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If \( M \) is any \( n \) -manifold, a connected sum \( M\# {\mathbb{S}}^{n} \) is homeomorphic to \( M \), at least if we make our choices carefully (Fig. 6.6). | Let \( {B}_{2} \subseteq {\mathbb{S}}^{n} \) be the open lower hemisphere, so \( {\left( {\mathbb{S}}^{n}\right) }^{\prime } = {\mathbb{S}}^{n} \smallsetminus {B}_{2} \) is the closed upper hemisphere, which is homeomorphic to a closed ball. Then \( M\# {\mathbb{S}}^{n} \) is obtained from \( M \) by cutting out the op... | Yes |
Proposition 6.12. Let \( {M}_{1} \) and \( {M}_{2} \) be surfaces that admit presentations \( \left\langle {{S}_{1} \mid {W}_{1}}\right\rangle \) and \( \left\langle {{S}_{2} \mid {W}_{2}}\right\rangle \), respectively, in which \( {S}_{1} \) and \( {S}_{2} \) are disjoint sets and each presentation has a single face. ... | Proof. Consider the presentation \( \left\langle {{S}_{1}, a, b, c \mid {W}_{1}{c}^{-1}{b}^{-1}{a}^{-1},{abc}}\right\rangle \) (pictured in the left half of Fig. 6.16). Pasting along \( a \) and folding twice, we see that this presentation is equivalent to \( \left\langle {{S}_{1} \mid {W}_{1}}\right\rangle \) and ther... | Yes |
Every compact surface admits a polygonal presentation. | Proof. Let \( M \) be a compact surface. It follows from Theorem 5.36 that \( M \) is homeomorphic to the polyhedron of a 2-dimensional simplicial complex \( K \), in which each 1-simplex is a face of exactly two 2-simplices.\n\nFrom this complex, we can construct a surface presentation \( \mathcal{P} \) with one word ... | No |
Lemma 6.16. The Klein bottle is homeomorphic to \( {\mathbb{P}}^{2}\# {\mathbb{P}}^{2} \) . | Proof. By a sequence of elementary transformations, we find that the Klein bottle has the following presentations (see Fig. 6.18):\n\n\[ \left\langle {a, b \mid {aba}{b}^{-1}}\right\rangle \]\n\n\[ \approx \left\langle {a, b, c \mid {abc},{c}^{-1}a{b}^{-1}}\right\rangle \;\text{(cut along}c\text{)} \]\n\n\[ \approx \le... | Yes |
Lemma 6.17. The connected sum \( {\mathbb{T}}^{2}\# {\mathbb{P}}^{2} \) is homeomorphic to \( {\mathbb{P}}^{2}\# {\mathbb{P}}^{2}\# {\mathbb{P}}^{2} \) . | Proof. Start with \( \left\langle {a, b, c \mid {aba}{b}^{-1}{cc}}\right\rangle \) (Fig. 6.19), which is a presentation of \( K\# {\mathbb{P}}^{2} \) , and therefore by the preceding lemma is a presentation of \( {\mathbb{P}}^{2}\# {\mathbb{P}}^{2}\# {\mathbb{P}}^{2} \) . Following Fig. 6.19, we cut along \( d \), past... | Yes |
Proposition 6.18. The Euler characteristic of a polygonal presentation is unchanged by elementary transformations. | Proof. It is immediate that relabeling, rotating, and reflecting do not change the Euler characteristic of a presentation, because they leave the numbers of 0 -cells, 1-cells, and 2-cells individually unchanged. For the other transformations, we need only check that the changes to these three numbers cancel out. Subdiv... | Yes |
Proposition 6.19 (Euler Characteristics of Compact Surfaces). The Euler characteristic of a standard surface presentation is equal to\n\n(a) 2 for the sphere,\n\n(b) \( 2 - {2n} \) for the connected sum of \( n \) tori,\n\n(c) \( 2 - n \) for the connected sum of \( n \) projective planes. | Proof. Just compute. | No |
Proposition 6.20. A compact surface is orientable if and only if it is homeomorphic to the sphere or a connected sum of one or more tori. | Proof. The standard presentations of the sphere and the connected sums of tori are oriented, so these surfaces are certainly orientable. To show that an orientable surface is homeomorphic to one of these, let \( M \) be any surface that admits at least one orientable presentation. Starting with that presentation, follo... | Yes |
Proposition 7.1. For any topological spaces \( X \) and \( Y \), homotopy is an equivalence relation on the set of all continuous maps from \( X \) to \( Y \) . | Proof. Any map \( f \) is homotopic to itself via the trivial homotopy \( H\left( {x, t}\right) = f\left( x\right) \) , so homotopy is reflexive. Similarly, if \( H : f \simeq g \), then a homotopy from \( g \) to \( f \) is given by \( \widetilde{H}\left( {x, t}\right) = H\left( {x,1 - t}\right) \), so homotopy is sym... | Yes |
