}}_{e}^{\prime } \) (by uniqueness of lifts), and \( e \cdot \left\lbrack f\right\rbrack = e \cdot \left\lbrack {f}^{\prime }\right\rbrack = {\widetilde{f}}_{e}^{\prime }\left( 1\right) = g\left( 1\right) = e \) , which means that \( \left\lbrack f\right\rbrack \) is in the isotropy group of \( e \) .
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Yes
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Corollary 11.30. Suppose \( q : E \rightarrow X \) is a covering map. The monodromy action is free on each fiber of \( q \) if and only if \( E \) is simply connected.
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Proof. The action is free if and only if each isotropy group is trivial, which by Theorem 11.29 is equivalent to \( {q}_{ * }{\pi }_{1}\left( {E, e}\right) \) being the trivial group for each \( e \) in the fiber. Since \( {q}_{ * } \) is injective, this is true if and only if \( E \) is simply connected.
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Yes
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Corollary 11.31. Suppose \( q : E \rightarrow X \) is a covering map and \( E \) is simply connected. Then each fiber of \( q \) has the same cardinality as the fundamental group of \( X \) .
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Proof. By the previous corollary, the monodromy action is free. Choose a base point \( x \in X \) and a point \( e \) in the fiber over \( x \), and consider the map \( {\pi }_{1}\left( {X, x}\right) \rightarrow {q}^{-1}\left( x\right) \) given by \( \left\lbrack f\right\rbrack \mapsto e \cdot \left\lbrack f\right\rbrack \) . It is surjective because the monodromy action is transitive, and it is injective because the action is free.
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Yes
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Corollary 11.33 (Coverings of Simply Connected Spaces). If \( X \) is a simply connected space, every covering map \( q : E \rightarrow X \) is a homeomorphism.
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Proof. The injectivity theorem shows that \( E \) is also simply connected. Then Corollary 11.31 shows that the cardinality of the fibers is 1, so \( q \) is injective. Thus it is a homeomorphism by Proposition 11.1(b).
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Yes
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Theorem 11.34 (Conjugacy Theorem). Let \( q : E \rightarrow X \) be a covering map. For any \( x \in X \), as e varies over the fiber \( {q}^{-1}\left( x\right) \), the set of induced subgroups \( {q}_{ * }{\pi }_{1}\left( {E, e}\right) \) is exactly one conjugacy class in \( {\pi }_{1}\left( {X, x}\right) \) .
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Proof. Given \( x \in X \), Theorem 11.29 shows that the set of subgroups \( {q}_{ * }{\pi }_{1}\left( {E, e}\right) \) as \( e \) varies over \( {q}^{-1}\left( x\right) \) is equal to the set of isotropy groups of points in \( {q}^{-1}\left( x\right) \) under the monodromy action. Then Proposition 11.23(b) shows that this set of isotropy groups is exactly one conjugacy class.
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Yes
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Proposition 11.35 (Characterizations of Normal Coverings). Suppose \( q : E \rightarrow \) \( X \) is a covering map. Then the following are equivalent:\n\n(a) The subgroup \( {q}_{ * }{\pi }_{1}\left( {E, e}\right) \) is normal for some \( e \in E \) (i.e., \( q \) is normal).\n\n(b) For some \( x \in X \), the subgroups \( {q}_{ * }{\pi }_{1}\left( {E, e}\right) \) are the same for all \( e \in {q}^{-1}\left( x\right) \) .\n\n(c) For every \( x \in X \), the subgroups \( {q}_{ * }{\pi }_{1}\left( {E, e}\right) \) are the same for all \( e \in {q}^{-1}\left( x\right) \) .\n\n(d) The subgroup \( {q}_{ * }{\pi }_{1}\left( {E, e}\right) \) is normal for every \( e \in E \) .
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Proof. Because a subgroup is normal if and only if it is the sole member of its conjugacy class, the implications \( \left( \mathrm{d}\right) \Rightarrow \left( \mathrm{c}\right) \Rightarrow \left( \mathrm{b}\right) \Rightarrow \left( \mathrm{a}\right) \) are easy consequences of the conjugacy theorem. Thus we need only prove (a) \( \Rightarrow \) (d).\n\nAssume that (a) holds, and let \( {e}_{0} \in E \) be a point such that \( {q}_{ * }{\pi }_{1}\left( {E,{e}_{0}}\right) \) is normal in \( {\pi }_{1}\left( {X,{x}_{0}}\right) \), where \( {x}_{0} = q\left( {e}_{0}\right) \) . Suppose \( e \) is any other point of \( E \), and let \( x = q\left( e\right) \) . Let \( h \) be a path in \( E \) from \( {e}_{0} \) to \( e \), and set \( g = q \circ h \), which is a path in \( X \) from \( {x}_{0} \) to \( x \) (Fig. 11.8). We have four maps \n\n(11.2)\n\nwhere \( {\Phi }_{h}\left\lbrack f\right\rbrack = \left\lbrack \bar{h}\right\rbrack \cdot \left\lbrack f\right\rbrack \cdot \left\lbrack h\right\rbrack \), and \( {\Phi }_{g} \) is defined similarly. The top and bottom rows are isomorphisms by Theorem 7.13, and the diagram commutes because\n\n\[ \n{q}_{ * }{\Phi }_{h}\left\lbrack f\right\rbrack = {q}_{ * }\left( {\left\lbrack \bar{h}\right\rbrack \cdot \left\lbrack f\right\rbrack \cdot \left\lbrack h\right\rbrack }\right) = \left\lbrack \bar{g}\right\rbrack \cdot {q}_{ * }\left\lbrack f\right\rbrack \cdot \left\lbrack g\right\rbrack = {\Phi }_{g}{q}_{ * }\left\lbrack f\right\rbrack .\n\]\n\nIt follows that \( {\Phi }_{g} \) takes \( {q}_{ * }{\pi }_{1}\left( {E,{e}_{0}}\right) \) to \( {q}_{ * }{\pi }_{1}\left( {E, e}\right) \) . Since an isomorphism takes normal subgroups to normal subgroups, (d) follows.\n\n\n\nFig. 11.8: Proof of Proposition 11.35.
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Yes
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Proposition 11.36 (Properties of Covering Homomorphisms). Let \( {q}_{1} : {E}_{1} \rightarrow X \) and \( {q}_{2} : {E}_{2} \rightarrow X \) be coverings of the same space \( X \). (a) If two covering homomorphisms from \( {q}_{1} \) to \( {q}_{2} \) agree at one point of \( {E}_{1} \), then they are equal.
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Proof. A covering homomorphism from \( {q}_{1} \) to \( {q}_{2} \) can also be viewed as a lift of \( {q}_{1} \):  (11.3) Thus (a) follows from the unique lifting property.
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Yes
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Theorem 11.37 (Covering Homomorphism Criterion). Let \( {q}_{1} : {E}_{1} \rightarrow X \) and \( {q}_{2} : {E}_{2} \rightarrow X \) be two coverings of the same space \( X \), and suppose \( {e}_{1} \in {E}_{1} \) and \( {e}_{2} \in {E}_{2} \) are base points such that \( {q}_{1}\left( {e}_{1}\right) = {q}_{2}\left( {e}_{2}\right) \). There exists a covering homomorphism from \( {q}_{1} \) to \( {q}_{2} \) taking \( {e}_{1} \) to \( {e}_{2} \) if and only if \( {q}_{1 * }{\pi }_{1}\left( {{E}_{1},{e}_{1}}\right) \subseteq {q}_{2 * }{\pi }_{1}\left( {{E}_{2},{e}_{2}}\right) \).
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Proof. Because a covering homomorphism from \( {q}_{1} \) to \( {q}_{2} \) is a lift of \( {q}_{1} \) as in (11.3), both the necessity and the sufficiency of the subgroup condition follow from the lifting criterion (Theorem 11.18).
