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Lemma 62.2 (Borsuk lemma). Let \( a \) and \( b \) be points of \( {S}^{2} \) . Let \( A \) be a compact space, and let \( f : A \rightarrow {S}^{2} - a - b \) be a continuous injective map. If \( f \) is nulhomotopic, then \( a \) and \( b \) lie in the same component of \( {S}^{2} - f\left( A\right) \) . | Proof. Because \( A \) is compact and \( {S}^{2} \) is Hausdorff, \( f\left( A\right) \) is a compact subspace of \( {S}^{2} \) that is homeomorphic to \( A \) . Because \( f \) is nulhomotopic, so is the inclusion mapping of \( f\left( A\right) \) into \( {S}^{2} - a - b \) . Hence it suffices to prove the lemma in th... | Yes |
Theorem 62.3 (Invariance of domain). If \( U \) is an open subset of \( {\mathbb{R}}^{2} \) and \( f : U \rightarrow \) \( {\mathbb{R}}^{2} \) is continuous and injective, then \( f\left( U\right) \) is open in \( {\mathbb{R}}^{2} \) and the inverse function \( {f}^{-1} : f\left( U\right) \rightarrow U \) is continuous... | Proof. As usual, we can replace \( {\mathbb{R}}^{2} \) by \( {S}^{2} \) . We show that if \( U \) is an open subset of \( {\mathbb{R}}^{2} \) and \( f : U \rightarrow {S}^{2} \) is continuous and injective, then \( f\left( U\right) \) is open in \( {S}^{2} \) and the inverse function is continuous.\n\nStep 1. We show t... | Yes |
Theorem 63.1. Let \( X \) be the union of two open sets \( U \) and \( V \), such that \( U \cap V \) can be written as the union of two disjoint open sets \( A \) and \( B \) . Assume that there is a path \( \alpha \) in \( U \) from a point \( a \) of \( A \) to a point \( b \) of \( B \), and that there is a path \(... | Proof. The proof is in many ways an imitation of the proof in \( §{54} \) that the fundamental group of the circle is infinite cyclic. As in that proof, the crucial step is to find an appropriate covering space \( E \) for the space \( X \) . Step 1. (Construction of \( E \) ). We construct \( E \) by pasting together ... | Yes |
Theorem 63.2 (A nonseparation theorem). Let \( D \) be an arc in \( {S}^{2} \) . Then \( D \) does not separate \( {S}^{2} \) . | Proof. We give two proofs of this theorem. The first uses the results of the preceding section, and the second does not.\n\nFirst proof. Because \( D \) is contractible, the identity map \( \mathrm{i} : D \rightarrow D \) is nulhomo-topic. Hence if \( a \) and \( b \) are any two points of \( {S}^{2} \) not in \( D \),... | Yes |
Lemma 64.1. Let \( X \) be a theta space that is a subspace of \( {S}^{2} \) ; let \( A, B \), and \( C \) be the arcs whose union is \( X \) . Then \( X \) separates \( {S}^{2} \) into three components, whose boundaries are \( A \cup B, B \cup C \), and \( A \cup C \), respectively. The component having \( A \cup B \)... | Proof. Let \( a \) and \( b \) be the end points of the arcs \( A, B \), and \( C \) . Consider the simple closed curve \( A \cup B \) ; it separates \( {S}^{2} \) into two components \( U \) and \( {U}^{\prime } \), each of which is open in \( {S}^{2} \) and has boundary \( A \cup B \) . See Figure 64.3.\n\n![f9109c85... | Yes |