Proposition 7.2. The homotopy relation is preserved by composition: if\n\n\[ \n{f}_{0},{f}_{1} : X \rightarrow Y\;\text{ and }\;{g}_{0},{g}_{1} : Y \rightarrow Z \n\]\n\nare continuous maps with \( {f}_{0} \simeq {f}_{1} \) and \( {g}_{0} \simeq {g}_{1} \), then \( {g}_{0} \circ {f}_{0} \simeq {g}_{1} \circ {f}_{1} \) ... | Proof. Suppose \( F : {f}_{0} \simeq {f}_{1} \) and \( G : {g}_{0} \simeq {g}_{1} \) are homotopies. Define \( H : X \times I \rightarrow \) \( Z \) by \( H\left( {x, t}\right) = G\left( {F\left( {x, t}\right), t}\right) \) . At \( t = 0, H\left( {x,0}\right) = G\left( {{f}_{0}\left( x\right) ,0}\right) = {g}_{0}\left(... | Yes |
Define maps \( f, g : \mathbb{R} \rightarrow {\mathbb{R}}^{2} \) by\n\n\[ f\left( x\right) = \left( {x,{x}^{2}}\right) ;\;g\left( x\right) = \left( {x, x}\right) . \]\n\nThen the map \( H\left( {x, t}\right) = \left( {x,{x}^{2} - t{x}^{2} + {tx}}\right) \) is a homotopy from \( f \) to \( g \) . | The map \( H\left( {x, t}\right) = \left( {x,{x}^{2} - t{x}^{2} + {tx}}\right) \) is a homotopy from \( f \) to \( g \). When \( t = 0 \), \( H(x, 0) = (x, x^2) \), which is \( f(x) \). When \( t = 1 \), \( H(x, 1) = (x, x) \), which is \( g(x) \). As \( t \) varies from 0 to 1, the second component of \( H(x, t) \) sm... | Yes |
Let \( B \subseteq {\mathbb{R}}^{n} \) and let \( X \) be any topological space. Suppose \( f, g : X \rightarrow \) \( B \) are any two continuous maps with the property that for all \( x \in X \), the line segment from \( f\left( x\right) \) to \( g\left( x\right) \) lies in \( B \) . This is the case, for example, if... | \[ H\left( {x, t}\right) = f\left( x\right) + t\left( {g\left( x\right) - f\left( x\right) }\right) . \] | Yes |
Consider the path \( f : I \rightarrow \mathbb{C} \smallsetminus \{ 0\} \) defined (in complex notation) by\n\n\[ f\left( s\right) = {e}^{2\pi is} \]\n\nand the map \( H : I \times I \rightarrow \mathbb{C} \smallsetminus \{ 0\} \) by\n\n\[ H\left( {s, t}\right) = {e}^{2\pi ist}. \]\n\nAt each time \( t,{H}_{t} \) is a ... | This last example shows that the circular path around the origin is homotopic in \( {\mathbb{R}}^{2} \smallsetminus \{ 0\} \) to a constant path, so that simply asking whether a closed path is homotopic to a constant is not sufficient to detect holes. To remedy this, we need to consider homotopies of paths throughout w... | Yes |
Lemma 7.9. Any reparametrization of a path \( f \) is path-homotopic to \( f \) . | Proof. Suppose \( f \circ \varphi \) is a reparametrization of \( f \), and let \( H : I \times I \rightarrow I \) denote the straight-line homotopy from the identity map to \( \varphi \) . Then \( f \circ H \) is a path homotopy from \( f \) to \( f \circ \varphi \) . | Yes |
Proposition 7.10 (Homotopy Invariance of Path Multiplication). The operation of path multiplication is well defined on path classes. More precisely, if \( {f}_{0} \sim {f}_{1} \) and \( {g}_{0} \sim {g}_{1} \), and if \( {f}_{0} \) and \( {g}_{0} \) are composable, then \( {f}_{1} \) and \( {g}_{1} \) are composable an... | Proof. Let \( F : {f}_{0} \sim {f}_{1} \) and \( G : {g}_{0} \sim {g}_{1} \) be path homotopies (Fig. 7.3). The required homotopy \( H : {f}_{0} \cdot {g}_{0} \sim {f}_{1} \cdot {g}_{1} \) is given by\n\n\[ H\left( {s, t}\right) = \left\{ \begin{array}{ll} F\left( {{2s}, t}\right) ; & 0 \leq s \leq \frac{1}{2},0 \leq t... | Yes |
Theorem 7.11 (Properties of Path Class Products). Let \( f \) be any path from p to \( q \) in a space \( X \), and let \( g \) and \( h \) be any paths in \( X \) . Path multiplication satisfies the following properties:\n\n(a) \( \left\lbrack {c}_{p}\right\rbrack \cdot \left\lbrack f\right\rbrack = \left\lbrack f\rig... | Proof. For (a), let us show that \( {c}_{p} \cdot f \sim f \) ; the product the other way works similarly. Define \( H : I \times I \rightarrow X \) (Fig. 7.4) by\n\n\[ H\left( {s, t}\right) = \left\{ \begin{array}{ll} p, & t \geq {2s} \\ f\left( \frac{{2s} - t}{2 - t}\right) , & t \leq {2s} \end{array}\right. \]\n\nGe... | Yes |