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Yes
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Consider the following two coverings of \( {\mathbb{T}}^{2} \) : the first is \( {\varepsilon }^{2} : {\mathbb{R}}^{2} \rightarrow \) \( {\mathbb{T}}^{2} \), the covering of Example 11.5 (the product of two copies of \( \varepsilon : \mathbb{R} \rightarrow {\mathbb{S}}^{1} \) ); and the second is the map \( q : {\mathbb{S}}^{1} \times \mathbb{R} \rightarrow {\mathbb{T}}^{2} \) given by \( q\left( {z, e}\right) = \left( {z,\varepsilon \left( e\right) }\right) \) . Identifying \( {\pi }_{1}\left( {\mathbb{T}}^{2}\right) \) with \( \mathbb{Z} \times \mathbb{Z} \), we see that \( {\left( {\varepsilon }^{2}\right) }_{ * }{\pi }_{1}\left( {\mathbb{R}}^{2}\right) \) is trivial, while \( {q}_{ * }{\pi }_{1}\left( {{\mathbb{S}}^{1} \times \mathbb{R}}\right) = \) \( \mathbb{Z} \times \{ 0\} \) . Therefore, there exists a covering homomorphism from \( {\varepsilon }^{2} \) to \( q \) . (Why do the base points not matter?)
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It is easy to check that \( \varphi \left( {x, e}\right) = \left( {\varepsilon \left( x\right), e}\right) \) is such a homomorphism.
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Yes
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Theorem 11.40 (Covering Isomorphism Criterion). Suppose \( {q}_{1} : {E}_{1} \rightarrow X \) and \( {q}_{2} : {E}_{2} \rightarrow X \) are two coverings of the same space \( X \). (a) Given \( {e}_{1} \in {E}_{1} \) and \( {e}_{2} \in {E}_{2} \) such that \( {q}_{1}\left( {e}_{1}\right) = {q}_{2}\left( {e}_{2}\right) \), there exists a (necessarily unique) covering isomorphism from \( {q}_{1} \) to \( {q}_{2} \) taking \( {e}_{1} \) to \( {e}_{2} \) if and only if \( {q}_{1 * }{\pi }_{1}\left( {{E}_{1},{e}_{1}}\right) = {q}_{2 * }{\pi }_{1}\left( {{E}_{2},{e}_{2}}\right) \) .
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Proof. First we prove (a). Suppose there exists a covering isomorphism \( \varphi : {E}_{1} \rightarrow \) \( {E}_{2} \) such that \( \varphi \left( {e}_{1}\right) = {e}_{2} \), and let \( x = {q}_{1}\left( {e}_{1}\right) = {q}_{2}\left( {e}_{2}\right) \) . By Proposition 11.36(b), \( \varphi \) restricts to a \( {\pi }_{1}\left( {X, x}\right) \) -isomorphism from \( {q}_{1}^{-1}\left( x\right) \) to \( {q}_{2}^{-1}\left( x\right) \) taking \( {e}_{1} \) to \( {e}_{2} \), so it follows from Proposition 11.26 that the isotropy groups of \( {e}_{1} \) and \( {e}_{2} \), namely \( {q}_{1 * }{\pi }_{1}\left( {{E}_{1},{e}_{1}}\right) \) and \( {q}_{2 * }{\pi }_{1}\left( {{E}_{2},{e}_{2}}\right) \), are equal.\n\nConversely, suppose \( {q}_{2 * }{\pi }_{1}\left( {{E}_{2},{e}_{2}}\right) = {q}_{1 * }{\pi }_{1}\left( {{E}_{1},{e}_{1}}\right) \) . Then by the covering homomorphism criterion, there exist covering homomorphisms \( \varphi : {E}_{1} \rightarrow {E}_{2} \) and \( \psi : {E}_{2} \rightarrow {E}_{1} \), with \( \varphi \left( {e}_{1}\right) = {e}_{2} \) and \( \psi \left( {e}_{2}\right) = {e}_{1} \) . The composite map \( \psi \circ \varphi \) is a covering homomorphism from \( {E}_{1} \) to itself that fixes \( {e}_{1} \), so it is the identity. Similarly, \( \varphi \circ \psi \) is the identity, so \( \varphi \) is the required covering isomorphism.
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Yes
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Proposition 11.41 (Universality of Simply Connected Coverings). (a) Let \( q : E \rightarrow X \) be a covering map with \( E \) simply connected. If \( {q}^{\prime } : {E}^{\prime } \rightarrow X \) is any covering, there exists a covering map \( Q : E \rightarrow {E}^{\prime } \) such that the following diagram commutes:
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Proof. Since the trivial subgroup is contained in every other subgroup, part (a) follows from the covering homomorphism criterion and the fact that every covering homomorphism is a covering map.
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No
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Theorem 11.43 (Existence of the Universal Covering Space). Every connected and locally simply connected topological space (in particular, every connected manifold) has a universal covering space.
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Proof. To get an idea how to proceed, suppose for a moment that \( X \) does have a universal covering \( q : \widetilde{X} \rightarrow X \) . The key fact is that once we choose base points \( {\widetilde{x}}_{0} \in \widetilde{X} \) and \( {x}_{0} = q\left( {\widetilde{x}}_{0}\right) \in X \), the fiber \( {q}^{-1}\left( x\right) \) over any \( x \in X \) is in one-to-one correspondence with path classes from \( {x}_{0} \) to \( x \) . To see why, define a map \( E \) from the set of such path classes to \( {q}^{-1}\left( x\right) \) by sending \( \left\lbrack f\right\rbrack \) to the terminal point of the lift of \( f \) starting at \( {\widetilde{x}}_{0} \) . Since lifts of homotopic paths have the same terminal point by the monodromy theorem, \( E \) is well defined. \( E \) is surjective, because given any \( \widetilde{x} \) in the fiber over \( x \), there is a path \( \widetilde{f} \) from \( {\widetilde{x}}_{0} \) to \( \widetilde{x} \), and then \( q \circ \widetilde{f} \) is a path from \( {x}_{0} \) to \( x \) whose lift ends at \( \widetilde{x} \) . Injectivity of \( E \) follows from the fact that \( \widetilde{X} \) is simply connected: if \( {f}_{1},{f}_{2} \) are two paths from \( {x}_{0} \) to \( x \) whose lifts \( {\widetilde{f}}_{1},{\widetilde{f}}_{2} \) end at the same point, then \( {\widetilde{f}}_{1} \) and \( {\widetilde{f}}_{2} \) are path-homotopic, and therefore so are \( {f}_{1} = q \circ {\widetilde{f}}_{1} \) and \( {f}_{2} = q \circ {\widetilde{f}}_{2} \)\n\nNow let \( X \) be any space satisfying the hypotheses of the theorem, and choose any base point \( {x}_{0} \in X \) . Guided by the observation in the preceding paragraph, we define \( \widetilde{X} \) to be the set of path classes of paths in \( X \) starting at \( {x}_{0} \), and define \( q : \widetilde{X} \rightarrow X \) by \( q\left( \left\lbrack f\right\rbrack \right) = f\left( 1\right) \) . We prove that \( \widetilde{X} \) has the required properties in a series of steps.\n\nSTEP 1: Topologize \( \widetilde{X} \) . We define a topology on \( \widetilde{X} \) by constructing a basis. For each \( \left\lbrack f\right\rbrack \in \widetilde{X} \) and each simply connected open subset \( U \subseteq X \) containing \( f\left( 1\right) \) , define the set \( \left\lbrack {f \cdot U}\right\rbrack \subseteq \widetilde{X} \) by\n\n\n\nFig. 11.10: Proof that the collection of sets \( \left\lbrack {f \cdot U}\right\rbrack \) is a basis.\n\n\[ \left\lbrack {f \cdot U}\right\rbrack = \{ \left\lbrack {f \cdot a}\right\rbrack : a\text{ is a path in }U\text{ starting at }f\left( 1\right) \} . \]\n\nLet \( \mathcal{B} \) denote the collection of all such sets \( \left\lbrack {f \cdot U}\right\rbrack \) ; w
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Yes
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Proposition 12.1 (Properties of the Automorphism Group). Let \( q : E \rightarrow X \) be a covering map.\n\n(a) If two automorphisms of \( q \) agree at one point, they are identical.\n\n(b) Given \( x \in X \), each covering automorphism restricts to a \( {\pi }_{1}\left( {X, x}\right) \) -automorphism of the fiber \( {q}^{-1}\left( x\right) \) (with respect to the monodromy action).\n\n(c) For any evenly covered open subset \( U \subseteq X \), each covering automorphism permutes the components of \( {q}^{-1}\left( U\right) \) .\n\n(d) The group \( {\operatorname{Aut}}_{q}\left( E\right) \) acts freely on \( E \) by homeomorphisms.