Theorem 64.2. Let \( X \) be the utilities graph. Then \( X \) cannot be imbedded in the plane. | Proof. If \( X \) can be imbedded in the plane, then it can be imbedded in \( {S}^{2} \) . So suppose \( X \) is a subspace of \( {S}^{2} \) . We derive a contradiction.\n\nWe use the notation of Example 2, where \( g, w, e,{h}_{1},{h}_{2} \), and \( {h}_{3} \) are the vertices of \( X \) . Let \( A, B \), and \( C \) ... | Yes |
Lemma 64.3. Let \( X \) be a subspace of \( {S}^{2} \) that is a complete graph on four vertices \( {a}_{1} \) , \( {a}_{2},{a}_{3} \), and \( {a}_{4} \) . Then \( X \) separates \( {S}^{2} \) into four components. The boundaries of these components are the sets \( {X}_{1},{X}_{2},{X}_{3} \), and \( {X}_{4} \), where \... | Proof. Let \( Y \) be the union of all the arcs of \( X \) different from the arc \( {a}_{2}{a}_{4} \) . Then we can write \( Y \) as a theta space by setting\n\n\[ A = {a}_{1}{a}_{2}{a}_{3} \]\n\n\[ B = {a}_{1}{a}_{3} \]\n\n\[ C = {a}_{1}{a}_{4}{a}_{3} \]\n\nSee Figure 64.5. The arcs \( A, B \), and \( C \) intersect ... | Yes |
Lemma 65.1. Let \( G \) be a subspace of \( {S}^{2} \) that is a complete graph on four vertices \( {a}_{1},\ldots ,{a}_{4} \). Let \( C \) be the subgraph \( {a}_{1}{a}_{2}{a}_{3}{a}_{4}{a}_{1} \), which is a simple closed curve. Let \( p \) and \( q \) be interior points of the edges \( {a}_{1}{a}_{3} \) and \( {a}_{... | Proof. (a) As in the proof of Lemma 64.3, the theta space \( C \cup {a}_{1}{a}_{3} \) separates \( {S}^{2} \) into three components \( U, V \), and \( W \). One of these, say \( W \), has \( C \) as its boundary; it is the only component whose boundary contains both \( {a}_{2} \) and \( {a}_{4} \). Therefore, \( {a}_{2... | Yes |
Lemma 66.1. Let \( f \) be a loop in \( {\mathbb{R}}^{2} - a \) .\n\n(a) If \( \bar{f} \) is the reverse of \( f \), then \( n\left( {\bar{f}, a}\right) = - n\left( {f, a}\right) \) . | Proof. (a) To compute \( n\left( {\widetilde{f}, a}\right) \), one replace \( s \) by \( 1 - s \) throughout the definition. This has the effect of changing \( \widetilde{g}\left( 1\right) - \widetilde{g}\left( 0\right) \) by a sign | No |
Theorem 66.2. Let \( f \) be a simple loop in \( {\mathbb{R}}^{2} \) . If a lies in the unbounded component of \( {\mathbb{R}}^{2} - f\left( I\right) \), then \( n\left( {f, a}\right) = 0 \) ; while if \( a \) lies in the bounded component, \( n\left( {f, a}\right) = \pm 1 \) . | Proof. Since \( n\left( {f, a}\right) = n\left( {f - a,\mathbf{0}}\right) \), we may restrict ourselves to the case \( a = \mathbf{0} \) . Furthermore, we may assume that the base point of \( f \) lies on the positive \( x \) -axis. For one can gradually rotate \( {\mathbb{R}}^{2} - \mathbf{0} \) until the base point o... | Yes |
Lemma 66.3. Let \( f \) be a piecewise-differentiable loop in the complex plane; let \( a \) be a point not in the image of \( f \) . Then\n\n\[ n\left( {f, a}\right) = \frac{1}{2\pi i}{\int }_{f}\frac{dz}{z - a}. \]\n\nThis equation is often used as the definition of the winding number of \( f \) . | Proof. The proof is a simple exercise in computation. Let \( p : \mathbb{R} \rightarrow {S}^{1} \) be the standard covering map. Let \( r\left( s\right) = \parallel f\left( s\right) - a\parallel \) and \( g\left( s\right) = \left\lbrack {f\left( s\right) - a}\right\rbrack /r\left( s\right) \) . Let \( \widetilde{g} \) ... | Yes |