Theorem 7.13 (Change of Base Point). Suppose \( X \) is path-connected, \( p, q \in X \) , and \( g \) is any path from \( p \) to \( q \) . The map \( {\Phi }_{g} : {\pi }_{1}\left( {X, p}\right) \rightarrow {\pi }_{1}\left( {X, q}\right) \) defined by\n\n\[ \n{\Phi }_{g}\left\lbrack f\right\rbrack = \left\lbrack \bar... | Proof. Before we begin, we should verify that \( {\Phi }_{g} \) makes sense (Fig. 7.6): since \( g \) goes from \( p \) to \( q \) and \( f \) goes from \( p \) to \( p \), paths in the class \( \left\lbrack \bar{g}\right\rbrack \cdot \left\lbrack f\right\rbrack \cdot \left\lbrack g\right\rbrack \) go from \( q \) to \... | Yes |
Lemma 7.17 (Square Lemma). Let \( F : I \times I \rightarrow X \) be a continuous map, and let \( f, g, h \), and \( k \) be the paths in \( X \) defined by\n\n\[ f\left( s\right) = F\left( {s,0}\right) \]\n\n\[ g\left( s\right) = F\left( {1, s}\right) \]\n\n\[ h\left( s\right) = F\left( {0, s}\right) \]\n\n\[ k\left( ... | Proof. See Problem 7-4. | No |
Lemma 7.19. Suppose \( M \) is a manifold of dimension \( n \geq 2 \). If \( f \) is a path in \( M \) from \( {p}_{1} \) to \( {p}_{2} \) and \( q \) is any point in \( M \) other than \( {p}_{1} \) or \( {p}_{2} \), then \( f \) is path-homotopic to a path that does not pass through \( q \). | Proof. Consider the open cover \( \{ U, V\} \) of \( M \), where \( U \) is a coordinate ball centered at \( q \) and \( V = M \smallsetminus \{ q\} \). If \( f : I \rightarrow M \) is any path from \( {p}_{1} \) to \( {p}_{2} \), then \( \left\{ {{f}^{-1}\left( U\right) ,{f}^{-1}\left( V\right) }\right\} \) is an open... | Yes |
Theorem 7.21. The fundamental group of a manifold is countable. | Proof. Let \( M \) be a manifold, and let \( \mathcal{U} \) be a countable cover of \( M \) by coordinate balls. For each \( U,{U}^{\prime } \in \mathcal{U} \) the intersection \( U \cap {U}^{\prime } \) has at most countably many components; choose a point in each such component and let \( X \) denote the (countable) ... | Yes |
Proposition 7.24. For any continuous map \( \varphi ,{\varphi }_{ * } \) is a group homomorphism. | Proof. Just note that\n\n\[{\varphi }_{ * }\left( {\left\lbrack f\right\rbrack \cdot \left\lbrack g\right\rbrack }\right) = {\varphi }_{ * }\left\lbrack {f \cdot g}\right\rbrack = \left\lbrack {\varphi \circ \left( {f \cdot g}\right) }\right\rbrack .\]\n\nThus it suffices to show that \( \varphi \circ \left( {f \cdot g... | Yes |
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 i... | 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 }_{ * } \... | Yes |
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) \rightarro... | 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. | Yes |
Corollary 7.29. A retract of a simply connected space is simply connected. | 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 \( {\pi }_{1}\left( {A, p}\right) \) . | Yes |
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. | 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 \( {\mathbb{R}}^{2} \). | No |
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. | 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) ... | Yes |
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 \( {\overline{\mathbb{B}}}^{n} \smallsetminus \{ 0\} \) . | 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 retracti... | Yes |
Proposition 7.37 showed that the circle is a strong deformation retract of \( {\mathbb{R}}^{2} \smallsetminus \{ 0\} \) . Therefore, inclusion \( {\mathbb{S}}^{1} \hookrightarrow {\mathbb{R}}^{2} \smallsetminus \{ 0\} \) induces an isomorphism of fundamental groups. | Once we show that \( {\pi }_{1}\left( {\mathbb{S}}^{1}\right) \) is infinite cyclic, this characterizes \( {\pi }_{1}\left( {{\mathbb{R}}^{2}\smallsetminus \{ 0\} }\right) \) as well. | No |
The figure-eight space \( \mathcal{E} \) of Example 7.32 and the theta space, defined by\n\n\[ \Theta = \left\{ {\left( {x, y}\right) \in {\mathbb{R}}^{2} : {x}^{2} + {y}^{2} = 4,\text{ or }y = 0\text{ and } - 2 \leq x \leq 2}\right\} ,\n\]\n\nare both strong deformation retracts of \( {\mathbb{R}}^{2} \) with the two ... | The deformation retractions, indicated schematically in Fig. 7.14, are defined by carving the space up into regions in which straight-line homotopies are easily defined; the resulting maps are continuous by the gluing lemma. Therefore, since homotopy equivalence is transitive, \( \mathcal{E} \) and \( \Theta \) are hom... | Yes |
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 \( {\P... | 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... | Yes |