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Proof. Parts (a) and (b) follow immediately from Proposition 11.36(a, b). To prove (c), let \( U \) be an evenly covered open subset, and let \( {U}_{\alpha } \) be a component of \( {q}^{-1}\left( U\right) \) . Since \( \varphi \left( {U}_{\alpha }\right) \) is a connected subset of \( {q}^{-1}\left( U\right) \), it must be contained in a single component; applying the same argument to \( {\varphi }^{-1} \) shows that \( \varphi \left( {U}_{\alpha }\right) \) is exactly a component. Finally, to prove (d), just note that the automorphism group acts by homeomorphisms by definition, and the fact that it acts freely follows from (a) by comparing \( \varphi \) with the identity.
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Yes
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For the covering \( \varepsilon : \mathbb{R} \rightarrow {\mathbb{S}}^{1} \), the integral translations \( x \mapsto x + k \) for \( k \in \mathbb{Z} \) are easily seen to be automorphisms. To see that every automorphism is of this form, let \( \varphi \in {\operatorname{Aut}}_{\varepsilon }\left( \mathbb{R}\right) \) be arbitrary. If we set \( n = \varphi \left( 0\right) \), then \( \varphi \) and the translation \( x \mapsto x + n \) are both covering automorphisms taking 0 to \( n \), so they are equal by Proposition 12.1(a). Thus the automorphism group of \( \varepsilon : \mathbb{R} \rightarrow {\mathbb{S}}^{1} \) is isomorphic to \( \mathbb{Z} \), acting on \( \mathbb{R} \) by integral translations.
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A similar argument shows that the automorphism group of \( {\varepsilon }^{n} : {\mathbb{R}}^{n} \rightarrow {\mathbb{T}}^{n} \) is isomorphic to \( {\mathbb{Z}}^{n} \) acting by \( \left( {{x}_{1},\ldots ,{x}_{n}}\right) \cdot \left( {{k}_{1},\ldots ,{k}_{n}}\right) = \left( {{x}_{1} + {k}_{1},\ldots ,{x}_{n} + {k}_{n}}\right) \).
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Yes
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Theorem 12.4 (Orbit Criterion for Covering Automorphisms). Let \( q : E \rightarrow X \) be a covering map. If \( {e}_{1},{e}_{2} \in E \) are two points in the same fiber \( {q}^{-1}\left( x\right) \), there exists a covering automorphism taking \( {e}_{1} \) to \( {e}_{2} \) if and only if the induced subgroups \( {q}_{ * }{\pi }_{1}\left( {E,{e}_{1}}\right) \) and \( {q}_{ * }{\pi }_{1}\left( {E,{e}_{2}}\right) \) of \( {\pi }_{1}\left( {X, x}\right) \) are equal.
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Proof. This follows immediately from the covering isomorphism criterion (Theorem 11.40).
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No
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Corollary 12.5 (Normal Coverings Have Transitive Automorphism Groups). If \( q : E \rightarrow X \) is a covering map, then \( {\operatorname{Aut}}_{q}\left( E\right) \) acts transitively on each fiber if and only if \( q \) is a normal covering.
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Proof. Let \( q : E \rightarrow X \) be a covering map, and let \( x \) be an arbitrary point of \( X \) . By virtue of Proposition 11.35 and Theorem 12.4, we have the following equivalences:\n\n\[ \n{\operatorname{Aut}}_{q}\left( E\right) \text{acts transitively on}{q}^{-1}\left( x\right) \n\]\n\n\[ \n\Leftrightarrow \text{the subgroups}{q}_{ * }{\pi }_{1}\left( {E, e}\right) \text{are the same for all}e \in {q}^{-1}\left( x\right) \n\]\n\n\( \Leftrightarrow q \) is a normal covering.\n\nBecause of the preceding corollary, some authors define a normal covering to be one whose covering automorphism group acts transitively on fibers. The two characterizations can be used interchangeably.
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Yes
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Theorem 12.6. Suppose \( q : E \rightarrow X \) is a covering map and \( x \) is any point in \( X \) . The restriction map \( \varphi \mapsto {\left. \varphi \right| }_{{q}^{-1}\left( x\right) } \) is a group isomorphism between \( {\operatorname{Aut}}_{q}\left( E\right) \) and the group \( {\operatorname{Aut}}_{{\pi }_{1}\left( {X, x}\right) }\left( {{q}^{-1}\left( x\right) }\right) \) of \( {\pi }_{1}\left( {X, x}\right) \) -automorphisms of \( {q}^{-1}\left( x\right) \) .
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Proof. Proposition 12.1(b) shows that each covering automorphism restricts to a \( {\pi }_{1}\left( {X, x}\right) \) -automorphism of \( {q}^{-1}\left( x\right) \) . Since \( {\left. \left( {\varphi }_{1} \circ {\varphi }_{2}\right) \right| }_{{q}^{-1}\left( x\right) } = {\left. {\left. {\varphi }_{1}\right| }_{{q}^{-1}\left( x\right) } \circ {\varphi }_{2}\right| }_{{q}^{-1}\left( x\right) } \) , the restriction map is a group homomorphism.\n\nProposition 12.1(a) shows that two covering automorphisms whose restrictions to \( {q}^{-1}\left( x\right) \) agree must be identical, so the restriction homomorphism is injective. To see that it is surjective, suppose \( \eta : {q}^{-1}\left( x\right) \rightarrow {q}^{-1}\left( x\right) \) is any \( {\pi }_{1}\left( {X, x}\right) \) -automorphism of the fiber. If \( {e}_{1} \) is any point in \( {q}^{-1}\left( x\right) \) and \( {e}_{2} = \eta \left( {e}_{1}\right) \), then the orbit criterion for \( G \) - automorphisms (Proposition 11.27) shows that the isotropy groups of \( {e}_{1} \) and \( {e}_{2} \) are the same. Since these isotropy groups are exactly \( {q}_{ * }{\pi }_{1}\left( {E,{e}_{1}}\right) \) and \( {q}_{ * }{\pi }_{1}\left( {E,{e}_{2}}\right) \), the orbit criterion for covering automorphisms shows that there exists \( \varphi \in {\operatorname{Aut}}_{q}\left( E\right) \) such that \( \varphi \left( {e}_{1}\right) = {e}_{2} \) . Then \( {\left. \eta \text{and}\varphi \right| }_{{q}^{-1}\left( x\right) } \) are both \( {\pi }_{1}\left( {X, x}\right) \) -isomorphisms of \( {q}^{-1}\left( x\right) \) that agree at one point, so they are equal.
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Yes
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Theorem 12.7 (Covering Automorphism Group Structure Theorem). Suppose \( q : E \rightarrow X \) is a covering map, \( e \in E \), and \( x = q\left( e\right) \) . Let \( G = {\pi }_{1}\left( {X, x}\right) \) and \( H = \) \( {q}_{ * }{\pi }_{1}\left( {E, e}\right) \subseteq {\pi }_{1}\left( {X, x}\right) \) . For each path class \( \gamma \in {N}_{G}\left( H\right) \) (the normalizer of \( H \) in \( G \) ), there is a unique covering automorphism \( {\varphi }_{\gamma } \in {\operatorname{Aut}}_{q}\left( E\right) \) that satisfies \( {\varphi }_{\gamma }\left( e\right) = e \cdot \gamma \) . The map \( \gamma \mapsto {\varphi }_{\gamma } \) is a surjective group homomorphism from \( {N}_{G}\left( H\right) \) to \( {\operatorname{Aut}}_{q}\left( E\right) \) with kernel equal to \( H \), so it descends to an isomorphism from \( {N}_{G}\left( H\right) /H \) to \( {\operatorname{Aut}}_{q}\left( E\right) \) :
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Proof. We have two isomorphisms:\n\n\[ {N}_{G}\left( H\right) /H\overset{ \cong }{ \rightarrow }{\operatorname{Aut}}_{G}\left( {{q}^{-1}\left( x\right) }\right) \overset{ \cong }{ \rightarrow }{\operatorname{Aut}}_{q}\left( E\right) . \]\n\nThe first isomorphism is induced by the map of Theorem 11.28, which sends an element \( \gamma \in {N}_{G}\left( H\right) \) to the unique \( G \) -automorphism of \( {q}^{-1}\left( x\right) \) taking \( e \) to \( e \cdot \gamma \) ; and the second is the inverse of the restriction map \( {\left. \varphi \mapsto \varphi \right| }_{{q}^{-1}\left( x\right) } \), which is an isomorphism by Theorem 12.6. The map \( \gamma \mapsto {\varphi }_{\gamma } \) described in the statement of the theorem is exactly the composition of these two maps.