Theorem 66.4 (Cauchy integral formula-classical version). Let \( C \) be a simple closed piecewise-differentiable curve in the complex plane. Let \( B \) be the bounded component of \( {\mathbb{R}}^{2} - C \) . If \( F\left( z\right) \) is analytic in an open set \( \Omega \) that contains \( B \) and \( C \), then for... | Proof. We derive this formula from the version of it proved in Ahlfors [A], which is the following:\n\nLet \( F \) be analytic in a region \( \Omega \) . Let \( f \) be a piecewise-differentiable loop in \( \Omega \) . Assume that \( n\left( {f, b}\right) = 0 \) for each \( b \) not in \( \Omega \) . If \( a \in \Omega... | Yes |
Lemma 67.1. Let \( G \) be an abelian group; let \( \left\{ {G}_{\alpha }\right\} \) be a family of subgroups of \( G \) . If \( G \) is the direct sum of the groups \( {G}_{\alpha } \), then \( G \) satisfies the following condition:\n\nGiven any abelian group \( H \) and any family of homomorphisms\n\n(*) \n\n\( {h}_... | Proof. We show first that if \( G \) has the stated extension property, then \( G \) is the direct sum of the \( {G}_{\alpha } \) . Suppose \( x = \sum {x}_{\alpha } = \sum {y}_{\alpha } \) ; we show that for any particular index \( \beta \) , we have \( {x}_{\beta } = {y}_{\beta } \) . Let \( H \) denote the group \( ... | Yes |
Corollary 67.2. Let \( G = {G}_{1} \oplus {G}_{2} \) . Suppose \( {G}_{1} \) is the direct sum of subgroups \( {H}_{\alpha } \) for \( \alpha \in J \), and \( {G}_{2} \) is the direct sum of subgroups \( {H}_{\beta } \) for \( \beta \in K \), where the index sets \( J \) and \( K \) are disjoint. Then \( G \) is the di... | Proof. If \( {h}_{\alpha } : {H}_{\alpha } \rightarrow H \) and \( {h}_{\beta } : {H}_{\beta } \rightarrow H \) are families of homomorphisms, they extend to homomorphisms \( {h}_{1} : {G}_{1} \rightarrow H \) and \( {h}_{2} : {G}_{2} \rightarrow H \) by the preceding lemma. Then \( {h}_{1} \) and \( {h}_{2} \) extend ... | No |
Corollary 67.3. If \( G = {G}_{1} \oplus {G}_{2} \), then \( G/{G}_{2} \) is isomorphic to \( {G}_{1} \) . | Proof. Let \( H = {G}_{1} \), let \( {h}_{1} : {G}_{1} \rightarrow H \) be the identity homomorphism, and let \( {h}_{2} : {G}_{2} \rightarrow H \) be the trivial homomorphism. Let \( h : G \rightarrow H \) be their extension to \( G \) . Then \( h \) is surjective with kernel \( {G}_{2} \) . | Yes |
Theorem 67.4. Given a family of abelian groups \( {\left\{ {G}_{\alpha }\right\} }_{\alpha \in J} \), there exists an abelian group \( G \) and a family of monomorphisms \( {i}_{\alpha } : {G}_{\alpha } \rightarrow G \) such that \( G \) is the direct sum of the groups \( {\mathrm{i}}_{\alpha }\left( {G}_{\alpha }\righ... | Proof. Consider first the cartesian product\n\n\[ \mathop{\prod }\limits_{{\alpha \in J}}{G}_{\alpha } \]\n\n it is an abelian group if we add two \( J \) -tuples by adding them coordinate-wise. Let \( G \) denote the subgroup of the cartesian product consisting of those tuples \( {\left( {x}_{\alpha }\right) }_{\alpha... | Yes |
Lemma 67.5. Let \( {\left\{ {G}_{\alpha }\right\} }_{\alpha \in J} \) be an indexed family of abelian groups; let \( G \) be an abelian group; let \( {\mathrm{i}}_{\alpha } : {G}_{\alpha } \rightarrow G \) be a family of homomorphisms. If each \( {\mathrm{i}}_{\alpha } \) is a monomorphism and \( G \) is the direct sum... | Proof. The only part that requires proof is the statement that if the extension condition holds, then each \( {i}_{\alpha } \) is a monomorphism. That is proved as follows. Given an index \( \beta \), set \( H = {G}_{\beta } \) and let \( {h}_{\alpha } : {G}_{\alpha } \rightarrow H \) be the identity homomorphism if \(... | Yes |