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. | 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 \( \widetilde{Y} \) i... | Yes |
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 identi... | 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. | No |
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 \( \mathrm{D} \) . | Proof. The proof is exactly the same as the proof for the fundamental group functor (Corollary 7.26). | No |
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(... | 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}^{\pri... | Yes |
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... | 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}... | Yes |
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... | 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... | Yes |
Theorem 8.3 (Unique Lifting Property of the Circle). Let B be a connected topological space. Suppose \( \varphi : B \rightarrow {\mathbb{S}}^{1} \) is continuous, and \( {\widetilde{\varphi }}_{1},{\widetilde{\varphi }}_{2} : B \rightarrow \mathbb{R} \) are lifts of \( \varphi \) that agree at some point of \( B \) . T... | Proof. Let \( \mathcal{A} = \left\{ {b \in B : {\widetilde{\varphi }}_{1}\left( b\right) = {\widetilde{\varphi }}_{2}\left( b\right) }\right\} \) . By hypothesis \( \mathcal{A} \) is not empty. Since \( B \) is connected, if we can show that \( \mathcal{A} \) is open and closed in \( B \), it must be all of \( B \) . T... | Yes |
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 ... | 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... | Yes |
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 \( {f}_{1} \) wit... | 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... | Yes |
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. | 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... | Yes |
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 \( \left\lbrack \omega \right\rbrack \) . | 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\... | Yes |
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 cop... | Proof. This is a direct consequence of Theorem 8.9 and Proposition 7.34. | No |
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}}^{... | 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 tha... | Yes |
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. | 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 \( \psi \circ \omega \)... | Yes |
Theorem 8.18. Let \( \varphi : {\mathbb{S}}^{1} \rightarrow {\mathbb{S}}^{1} \) be continuous. If \( \deg \varphi \neq 0 \), then \( \varphi \) is surjective. | 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 stereogr... | Yes |
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. | 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 anti... | Yes |
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. | 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}^... | Yes |
Proposition 9.2. Every element of \( { * }_{\alpha \in A}{G}_{\alpha } \) is represented by a unique reduced word. | Proof. We showed above that every equivalence class contains a reduced word, so we need only check that two reduced words representing the same equivalence class must be equal. The proof amounts to constructing a \ | No |
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 * \math... | 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 prod... | Yes |
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 : { * }_{\alph... | 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\rig... | Yes |
Corollary 9.6. The free product is the coproduct in the category of groups. | Proof. The characteristic property is exactly the defining property of the coproduct in a category. | No |
The free product is the unique group (up to isomorphism) satisfying the characteristic property. | Proof. Theorem 7.57 shows that coproducts in any category are unique up to isomorphism. | No |
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 \( \varphi \). | 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... | No |
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} \... | Proof. See Problem 9-3. | No |