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Yes
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Theorem 12.14 (Covering Space Quotient Theorem). Let \( E \) be a connected, locally path-connected space, and suppose we are given an effective action of a group \( \Gamma \) on \( E \) by homeomorphisms. Then the quotient map \( q : E \rightarrow E/\Gamma \) is a covering map if and only if the action is a covering space action. In this case, \( q \) is a normal covering map, and \( {\operatorname{Aut}}_{q}\left( E\right) = \Gamma \), considered as a group of homeomorphisms of \( E \) .
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Proof. Assume first that \( q \) is a covering map. Then the action of each \( g \in \Gamma \) is an automorphism of the covering, because it is a homeomorphism satisfying \( q\left( {g \cdot e}\right) = \) \( q\left( e\right) \), so we can identify \( \Gamma \) with a subgroup of \( {\operatorname{Aut}}_{q}\left( E\right) \) . Exercise 12.12 shows that the action of \( {\operatorname{Aut}}_{q}\left( E\right) \) is a covering space action, and then Exercise 12.13 shows that the action of \( \Gamma \) is too.\n\nConversely, suppose the action is a covering space action. Clearly, the quotient map \( q \) is surjective and continuous. In addition, it is an open map by the result of Problem 3-22. To show that \( q \) is a covering map, suppose \( x \in E/\Gamma \) is arbitrary. Choose \( e \in {q}^{-1}\left( x\right) \), and let \( U \) be a neighborhood of \( e \) satisfying (12.1). Since \( E \) is locally path-connected, by passing to the component of \( U \) containing \( e \), we can assume that \( U \) is path-connected. Let \( V = q\left( U\right) \), which is a path-connected neighborhood of \( x \) .\n\nNow, \( {q}^{-1}\left( V\right) \) is equal to the union of the disjoint connected open subsets \( g \) . \( U \) for \( g \in \Gamma \), so to show that \( q \) is a covering it remains only to show that \( q \) is a homeomorphism from each such set onto \( V \) . For each \( g \in \Gamma \), the restricted map \( g : U \rightarrow g \cdot U \) is a homeomorphism, and the diagram\n\n\n\ncommutes; thus it suffices to show that \( {\left. q\right| }_{U} : U \rightarrow V \) is a homeomorphism. It is surjective, continuous, and open; and it is injective because \( q\left( e\right) = q\left( {e}^{\prime }\right) \) for \( e,{e}^{\prime } \in \) \( U \) implies \( {e}^{\prime } = g \cdot e \) for some \( g \in \Gamma \), so \( e = {e}^{\prime } \) because of (12.1). This proves that \( q \) is a covering map.\n\nTo prove the final statement of the theorem, suppose the action is a covering space action. As noted above, each map \( e \mapsto g \cdot e \) is a covering automorphism, so \( \Gamma \subseteq {\operatorname{Aut}}_{q}\left( E\right) \) . By construction, \( \Gamma \) acts transitively on each fiber, so \( {\operatorname{Aut}}_{q}\left( E\right) \) does too, and thus \( q \) is a normal covering. If \( \varphi \) is any covering automorphism, choose \( e \in E \) and let \( {e}^{\prime } = \varphi \left( e\right) \) . Then there is some \( g \in \Gamma \) such that \( g \cdot e = {e}^{\prime } \) ; since \( \varphi \) and \( x \mapsto g \cdot x \) are covering automorphisms that agree at a point, they are equal. Thus \( \Gamma \) is the full automorphism group.
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No
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Proposition 12.15. Let \( \Gamma \) be a discrete subgroup of a connected and locally path-connected topological group \( G \) . Then the action of \( \Gamma \) on \( G \) by right translations is a covering space action, so the quotient map \( q : G \rightarrow G/\Gamma \) is a normal covering map.
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Proof. Because \( \Gamma \) is discrete, there is a neighborhood \( V \) of 1 in \( G \) such that \( V \cap \) \( \Gamma = \{ 1\} \) . Consider the continuous map \( F : G \times G \rightarrow G \) given by \( F\left( {g, h}\right) = {g}^{-1}h \) . Since \( {F}^{-1}\left( V\right) \) is a neighborhood of \( \left( {1,1}\right) \), there is a product open subset \( {U}_{1} \times {U}_{2} \subseteq \) \( G \times G \) such that \( \left( {1,1}\right) \in {U}_{1} \times {U}_{2} \subseteq {F}^{-1}\left( V\right) \) . If we set \( U = {U}_{1} \cap {U}_{2} \), this means that \( g, h \in U \) implies \( {g}^{-1}h \in V \) . We complete the proof by showing that \( U \) satisfies (12.1) (or rather, the analogous statement with \( g \cdot U \) replaced by \( U \cdot g \), because \( \Gamma \) acts on the right).\n\nSuppose \( g \) is an element of \( \Gamma \) such that \( U \cap \left( {U \cdot g}\right) \neq \varnothing \) . This means there exists \( h \in U \) such that \( {hg} \in U \) . By our construction of \( U \), it follows that \( g = {h}^{-1}\left( {hg}\right) \in \) \( V \cap \Gamma \), which implies that \( g = 1 \) as claimed.
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Yes
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Corollary 12.16. Suppose \( G \) and \( H \) are connected and locally path-connected topological groups, and \( \varphi : G \rightarrow H \) is a surjective continuous homomorphism with discrete kernel. If \( \varphi \) is an open or closed map, then it is a normal covering map.
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Proof. Let \( \Gamma = \operatorname{Ker}\varphi \) . By the preceding proposition, the quotient map \( q : G \rightarrow \) \( G/\Gamma \) is a normal covering map. The assumption that \( \varphi \) is either open or closed implies that it is a quotient map, and by the first isomorphism theorem the identifications made by \( \varphi \) are precisely those made by \( q \) . Thus the result follows from the uniqueness of quotient spaces.
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Yes
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For any integers \( a, b, c, d \) such that \( {ad} - {bc} \neq 0 \), consider the map \( q : {\mathbb{T}}^{2} \rightarrow {\mathbb{T}}^{2} \) given by \( q\left( {z, w}\right) = \left( {{z}^{a}{w}^{b},{z}^{c}{w}^{d}}\right) \). This is easily seen to be a surjective continuous homomorphism, and it is a closed map by the closed map lemma. Once we show that it has discrete kernel, it follows from the preceding corollary that it is a normal covering map.
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Let \( A \) denote the invertible linear transformation of \( {\mathbb{R}}^{2} \) whose matrix is \( \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \). Then we have a commutative diagram\n\n\n\n(12.2)\n\nwhere \( {\varepsilon }^{2}\left( {x, y}\right) = \left( {{e}^{2\pi ix},{e}^{2\pi iy}}\right) \) is the universal covering map of the torus. To identify \( \operatorname{Ker}q \), note that\n\n\[ q \circ {\varepsilon }^{2}\left( {x, y}\right) = \left( {1,1}\right) \Leftrightarrow {\varepsilon }^{2} \circ A\left( {x, y}\right) = \left( {1,1}\right) \]\n\n\[ \Leftrightarrow A\left( {x, y}\right) \in {\mathbb{Z}}^{2} \]\n\n\[ \Leftrightarrow \left( {x, y}\right) \in {A}^{-1}\left( {\mathbb{Z}}^{2}\right) \]\n\nwhere \( {A}^{-1}\left( {\mathbb{Z}}^{2}\right) \) denotes the additive subgroup \( \left\{ {{A}^{-1}\left( {m, n}\right) : \left( {m, n}\right) \in {\mathbb{Z}}^{2}}\right\} \) of \( {\mathbb{R}}^{2} \). Because \( {\varepsilon }^{2} \) is surjective, this shows that \( \operatorname{Ker}q = {\varepsilon }^{2} \circ {A}^{-1}\left( {\mathbb{Z}}^{2}\right) \).\n\nSince \( {A}^{-1} \) has rational entries, it follows easily that each element of \( \operatorname{Ker}q \) is a torsion element of \( {\mathbb{T}}^{2} \). Moreover, since \( {\mathbb{Z}}^{2} \) is generated (as a group) by the two elements \( \left( {1,0}\right) \) and \( \left( {0,1}\right) \), Ker \( q \) is generated by their images under \( {\varepsilon }^{2} \circ {A}^{-1} \). An abelian group that is generated by finitely many torsion elements is easily seen to be finite; in particular, it is discrete.