Theorem 67.6 (Uniqueness of direct sums). Let \( {\left\{ {G}_{\alpha }\right\} }_{\alpha \in J} \) be a family of abelian groups Suppose \( G \) and \( {G}^{\prime } \) are abelian groups and \( {i}_{\alpha } : {G}_{\alpha } \rightarrow G \) and \( {i}_{\alpha }^{\prime } : {G}_{\alpha } \rightarrow {G}^{\prime } \) a... | Proof. We apply the preceding lemma (four times!). Since \( G \) is the external direct sum of the \( {G}_{\alpha } \) and \( \left\{ {i}_{\alpha }^{\prime }\right\} \) is a family of homomorphisms, there exists a unique homomorphism \( \phi : G \rightarrow {G}^{\prime } \) such that \( \phi \circ {\mathrm{i}}_{\alpha ... | Yes |
Lemma 67.7. Let \( G \) be an abelian group; let \( {\left\{ {a}_{\alpha }\right\} }_{\alpha \in J} \) be a family of elements of \( G \) that generates \( G \) . Then \( G \) is a free abelian group with basis \( \left\{ {a}_{\alpha }\right\} \) if and only if for any abelian group \( H \) and any family \( \left\{ {y... | Proof. Let \( {G}_{\alpha } \) denote the subgroup of \( G \) generated by \( {a}_{\alpha } \) . Suppose first that the extension property holds. We show first that each group \( {G}_{\alpha } \) is infinite cyclic. Suppose that for some index \( \beta \), the element \( {a}_{\beta } \) generates a finite cyclic subgro... | Yes |
Theorem 67.8. If \( G \) is a free abelian group with basis \( \left\{ {{a}_{1},\ldots ,{a}_{n}}\right\} \), then \( n \) is uniquely determined by \( G \) . | Proof. The group \( G \) is isomorphic to the \( n \) -fold product \( \mathbb{Z} \times \cdots \times \mathbb{Z} \) ; the subgroup \( {2G} \) corresponds to the product \( \left( {2\mathbb{Z}}\right) \times \cdots \times \left( {2\mathbb{Z}}\right) \) . Then the quotient group \( G/{2G} \) is in bijective corresponden... | Yes |
Theorem 68.2. Given a family \( {\left\{ {G}_{\alpha }\right\} }_{\alpha \in J} \) of groups, there exists a group \( G \) and a family of monomorphisms \( {i}_{\alpha } : {G}_{\alpha } \rightarrow G \) such that \( G \) is the free product of the groups \( {i}_{\alpha }\left( {G}_{\alpha }\right) \) . | Proof. For convenience, we assume that the groups \( {G}_{\alpha } \) are disjoint as sets. (This can be accomplished by replacing \( {G}_{\alpha } \) by \( {G}_{\alpha } \times \{ \alpha \} \) for each index \( \alpha \), if necessary.)\n\nThen as before, we define a word (of length \( n \) ) in the elements of the gr... | No |
Lemma 68.5. Let \( {\left\{ {G}_{\alpha }\right\} }_{\alpha \in J} \) be a family of groups; let \( G \) be a group; let \( {i}_{\alpha } : {G}_{\alpha } \rightarrow G \) be a family of homomorphisms. If the extension condition of Lemma 68.3 holds, then each \( {\mathrm{i}}_{\alpha } \) is a monomorphism and \( G \) is... | Proof. We first show that each \( {i}_{\alpha } \) is a monomorphism. Given an index \( \beta \), let us set \( H = {G}_{\beta } \) . Let \( {h}_{\alpha } : {G}_{\alpha } \rightarrow H \) be the identity if \( \alpha = \beta \), and the trivial homomorphism if \( \alpha \neq \beta \) Let \( h : G \rightarrow H \) be th... | Yes |