Lemma 9.18. If an abelian group \( G \) has a finite basis, then every finite basis has the same number of elements. | 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 ... | Yes |
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 \( G \) . | 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 an... | Yes |
Proposition 9.21. Any abelian group that is finitely generated and torsion-free is free abelian of finite rank. | 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 \( n \) an... | Yes |
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 ,{\al... | Proof. This is an immediate consequence of Problems 9-4(b) and 9-5. | No |
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 \( {X}_{1} \vee \cdots \vee {X}_{n} \) . | 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 \( {\... | Yes |
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\ninduce... | 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} ... | No |
Proposition 10.11. Every finite connected graph contains a spanning tree. | 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, st... | Yes |
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... | 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 = \operato... | Yes |
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. | 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. | No |
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}... | \[ {\pi }_{1}\left( {X, v}\right) \cong \left\langle {{\beta }_{1},\ldots ,{\beta }_{n} \mid {\sigma }_{1},\ldots ,{\sigma }_{k}}\right\rangle . \] Proof. This follows immediately by induction from the two preceding propositions, using the result of Exercise 5.19. | Yes |
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 present... | 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 ... | Yes |
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}\# \c... | 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 eas... | Yes |
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 \( {\mathbb{P}}^{2}\# \cdots \# {\mathbb{P}}^{2} \) . | 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 ... | Yes |
Corollary 10.23. A connected sum of projective planes is not orientable. | 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. | No |
Corollary 10.24. Orientability of a compact surface is a topological invariant. | 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. | Yes |
Corollary 10.25. The Euler characteristic of a surface presentation is a topological invariant. | 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 diffe... | Yes |
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. | this is the content of Proposition 8.1. | No |
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. | 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 \( {p}_{n} \) ont... | Yes |
Define \( {\varepsilon }^{n} : {\mathbb{R}}^{n} \rightarrow {\mathbb{T}}^{n} \) by\n\n\[ \n{\varepsilon }^{n}\left( {{x}_{1},\ldots ,{x}_{n}}\right) = \left( {\varepsilon \left( {x}_{1}\right) ,\ldots ,\varepsilon \left( {x}_{n}\right) }\right) ,\n\]\n\nwhere \( \varepsilon \) is the exponential quotient map of Example... | Since a product of covering maps is a covering map (Proposition 11.1(c)), \( {\varepsilon }^{n} \) is a covering map. | Yes |
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}_... | 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 \( \sigma = {\left( {\left. q\right| }_{{\widetilde{U}}_{0}}\right) }^{-1} \) . | Yes |
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. | 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}... | No |
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) \( {\widetild... | 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 h... | Yes |
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. | 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 m... | Yes |
Let \( q : {X}_{3} \rightarrow {X}_{2} \) be the covering map of Exercise 11.7, and choose 1 as base point in both \( {X}_{3} \) and \( {X}_{2} \). To compute the subgroup induced by \( q \), we need to compute the action of \( q \) on the generators of \( {\pi }_{1}\left( {{X}_{3},1}\right) \). Let \( a, b, c \) be lo... | The images of these generators of \( {\pi }_{1}\left( {{X}_{3},1}\right) \) under \( {q}_{ * } \) are \( \left\lbrack a\right\rbrack ,{\left\lbrack b\right\rbrack }^{2} \), and \( \left\lbrack b\right\rbrack \cdot \left\lbrack a\right\rbrack \cdot {\left\lbrack b\right\rbrack }^{-1} \), so the subgroup induced by \( q ... | Yes |