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Yes
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Proposition 12.21 (Hausdorff Criterion for Orbit Spaces). Suppose \( E \) is a topological space and \( \Gamma \) is a group acting \( E \) by homeomorphisms. Then \( E/\Gamma \) is Hausdorff if and only if the action satisfies the following condition:\n\n\[ \text{if}e,{e}^{\prime } \in E\text{lie in different orbits, there exist neighborhoods}V \]\n\n(12.4)\n\n\[ \text{ofe and}{V}^{\prime }\text{of}{e}^{\prime }\text{such that}V \cap \left( {g \cdot {V}^{\prime }}\right) = \varnothing \text{for all}g \in \Gamma \text{.} \]
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Proof. Let \( q : E \rightarrow E/\Gamma \) denote the quotient map. If \( E/\Gamma \) is Hausdorff, then given \( e,{e}^{\prime } \) in different orbits, there are disjoint neighborhoods \( U \) of \( q\left( e\right) \) and \( {U}^{\prime } \) of \( q\left( {e}^{\prime }\right) \) , and then \( V = {q}^{-1}\left( U\right) \) and \( {V}^{\prime } = {q}^{-1}\left( {U}^{\prime }\right) \) satisfy (12.4).\n\nConversely, suppose the action satisfies (12.4). Given distinct points \( x,{x}^{\prime } \in \) \( E/\Gamma \), choose \( e,{e}^{\prime } \in E \) such that \( q\left( e\right) = x \) and \( q\left( {e}^{\prime }\right) = {x}^{\prime } \), and let \( V,{V}^{\prime } \) be neighborhoods satisfying the condition described in (12.4). Because \( q \) is an open map, \( q\left( V\right) \) and \( q\left( {V}^{\prime }\right) \) are neighborhoods of \( x \) and \( {x}^{\prime } \), respectively; and (12.4) implies that they are disjoint.
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Yes
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Proposition 12.22. Every continuous action of a compact topological group on a Hausdorff space is proper.
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Proof. Suppose \( G \) is a compact group acting continuously on a Hausdorff space \( E \), and let \( \Theta : G \times E \rightarrow E \times E \) be the map defined by (12.5). Given a compact set \( L \subseteq E \times E \), let \( K = {\pi }_{2}\left( L\right) \), where \( {\pi }_{2} : E \times E \rightarrow E \) is the projection on the second factor. Because \( E \times E \) is Hausdorff, \( L \) is closed in \( E \times E \) . Thus \( {\Theta }^{-1}\left( L\right) \) is a closed subset of the compact set \( G \times K \), hence compact.
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Yes
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Proposition 12.23. Suppose we are given a continuous action of a topological group \( G \) on a Hausdorff space \( E \). The action is proper if and only if for every compact subset \( K \subseteq E \), the set \( {G}_{K} = \{ g \in G : \left( {g \cdot K}\right) \cap K \neq \varnothing \} \) is compact.
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Proof. Let \( \Theta : G \times E \rightarrow E \times E \) be the map defined by (12.5). Suppose first that \( \Theta \) is proper. Then for any compact set \( K \subseteq E \), we have\n\n\[ \n{G}_{K} = \{ g \in G : \text{ there exists }e \in K\text{ such that }g \cdot e \in K\} \n\]\n\n\[ \n= \{ g \in G\text{ : there exists }e \in E\text{ such that }\Theta \left( {g, e}\right) \in K \times K\} \n\]\n\n(12.6)\n\n\[ \n= {\pi }_{G}\left( {{\Theta }^{-1}\left( {K \times K}\right) }\right) \n\]\n\nwhere \( {\pi }_{G} : G \times E \rightarrow G \) is the projection (Fig. 12.1). Thus \( {G}_{K} \) is compact.\n\nConversely, suppose \( {G}_{K} \) is compact for every compact set \( K \subseteq E \) . Given a compact subset \( L \subseteq E \times E \), let \( K = {\pi }_{1}\left( L\right) \cup {\pi }_{2}\left( L\right) \subseteq E \), where \( {\pi }_{1},{\pi }_{2} : E \times E \rightarrow E \) are the projections onto the first and second factors, respectively. Then\n\n\[ \n{\Theta }^{-1}\left( L\right) \subseteq {\Theta }^{-1}\left( {K \times K}\right) = \{ \left( {g, e}\right) : g \cdot e \in K\text{ and }e \in K\} \subseteq {G}_{K} \times K. \n\]\n\nSince \( E \times E \) is Hausdorff, \( L \) is closed in \( E \times E \), and so \( {\Theta }^{-1}\left( L\right) \) is closed in \( G \times E \) by continuity. Thus \( {\Theta }^{-1}\left( L\right) \) is a closed subset of the compact set \( {G}_{K} \times K \) and is therefore compact.
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Yes
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Proposition 12.24. If a topological group \( G \) acts continuously and properly on a locally compact Hausdorff space \( E \), then the orbit space \( E/G \) is Hausdorff.
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Proof. Let \( \mathcal{O} \subseteq E \times E \) be the orbit relation defined in Problem 3-22. By the result of that problem, the orbit space is Hausdorff if and only if \( \mathcal{O} \) is closed in \( E \times E \) . But \( \mathcal{O} \) is just the image of the map \( \Theta : G \times E \rightarrow E \times E \) defined by (12.5). Since \( E \) is a locally compact Hausdorff space, the same is true of \( E \times E \), so it follows from Theorem 4.95 that \( \Theta \) is a closed map. Thus the orbit relation is closed and \( E/G \) is Hausdorff.
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No
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Proposition 12.25. Suppose we are given a covering space action of a group \( \Gamma \) on a topological space \( E \), and \( E/\Gamma \) is Hausdorff. Then with the discrete topology, \( \Gamma \) acts properly on \( E \) .
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Proof. For convenience, write \( X = E/\Gamma \), and let \( q : E \rightarrow X \) be the quotient map, which is a normal covering map by the covering space quotient theorem. It follows from Proposition 3.57 and Problem 3-22 that the orbit relation \( \mathcal{O} \) defined by (3.6) is closed in \( E \times E \) . Also, Problem 11-1(a) shows that \( E \) is Hausdorff. We use Proposition 12.23 to show that the action is proper.\n\nSuppose \( K \subseteq E \times E \) is compact, and assume for the sake of contradiction that \( {\Gamma }_{K} \) is not compact; this means in particular that \( {\Gamma }_{K} \) is infinite. For each \( g \in {\Gamma }_{K} \), there is a point \( e \in K \) such that \( g \cdot e \in K \) . Define a map \( F : {\Gamma }_{K} \rightarrow K \times K \) by choosing one such point \( {e}_{g} \) for each \( g \), and letting \( F\left( g\right) = \left( {g \cdot {e}_{g},{e}_{g}}\right) \) . The fact that \( \Gamma \) acts freely implies that \( F \) is injective, so \( F\left( {\Gamma }_{K}\right) \) is an infinite subset of \( K \times K \) . It follows that \( F\left( {\Gamma }_{K}\right) \) has a limit point \( \left( {{x}_{0},{y}_{0}}\right) \in K \times K \) . Moreover, since \( F\left( {\Gamma }_{K}\right) \subseteq \mathcal{O} \), which is closed in \( E \times E \), we have \( \left( {{x}_{0},{y}_{0}}\right) \in \mathcal{O} \) as well, which means that there exists \( {g}_{0} \in \Gamma \) such that \( {x}_{0} = {g}_{0} \cdot {y}_{0} \) .\n\nNow let \( U \) be a neighborhood of \( {y}_{0} \) satisfying (12.1), and set \( V = {g}_{0} \cdot U \), which is a neighborhood of \( {x}_{0} \) . The fact that \( \left( {{x}_{0},{y}_{0}}\right) \) is a limit point in the Hausdorff space \( E \times E \) means that \( V \times U \) must contain infinitely many points of \( F\left( {\Gamma }_{K}\right) \) . But for each \( g \in {\Gamma }_{K} \) such that \( F\left( g\right) = \left( {g \cdot {e}_{g},{e}_{g}}\right) \in V \times U \), we have \( g \cdot {e}_{g} \in V \cap \left( {g \cdot U}\right) = \) \( \left( {{g}_{0} \cdot U}\right) \cap \left( {g \cdot U}\right) \), which implies that \( g = {g}_{0} \) . This contradicts the fact that there are infinitely many such \( g \) .