Corollary 68.6. Let \( G = {G}_{1} * {G}_{2} \), where \( {G}_{1} \) is the free product of the subgroups \( {\left\{ {H}_{\alpha }\right\} }_{\alpha \in J} \) and \( {G}_{2} \) is the free product of the subgroups \( {\left\{ {H}_{\beta }\right\} }_{\beta \in K} \) . If the index sets \( J \) and \( K \) are disjoint,... | Proof. The proof is almost a copy of the proof of Corollary 67.2. | No |
Theorem 68.7. Let \( G = {G}_{1} * {G}_{2} \) . Let \( {N}_{i} \) be a normal subgroup of \( {G}_{i} \), for \( i = 1,2 \) . If \( N \) is the least normal subgroup of \( G \) that contains \( {N}_{1} \) and \( {N}_{2} \), then\n\n\[ G/N \cong \left( {{G}_{1}/{N}_{1}}\right) * \left( {{G}_{2}/{N}_{2}}\right) \] | Proof. The composite of the inclusion and projection homomorphisms\n\n\[ {G}_{1} \rightarrow {G}_{1} * {G}_{2} \rightarrow \left( {{G}_{1} * {G}_{2}}\right) /N \]\ncarries \( {N}_{1} \) to the identity element, so that it induces a homomorphism\n\n\[ {i}_{1} : {G}_{1}/{N}_{1} \rightarrow \left( {{G}_{1} * {G}_{2}}\righ... | Yes |
Lemma 68.9. Let \( S \) be a subset of the group \( G \) . If \( N \) is the least normal subgroup of \( G \) containing \( S \), then \( N \) is generated by all conjugates of elements of \( S \) . | Proof. Let \( {N}^{\prime } \) be the subgroup of \( G \) generated by all conjugates of elements of \( S \) . We know that \( {N}^{\prime } \subset N \) ; to verify the reverse inclusion, we need merely show that \( {N}^{\prime } \) is normal in \( G \) . Given \( x \in {N}^{\prime } \) and \( c \in G \), we show that... | Yes |
Lemma 69.1. Let \( G \) be a group; let \( {\left\{ {a}_{\alpha }\right\} }_{\alpha \in J} \) be a family of elements of \( G \). If \( G \) is a free group with system of free generators \( \left\{ {a}_{\alpha }\right\} \), then \( G \) satisfies the following condition:\n\nGiven any group \( H \) and any family \( \l... | Proof. If \( G \) is free, then for each \( \alpha \), the group \( {G}_{\alpha } \) generated by \( {a}_{\alpha } \) is infinte cyclic, so there is a homomorphism \( {h}_{\alpha } : {G}_{\alpha } \rightarrow H \) with \( {h}_{\alpha }\left( {a}_{\alpha }\right) = {y}_{\alpha } \). Then Lemma 68.1 applies. To prove the... | Yes |
Lemma 69.3. Given \( G \), the subgroup \( \left\lbrack {G, G}\right\rbrack \) is a normal subgroup of \( G \) and the quotient group \( G/\left\lbrack {G, G}\right\rbrack \) is abelian. If \( h : G \rightarrow H \) is any homomorphism from \( G \) to an abelian group \( H \), then the kernel of \( h \) contains \( \le... | Proof. Step 1. First we show that any conjugate of a commutator is in \( \left\lbrack {G, G}\right\rbrack \) . We compute as follows:\n\n\[ g\left\lbrack {x, y}\right\rbrack {g}^{-1} = g\left( {{xy}{x}^{-1}{y}^{-1}}\right) {g}^{-1} \]\n\n\[ = \left( {{gxy}{x}^{-1}}\right) \left( 1\right) \left( {{y}^{-1}{g}^{-1}}\right... | Yes |