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 \le... | 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. 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{... | 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... | Yes |
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... | 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\[ \n= \left\{ {{g}^{\prime } \in G : s \cdot \left( {g{g}^{\prime }{g}^{-1}}\right) = s}\right\} \n\]\... | Yes |
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. | 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 \( g \in G \)... | Yes |
Proposition 11.26 ( \( G \) -Set Isomorphism Criterion). Suppose \( {S}_{1} \) and \( {S}_{2} \) are transitive right \( G \) -sets.\n\n(a) Given \( {s}_{1} \in {S}_{1} \) and \( {s}_{2} \in {S}_{2} \), there exists a (necessarily unique) G-isomorphism \( \varphi : {S}_{1} \rightarrow {S}_{2} \) taking \( {s}_{1} \) to... | Proof. To prove (a), suppose first that \( \varphi : {S}_{1} \rightarrow {S}_{2} \) is a \( G \) -isomorphism taking \( {s}_{1} \) to \( {s}_{2} \) . Proposition 11.24 applied to \( \varphi \) shows that \( {G}_{{s}_{1}} \subseteq {G}_{{s}_{2}} \), and the same result applied to \( {\varphi }^{-1} \) shows the reverse ... | Yes |
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... | Proof. This is an immediate consequence of Proposition 11.26(a). | No |
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{... | 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 \(... | Yes |
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}\ri... | 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 rep... | Yes |
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. | 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. | Yes |
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 \( X \) . | 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\rbra... | Yes |
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. | 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). | Yes |
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 \( {\pi }_{1}\left( {X, x}\right) \) . | 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 ... | Yes |
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 subgro... | 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 onl... | Yes |
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. | 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. | Yes |
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( {... | 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). | Yes |
Let \( {p}_{n} : {\mathbb{S}}^{1} \rightarrow {\mathbb{S}}^{1} \) be the \( n \) th power map defined in Example 11.4. The subgroup of \( {\pi }_{1}\left( {{\mathbb{S}}^{1},1}\right) \) induced by \( {p}_{n} \) is the cyclic subgroup generated by \( {\left\lbrack \omega \right\rbrack }^{n} \) (Example 11.21). | By the covering homomorphism criterion, there is a covering homomorphism from \( {p}_{m} \) to \( {p}_{n} \) if and only if \( m \) is divisible by \( n \) ; the homomorphism in that case is just \( {p}_{m/n} \) . | No |
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 : {\mathb... | It is easy to check that \( \varphi \left( {x, e}\right) = \left( {\varepsilon \left( x\right), e}\right) \) is such a homomorphism. | Yes |
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)... | 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 ... | Yes |
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 \... | 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 m... | Yes |
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 \( \varphi \in {\operatorname{Aut}}_{\varepsilon }\left( \mathbb{R}\right) \) ... | 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{... | Yes |
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... | Proof. This follows immediately from the covering isomorphism criterion (Theorem 11.40). | No |
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. | 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 ... | Yes |
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