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Yes
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Theorem 12.26. Suppose \( E \) is a connected, locally path-connected, and locally compact Hausdorff space, and a discrete group \( \Gamma \) acts continuously, freely, and properly on \( E \) . Then the action is a covering space action, \( E/\Gamma \) is Hausdorff, and the quotient map \( q : E \rightarrow E/\Gamma \) is a normal covering map.
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Proof. We need only show that the action is a covering space action, for then Proposition 12.24 shows that \( E/\Gamma \) is Hausdorff, and the covering space quotient theorem shows that \( q \) is a normal covering map.\n\nSuppose \( {e}_{0} \in E \) is arbitrary. Because \( E \) is locally compact, \( {e}_{0} \) has a neighborhood \( V \) contained in a compact set \( K \) . By Proposition 12.23, the set \( {\Gamma }_{K} = \{ g \in \Gamma \) : \( K \cap \left( {g \cdot K}\right) \neq \varnothing \} \) is compact. Because \( \Gamma \) has the discrete topology, this means \( {\Gamma }_{K} \) is finite; let us write \( {\Gamma }_{K} = \left\{ {1,{g}_{1},\ldots ,{g}_{m}}\right\} \) . Since the action is free and \( E \) is Hausdorff, for each \( {g}_{i} \) there are disjoint neighborhoods \( {W}_{i} \) of \( {e}_{0} \) and \( {W}_{i}^{\prime } \) of \( {g}_{i} \cdot {e}_{0} \) . Let\n\n\[ U = V \cap {W}_{1} \cap \left( {{g}_{1}^{-1} \cdot {W}_{1}^{\prime }}\right) \cap \cdots \cap {W}_{m} \cap \left( {{g}_{m}^{-1} \cdot {W}_{m}^{\prime }}\right) . \]\n\nWe will show that \( U \) satisfies (12.1).\n\nFirst consider \( g = {g}_{i} \) for some \( i \) . If \( e \in U \subseteq {g}_{i}^{-1} \cdot {W}_{i}^{\prime } \), then \( {g}_{i} \cdot e \in {W}_{i}^{\prime } \), which is disjoint from \( {W}_{i} \) and therefore from \( U \) . Thus \( U \cap \left( {{g}_{i} \cdot U}\right) = \varnothing \) . On the other hand, if \( g \in \Gamma \) is not the identity and not one of the \( {g}_{i} \) ’s, then for any \( e \in U \subseteq V \subseteq K \), we have \( g \cdot e \in g \cdot K \), which is disjoint from \( K \) and therefore also from \( U \) . Thus once again we have \( U \cap \left( {g \cdot U}\right) = \varnothing \) .
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Yes
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Theorem 12.29. Let \( M \) be a compact surface. The universal covering space of \( M \) is homeomorphic to\n\n(a) \( {\mathbb{S}}^{2} \) if \( M \approx {\mathbb{S}}^{2} \) or \( {\mathbb{P}}^{2} \) ,\n\n(b) \( {\mathbb{R}}^{2} \) if \( M \approx {\mathbb{T}}^{2} \) or \( {\mathbb{P}}^{2}\# {\mathbb{P}}^{2} \) ,\n\n(c) \( {\mathbb{B}}^{2} \) if \( M \) is any other surface.
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Proof. Because \( {\mathbb{S}}^{2} \) is simply connected, it is its own universal covering space. It was shown in Example 11.42 that the universal covering space of \( {\mathbb{T}}^{2} \) is \( {\mathbb{R}}^{2} \), and that of \( {\mathbb{P}}^{2} \) is \( {\mathbb{S}}^{2} \) . If \( M \) is a connected sum of \( n \geq 2 \) projective planes, then by the result of Problem 11-6, \( M \) has a two-sheeted covering by a manifold \( N \), which is a connected sum of \( n - 1 \) tori. If \( \widetilde{M} \) is the universal covering space of \( M \), then \( \widetilde{M} \) also covers \( N \) by Proposition 11.41(a), so \( M \) and \( N \) have the same universal covering space. Thus to complete the proof of the theorem, it suffices to show that every connected sum of \( n \geq 2 \) tori is covered by \( {\mathbb{B}}^{2} \) . This is the result of Theorem 12.30 below.
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Yes
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Lemma 13.1. If \( c \) is a singular chain, then \( \partial \left( {\partial c}\right) = 0 \) .
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Proof. Since each chain group \( {C}_{p}\left( X\right) \) is generated by singular simplices, it suffices to show this in the case in which \( c = \sigma \) is a singular \( p \) -simplex.\n\nFirst we note that the face maps satisfy the commutation relation\n\n\[ \n{F}_{i, p} \circ {F}_{j, p - 1} = {F}_{j, p} \circ {F}_{i - 1, p - 1}\;\text{ when }i > j, \n\]\n\nas can be seen immediately by observing that the vertices of \( {\Delta }_{p - 2} \) are mapped according to the following chart:\n\n\[ \n\begin{matrix} {F}_{j, p - 1} & {F}_{i, p} & {F}_{i - 1, p - 1} & {F}_{j, p} \\ {e}_{0} \mapsto {e}_{0} & \mapsto {e}_{0} & {e}_{0} \mapsto {e}_{0} & \mapsto {e}_{0} \end{matrix} \n\]\n\n\[ \n\ldots \;\ldots \;\ldots \;\ldots \;\ldots \n\]\n\n\[ \n{e}_{j - 1} \mapsto {e}_{j - 1} \mapsto {e}_{j - 1}\;{e}_{j - 1} \mapsto {e}_{j - 1} \mapsto {e}_{j - 1} \n\]\n\n\[ \n\begin{array}{ll} {e}_{j} \mapsto {e}_{j + 1} \mapsto {e}_{j + 1} & {e}_{j} \mapsto {e}_{j} \mapsto {e}_{j + 1} \end{array} \n\]\n\n\[ \n{e}_{i - 2} \mapsto {e}_{i - 1} \mapsto {e}_{i - 1}\;{e}_{i - 2} \mapsto {e}_{i - 2} \mapsto {e}_{i - 1} \n\]\n\n\[ \n\begin{array}{l} {e}_{i - 1} \mapsto {e}_{i}\; \mapsto {e}_{i + 1}\;{e}_{i - 1} \mapsto {e}_{i}\; \mapsto {e}_{i + 1} \\ \end{array} \n\]\n\n\[ \n\cdots \;\cdots \;\cdots \;\cdots \;\cdots \n\]\n\n\[ \n{e}_{p - 2} \mapsto {e}_{p - 1} \mapsto {e}_{p}\;{e}_{p - 2} \mapsto {e}_{p - 1} \mapsto {e}_{p} \n\]\n\nIn other words, both \( {F}_{i, p} \circ {F}_{j, p - 1} \) and \( {F}_{j, p} \circ {F}_{i - 1, p - 1} \) are equal to the affine simplex \( A\left( {{e}_{0},\ldots ,{\widehat{e}}_{j},\ldots ,{\widehat{e}}_{i},\ldots ,{e}_{p}}\right) \) . Using this, we compute\n\n\[ \n\partial \left( {\partial \sigma }\right) = \mathop{\sum }\limits_{{j = 0}}^{{p - 1}}\mathop{\sum }\limits_{{i = 0}}^{p}{\left( -1\right) }^{i + j}\sigma \circ {F}_{i, p} \circ {F}_{j, p - 1} \n\]\n\n\[ \n= \mathop{\sum }\limits_{{0 \leq j < i \leq p}}{\left( -1\right) }^{i + j}\sigma \circ {F}_{i, p} \circ {F}_{j, p - 1} + \mathop{\sum }\limits_{{0 \leq i \leq j \leq p - 1}}{\left( -1\right) }^{i + j}\sigma \circ {F}_{i, p} \circ {F}_{j, p - 1}. \n\]\n\nMaking the substitutions \( i = {j}^{\prime }, j = {i}^{\prime } - 1 \) into the second sum and using (13.1), we see that the sums cancel term by term.