Theorem 69.4. If \( G \) is a free group with free generators \( {a}_{\alpha } \), then \( G/\left\lbrack {G, G}\right\rbrack \) is a free abelian group with basis \( \left\lbrack {a}_{\alpha }\right\rbrack \), where \( \left\lbrack {a}_{\alpha }\right\rbrack \) denotes the coset of \( {a}_{\alpha } \) in \( G/\left\lb... | Proof. We apply Lemma 67.7. Given any family \( \left\{ {y}_{\alpha }\right\} \) of elements of the abelian group \( H \), there exists a homomorphism \( h : G \rightarrow H \) such that \( h\left( {a}_{\alpha }\right) = {y}_{\alpha } \) for each \( \alpha \) . Because \( H \) is abelian, the kernel of \( h \) contains... | Yes |
Corollary 69.5. If \( G \) is a free group with \( n \) free generators, then any system of free generators for \( G \) has \( n \) elements. | Proof. The free abelian group \( G/\left\lbrack {G, G}\right\rbrack \) has rank \( n \) . | No |
Theorem 70.1 (Seifert-van Kampen theorem). Let \( X = U \cup V \), where \( U \) and \( V \) are open in \( X \) ; assume \( U, V \), and \( U \cap V \) are path connected; let \( {x}_{0} \in U \cap V \) . Let \( H \) be a group, and let\n\n\[{\phi }_{1} : {\pi }_{1}\left( {U,{x}_{0}}\right) \rightarrow H\;\text{ and }... | Proof. Uniqueness is easy. Theorem 59.1 tells us that \( {\pi }_{1}\left( {X,{x}_{0}}\right) \) is generated by the images of \( {j}_{1} \) and \( {j}_{2} \) . The value of \( \Phi \) on the generator \( {j}_{1}\left( {g}_{1}\right) \) must equal \( {\phi }_{1}\left( {g}_{1}\right) \), and its value on \( {j}_{2}\left(... | Yes |
Theorem 70.2 (Seifert-van Kampen theorem, classical version). Assume the hypotheses of the preceding theorem. Let\n\n\[ j : {\pi }_{1}\left( {U,{x}_{0}}\right) * {\pi }_{1}\left( {V,{x}_{0}}\right) \rightarrow {\pi }_{1}\left( {X,{x}_{0}}\right) \]\n\nbe the homomorphism of the free product that extends the homomorphis... | Proof. The fact that \( {\pi }_{1}\left( {X,{x}_{0}}\right) \) is generated by the images of \( {j}_{1} \) and \( {j}_{2} \) implies that \( j \) is surjective.\n\nWe show that \( N \subset \ker j \) . Since \( \ker j \) is normal, it is enough to show that \( {i}_{1}{\left( g\right) }^{-1}{i}_{2}\left( g\right) \) bel... | Yes |
Theorem 71.1. Let \( X \) be the wedge of the circles \( {S}_{1},\ldots ,{S}_{n} \) ; let \( p \) be the common point of these circles. Then \( {\pi }_{1}\left( {X, p}\right) \) is a free group. If \( {f}_{i} \) is a loop in \( {S}_{i} \) that represents a generator of \( {\pi }_{1}\left( {{S}_{i}, p}\right) \), then t... | Proof. The result is immediate if \( n = 1 \) . We proceed by induction on \( n \) . The proof is similar to the one given in Example 1 of the preceding section.\n\nLet \( X \) be the wedge of the circles \( {S}_{1},\ldots ,{S}_{n} \), with \( p \) the common point of these circles. Choose a point \( {q}_{i} \) of \( {... | Yes |
Lemma 71.2. Let \( X \) be the wedge of the circles \( {S}_{\alpha } \), for \( \alpha \in J \) . Then \( X \) is normal. Furthermore, any compact subspace of \( X \) is contained in the union of finitely many circles \( {S}_{\alpha } \). | Proof. It is clear that one-point sets are closed in \( X \) . Let \( A \) and \( B \) be disjoint closed subsets of \( X \) ; assume that \( B \) does not contain \( p \) Choose disjoint subsets \( {U}_{\alpha } \) and \( {V}_{\alpha } \) of \( {S}_{\alpha } \) that are open in \( {S}_{\alpha } \) and contain \( \{ p\... | Yes |
Theorem 71.3. Let \( X \) be the wedge of the circles \( {S}_{\alpha } \), for \( \alpha \in J \) ; let \( p \) be the common point of these circles. Then \( {\pi }_{1}\left( {X, p}\right) \) is a free group. If \( {f}_{\alpha } \) is a loop in \( {S}_{\alpha } \) representing a generator of \( {\pi }_{1}\left( {{S}_{\... | Proof. Let \( {i}_{\alpha } : {\pi }_{1}\left( {{S}_{\alpha }, p}\right) \rightarrow {\pi }_{1}\left( {X, p}\right) \) be the homomorphism induced by inclusion; let \( {G}_{\alpha } \) be the image of \( {i}_{\alpha } . \n\nNote that if \( f \) is any loop in \( X \) based at \( p \), then the image set of \( f \) is c... | Yes |
Theorem 72.1. Let \( X \) be a Hausdorff space; let \( A \) be a closed path-connected subspace of \( X \) . Suppose that there is a continuous map \( h : {B}^{2} \rightarrow X \) that maps Int \( {B}^{2} \) bijectively onto \( X - A \) and maps \( {S}^{1} = \) Bd \( {B}^{2} \) into \( A \) . Let \( p \in {S}^{1} \) an... | Proof. Step 1. The origin \( \mathbf{0} \) is the center point of \( {B}^{2} \) ; let \( {x}_{0} \) be the point \( h\left( \mathbf{0}\right) \) of \( X \) . If \( U \) is the open set \( U = X - {x}_{0} \) of \( X \), we show that \( A \) is a deformation retract of \( U \) . See Figure 72.1. Let \( C = h\left( {B}^{2... | Yes |
Theorem 73.1. The fundamental group of the torus has a presentation consisting of two generators \( \alpha ,\beta \) and a single relation \( {\alpha \beta }{\alpha }^{-1}{\beta }^{-1} \) . | Proof. Let \( X = {S}^{1} \times {S}^{1} \) be the torus, and let \( h : {I}^{2} \rightarrow X \) be obtained by restricting the standard covering map \( p \times p : \mathbb{R} \times \mathbb{R} \rightarrow {S}^{1} \times {S}^{1} \) . Let \( p \) be the point \( \left( {0,0}\right) \) of Bd \( {I}^{2} \), let \( a = h... | Yes |
Lemma 73.3. Let \( \pi : E \rightarrow X \) be a closed quotient map. If \( E \) is normal, then so is \( X \) . | Proof Assume \( E \) is normal. One-point sets are closed in \( X \) because one-point sets are closed in \( E \) . Now let \( A \) and \( B \) be disjoint closed sets of \( X \) . Then \( {\pi }^{-1}\left( A\right) \) and \( {\pi }^{-1}\left( B\right) \) are disjoint closed sets of \( E \) . Choose disjoint open sets ... | Yes |
Theorem 73.4. The fundamental group of the \( n \) -fold dunce cap is a cyclic group of order \( n \) . | Proof. Let \( h : {B}^{2} \rightarrow X \) be the quotient map, where \( X \) is the \( n \) -fold dunce cap. Set \( A = h\left( {S}^{1}\right) \) . Let \( p = \left( {1,0}\right) \in {S}^{1} \) and let \( a = h\left( p\right) \) . Then \( h \) maps the arc \( C \) of \( {S}^{1} \) running from \( p \) to \( r\left( p\... | Yes |
Theorem 74.1. Let \( X \) be the space obtained from a finite collection of polygonal regions by pasting edges together according to some labelling scheme. Then \( X \) is a compact Hausdorff space. | Proof. For simplicity, we treat the case where \( X \) is formed from a single polygonal region. The general case is similar.\n\nIt is immediate that \( X \) is compact, since the quotient map is continuous. To show \( X \) is Hausdorff, it suffices to show that the quotient map \( \pi \) is a closed map. (See Lemma 73... | Yes |