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Yes
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Proposition 13.2 (Functorial Properties of Homology). Let \( X \) , \( Y \) , and \( Z \) be topological spaces.\n\n(a) The homomorphism \( {\left( {\operatorname{Id}}_{X}\right) }_{ * } : {H}_{p}\left( X\right) \rightarrow {H}_{p}\left( X\right) \) induced by the identity map of \( X \) is the identity of \( {H}_{p}\left( X\right) \) .\n\n(b) If \( f : X \rightarrow Y \) and \( g : Y \rightarrow Z \) are continuous maps, then\n\n\[ \n{\left( g \circ f\right) }_{ * } = {g}_{ * } \circ {f}_{ * } : {H}_{p}\left( X\right) \rightarrow {H}_{p}\left( Z\right) .\n\]\n\nThus the pth singular homology group defines a covariant functor from the category of topological spaces to the category of abelian groups.
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Proof. It is easy to check that both properties hold already for \( {f}_{\# } \) .
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No
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Proposition 13.5. Let \( X \) be a space, let \( {\left\{ {X}_{\alpha }\right\} }_{\alpha \in A} \) be the set of path components of \( X \), and let \( {\iota }_{\alpha } : {X}_{\alpha } \hookrightarrow X \) be inclusion. Then for each \( p \geq 0 \) the map\n\n\[ \n{\bigoplus }_{\alpha \in A}{H}_{p}\left( {X}_{\alpha }\right) \rightarrow {H}_{p}\left( X\right)\n\]\n\nwhose restriction to \( {H}_{p}\left( {X}_{\alpha }\right) \) is \( {\left( {\iota }_{\alpha }\right) }_{ * } : {H}_{p}\left( {X}_{\alpha }\right) \rightarrow {H}_{p}\left( X\right) \), is an isomorphism.
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Proof. Since the image of any singular simplex must lie entirely in one path component, the chain maps \( {\left( {\iota }_{\alpha }\right) }_{\# } : {C}_{p}\left( {X}_{\alpha }\right) \rightarrow {C}_{p}\left( X\right) \) already induce isomorphisms\n\n\[ \n{\bigoplus }_{\alpha \in A}{C}_{p}\left( {X}_{\alpha }\right) \rightarrow {C}_{p}\left( X\right)\n\]\n\nThe result for homology follows easily from this.
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Yes
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Proposition 13.6 (Zero-Dimensional Homology). For any topological space \( X \) , \( {H}_{0}\left( X\right) \) is a free abelian group with basis consisting of an arbitrary point in each path component.
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Proof. It suffices to show that \( {H}_{0}\left( X\right) \) is the infinite cyclic group generated by the class of any point when \( X \) is path-connected, for then in the general case Proposition 13.5 guarantees that \( {H}_{0}\left( X\right) \) is the direct sum of infinite cyclic groups, one for each path component.\n\nA singular 0-chain is a formal linear combination of points in \( X \) with integer coefficients: \( c = \mathop{\sum }\limits_{{i = 1}}^{m}{n}_{i}{x}_{i} \) . Because the boundary operator is the zero map in dimension 0 , every 0 -chain is a cycle.\n\nAssume that \( X \) is path-connected, and define a map \( \varepsilon : {C}_{0}\left( X\right) \rightarrow \mathbb{Z} \) by\n\n\[ \varepsilon \left( {\mathop{\sum }\limits_{{i = 1}}^{m}{n}_{i}{x}_{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{m}{n}_{i} \]\n\nIt is immediate from the definition that \( \varepsilon \) is a surjective homomorphism. We will show that \( \operatorname{Ker}\varepsilon = {B}_{0}\left( X\right) \), from which it follows by the first isomorphism theorem that \( \varepsilon \) induces an isomorphism \( {H}_{0}\left( X\right) \rightarrow \mathbb{Z} \) . Since \( \varepsilon \) takes any single point to 1, the result follows.\n\nIf \( \sigma \) is a singular 1 -simplex, then \( \partial \sigma = \sigma \left( 1\right) - \sigma \left( 0\right) \), so \( \varepsilon \left( {\partial \sigma }\right) = 1 - 1 = 0 \) . Therefore, \( {B}_{0}\left( X\right) \subseteq \operatorname{Ker}\varepsilon \).\n\nTo show that \( \operatorname{Ker}\varepsilon \subseteq {B}_{0}\left( X\right) \), choose any point \( {x}_{0} \in X \), and for each \( x \in X \) let \( \alpha \left( x\right) \) be a path from \( {x}_{0} \) to \( x \) . This is a singular 1 -simplex whose boundary is the 0 -chain \( x - {x}_{0} \) . Thus, for an arbitrary 0 -chain \( c = \mathop{\sum }\limits_{i}{n}_{i}{x}_{i} \) we compute\n\n\[ \partial \left( {\mathop{\sum }\limits_{i}{n}_{i}\alpha \left( {x}_{i}\right) }\right) = \mathop{\sum }\limits_{i}{n}_{i}{x}_{i} - \mathop{\sum }\limits_{i}{n}_{i}{x}_{0} = c - \varepsilon \left( c\right) {x}_{0}. \]\n\nIn particular, if \( \varepsilon \left( c\right) = 0 \), then \( c \in {B}_{0}\left( X\right) \).
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Yes
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Proposition 13.7 (Homology of a Discrete Space). If \( X \) is a discrete space, then \( {H}_{0}\left( X\right) \) is a free abelian group with one generator for each point of \( X \), and \( {H}_{p}\left( X\right) = 0 \) for \( p > 0 \) .
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Proof. The case \( p = 0 \) follows from the preceding proposition, so we concentrate on \( p > 0 \) . By Proposition 13.5, it suffices to show that \( {H}_{p}\left( *\right) = 0 \) when \( * \) is a one-point space. In that case, there is exactly one singular simplex in each dimension, namely the constant map \( {\sigma }_{p} : {\Delta }_{p} \rightarrow * \), so each chain group \( {C}_{p}\left( *\right) \) is the infinite cyclic group generated by \( {\sigma }_{p} \) . For \( p > 0 \), the boundary of \( {\sigma }_{p} \) is the alternating sum\n\n\[ \n\partial {\sigma }_{p} = \mathop{\sum }\limits_{{i = 0}}^{p}{\left( -1\right) }^{i}{\sigma }_{p} \circ {F}_{i, p} = \mathop{\sum }\limits_{{i = 0}}^{p}{\left( -1\right) }^{i}{\sigma }_{p - 1} = \left\{ \begin{array}{ll} 0 & \text{ if }p\text{ is odd,} \\ {\sigma }_{p - 1} & \text{ if }p\text{ is even. } \end{array}\right.\n\]\n\nThus \( \partial : {C}_{p}\left( *\right) \rightarrow {C}_{p - 1}\left( *\right) \) is an isomorphism when \( p \) is even and positive, and the zero map when \( p \) is odd:\n\n\[ \n\cdots \overset{ \cong }{ \rightarrow }{C}_{3}\left( *\right) \overset{0}{ \rightarrow }{C}_{2}\left( *\right) \overset{ \cong }{ \rightarrow }{C}_{1}\left( *\right) \overset{0}{ \rightarrow }{C}_{0}\left( *\right) \rightarrow 0.\n\]\n\nThis sequence is exact at each group except the last, so \( {H}_{p}\left( *\right) = 0 \) for \( p > 0 \) .
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Yes
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Theorem 13.8. If \( {f}_{0},{f}_{1} : X \rightarrow Y \) are homotopic maps, then for each \( p \geq 0 \) the induced homomorphisms \( {\left( {f}_{0}\right) }_{ * },{\left( {f}_{1}\right) }_{ * } : {H}_{p}\left( X\right) \rightarrow {H}_{p}\left( Y\right) \) are equal.
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Proof of Theorem 13.8. We begin by considering the special case in which \( Y = \) \( X \times I \) and \( {f}_{i} = {\iota }_{i} \), where \( {\iota }_{0},{\iota }_{1} : X \rightarrow X \times I \) are the maps\n\n\[ \n{\iota }_{0}\left( x\right) = \left( {x,0}\right) ,\;{\iota }_{1}\left( x\right) = \left( {x,1}\right) .\n\]\n\n(See Fig. 13.3.) Clearly, \( {\iota }_{0} \simeq {\iota }_{1} \) (the homotopy is the identity map of \( X \times I \) ). We will show below that \( {\left( {\iota }_{0}\right) }_{ * } = {\left( {\iota }_{1}\right) }_{ * } \) . As it turns out, this immediately implies the general case, as follows. Suppose \( {f}_{0},{f}_{1} : X \rightarrow Y \) are continuous maps and \( H : X \times I \rightarrow Y \) is a homotopy from \( {f}_{0} \) to \( {f}_{1} \) (Fig. 13.3). Then since \( H \circ {\iota }_{i} = {f}_{i} \), we have\n\n\[ \n{\left( {f}_{0}\right) }_{ * } = {\left( H \circ {\iota }_{0}\right) }_{ * } = {H}_{ * } \circ {\left( {\iota }_{0}\right) }_{ * } = {H}_{ * } \circ {\left( {\iota }_{1}\right) }_{ * } = {\left( H \circ {\iota }_{1}\right) }_{ * } = {\left( {f}_{1}\right) }_{ * }.\n\]\n\nTo prove \( {\left( {\iota }_{0}\right) }_{ * } = {\left( {\iota }_{1}\right) }_{ * } \), it would suffice to show that \( {\left( {\iota }_{0}\right) }_{\# }c \) and \( {\left( {\iota }_{1}\right) }_{\# }c \) differ by a boundary for each chain \( c \) . In fact, a little experimentation will probably convince you that this is usually false. But in fact all we need is that they differ by a boundary when \( c \) is a cycle. So we might try to define a map \( h : {Z}_{p}\left( X\right) \rightarrow {C}_{p + 1}\left( {X \times I}\right) \) such that\n\n\[ \n\partial h\left( c\right) = {\left( {\iota }_{1}\right) }_{\# }c - {\left( {\iota }_{0}\right) }_{\# }c.\n\]\n\n(13.2)\n\nIt turns out to be hard to define such a thing for cycles only. Instead, we define \( h\left( c\right) \) for all \( p \) -chains \( c \), and show that it satisfies a formula that implies (13.2) when \( c \) is a cycle.\n\nFor each \( p \geq 0 \), we will define a homomorphism \( h : {C}_{p}\left( X\right) \rightarrow {C}_{p + 1}\left( {X \times I}\right) \) such that the following identity is satisfied:\n\n\[ \nh \circ \partial + \partial \circ h = {\left( {\iota }_{1}\right) }_{\# } - {\left( {\iota }_{0}\right) }_{\# }.\n\]\n\n(13.3)\n\nFrom (13.3) it follows immediately that \( {\left( {\iota }_{1}\right) }_{\# }c - {\left( {\iota }_{0}\right) }_{\# }c = \partial h\left( c\right) \) whenever \( \partial c = 0 \) , and therefore \( {\left( {\iota }_{1}\right) }_{ * }\left\lbrack c\right\rbrack = {\left( {\iota }_{0}\right) }_{ * }\left\lbrack c\right\rbrack \) .\n\nThe construction of \( h \) is basically a \
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Yes
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Corollary 13.9 (Homotopy Invariance of Singular Homology). Suppose \( f : X \rightarrow Y \) is a homotopy equivalence. Then for each \( p \geq 0, {f}_{ * } : {H}_{p}\left( X\right) \rightarrow {H}_{p}\left( Y\right) \) is an isomorphism.
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Proof of Theorem 13.8. We begin by considering the special case in which \( Y = X \times I \) and \( {f}_{i} = {\iota }_{i} \), where \( {\iota }_{0},{\iota }_{1} : X \rightarrow X \times I \) are the maps \[ {\iota }_{0}\left( x\right) = \left( {x,0}\right) ,\;{\iota }_{1}\left( x\right) = \left( {x,1}\right) . \] (See Fig. 13.3.) Clearly, \( {\iota }_{0} \simeq {\iota }_{1} \) (the homotopy is the identity map of \( X \times I \) ). We will show below that \( {\left( {\iota }_{0}\right) }_{ * } = {\left( {\iota }_{1}\right) }_{ * } \). As it turns out, this immediately implies the general case, as follows. Suppose \( {f}_{0},{f}_{1} : X \rightarrow Y \) are continuous maps and \( H : X \times I \rightarrow Y \) is a homotopy from \( {f}_{0} \) to \( {f}_{1} \) (Fig. 13.3). Then since \( H \circ {\iota }_{i} = {f}_{i} \), we have \[ {\left( {f}_{0}\right) }_{ * } = {\left( H \circ {\iota }_{0}\right) }_{ * } = {H}_{ * } \circ {\left( {\iota }_{0}\right) }_{ * } = {H}_{ * } \circ {\left( {\iota }_{1}\right) }_{ * } = {\left( H \circ {\iota }_{1}\right) }_{ * } = {\left( {f}_{1}\right) }_{ * }. \] To prove \( {\left( {\iota }_{0}\right) }_{ * } = {\left( {\iota }_{1}\right) }_{ * } \), it would suffice to show that \( {\left( {\iota }_{0}\right) }_{\# }c \) and \( {\left( {\iota }_{1}\right) }_{\# }c \) differ by a boundary for each chain \( c \). In fact, a little experimentation will probably convince you that this is usually false. But in fact all we need is that they differ by a boundary when \( c \) is a cycle. So we might try to define a map \( h : {Z}_{p}\left( X\right) \rightarrow {C}_{p + 1}\left( {X \times I}\right) \) such that \[ \partial h\left( c\right) = {\left( {\iota }_{1}\right) }_{\# }c - {\left( {\iota }_{0}\right) }_{\# }c. \] (13.2) It turns out to be hard to define such a thing for cycles only. Instead, we define \( h\left( c\right) \) for all \( p \)-chains \( c \), and show that it satisfies a formula that implies (13.2) when \( c \) is a cycle. For each \( p \geq 0 \), we will define a homomorphism \( h : {C}_{p}\left( X\right) \rightarrow {C}_{p + 1}\left( {X \times I}\right) \) such that the following identity is satisfied: \[ h \circ \partial + \partial \circ h = {\left( {\iota }_{1}\right) }_{\# } - {\left( {\iota }_{0}\right) }_{\# }. \] (13.3) From (13.3) it follows immediately that \( {\left( {\iota }_{1}\right) }_{\# }c - {\left( {\iota }_{0}\right) }_{\# }c = \partial h\left( c\right) \) whenever \( \partial c = 0 \), and therefore \( {\left( {\iota }_{1}\right) }_{ * }\left\lbrack c\right\rbrack = {\left( {\iota }_{0}\right) }_{ * }\left\lbrack c\right\rbrack \). The construction of \( h \) is basically a \
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Yes
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Lemma 13.13. Suppose \( {f}_{0} \) and \( {f}_{1} \) are paths in \( X \), and \( {f}_{0} \sim {f}_{1} \) . Then, considered as a singular chain, \( {f}_{0} - {f}_{1} \) is a boundary.
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Proof. We must show there is a singular 2-chain whose boundary is the 1-chain \( {f}_{0} - {f}_{1} \) . Let \( H : {f}_{0} \sim {f}_{1} \), and let \( b : I \times I \rightarrow {\Delta }_{2} \) be the map\n\n\[ b\left( {x, y}\right) = \left( {x - {xy},{xy}}\right) ,\]\n\nwhich maps the square onto the triangle by sending each horizontal line segment linearly to a radial line segment (Fig. 13.5). Then \( b \) is a quotient map by the closed map lemma, and identifies the left-hand edge of the square to the origin. Since \( H \) respects the identifications made by \( b \), it passes to the quotient to yield a continuous map \( \sigma : {\Delta }_{2} \rightarrow X \) (i.e., a singular 2-simplex). From the definition of the boundary operator, \( \partial \sigma = {c}_{p} - {f}_{1} + {f}_{0} \), where \( p = {f}_{0}\left( 1\right) \) . Since \( {c}_{p} \) is the boundary of the constant 2-simplex that maps \( {\Delta }_{2} \) to \( p \), it follows that \( {f}_{0} - {f}_{1} \) is a boundary.
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Yes
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