Theorem 74.2. Let \( P \) be a polygonal region; let\n\n\[ w = {\left( {a}_{{i}_{1}}\right) }^{{\epsilon }_{1}}\cdots {\left( {a}_{{i}_{n}}\right) }^{{\epsilon }_{n}} \]\n\nbe a labelling scheme for the edges of \( P \) . Let \( X \) be the resulting quotient space; let \( \pi : P \rightarrow X \) be the quotient map. ... | Proof. The proof is similar to the proof we gave for the torus in \( §{73} \) . Because \( \pi \) maps all vertices of \( P \) to a single point of \( X \), the space \( A = \pi \left( {\operatorname{Bd}P}\right) \) is a wedge of \( k \) circles. For each \( i \), choose an edge of \( P \) that is labelled \( {a}_{i} \... | Yes |
Theorem 74.3. Let \( X \) denote the \( n \) -fold torus. Then \( {\pi }_{1}\left( {X,{x}_{0}}\right) \) is isomorphic to the quotient of the free group on the \( {2n} \) generators \( {\alpha }_{1},{\beta }_{1},\ldots ,{\alpha }_{n},{\beta }_{n} \) by the least normal subgroup containing the element \[ \left\lbrack {{... | Proof. In order to apply Theorem 74.2, one must show that under the labelling scheme for \( X \), all the vertices of the polygonal region belong to the same equivalence class. We leave this to you to check. | No |
Theorem 74.4. Let \( X \) denote the \( m \) -fold projective plane. Then \( {\pi }_{1}\left( {X,{x}_{0}}\right) \) is isomorphic to the quotient of the free group on \( m \) generators \( {\alpha }_{1},\ldots ,{\alpha }_{m} \) by the least normal subgroup containing the element\n\n\[{\left( {\alpha }_{1}\right) }^{2}{... | Proof. One needs only to check that under the labelling scheme for \( X \), all the vertices of the polygonal region belong to the same equivalence class. This we leave to you. I | No |
Theorem 75.1. Let \( F \) be a group; let \( N \) be a normal subgroup of \( F \) ; let \( q : F \rightarrow F/N \) be the projection. The projection homomorphism \[ p : F \rightarrow F/\left\lbrack {F, F}\right\rbrack \] induces an isomorphism \[ \phi : q\left( F\right) /\left\lbrack {q\left( F\right), q\left( F\right... | Proof. One has projection homomorphisms \( p, q, r, s \), as in the following diagram, where \( q\left( F\right) = F/N \) and \( p\left( F\right) = F/\left\lbrack {F, F}\right\rbrack \) .  Because \( r \circ p \) map... | Yes |
Corollary 75.2. Let \( F \) be a free group with free generators \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) ; let \( N \) be the least normal subgroup of \( F \) containing the element \( x \) of \( F \) ; let \( G = F/N \) . Let \( p.F \rightarrow F/\left\lbrack {F, F}\right\rbrack \) be projection. Then \( G/\left\lbr... | Proof. Note that because \( N \) is generated by \( x \) and all its conjugates, the group \( p\left( N\right) \) is generated by \( p\left( x\right) \) The corollary then follows from the preceding theorem. | No |
Theorem 75.3. If \( X \) is the \( n \) -fold connected sum of tori, then \( {H}_{1}\left( X\right) \) is a free abelian group of rank \( {2n} \) . | Proof. In view of the preceding corollary, Theorem 74.3 implies that \( {H}_{1}\left( X\right) \) is isomorphic to the quotient of the free abelian group \( {F}^{\prime } \) on the set \( {\alpha }_{1},{\beta }_{1},\ldots ,{\alpha }_{n},{\beta }_{n} \) by the subgroup generated by the element \( \left\lbrack {{\alpha }... | Yes |
Theorem 75.4. If \( X \) is the \( m \) -fold connected sum of projective planes, then the torsion subgroup \( T\left( X\right) \) of \( {H}_{1}\left( X\right) \) has order 2, and \( {H}_{1}\left( X\right) /T\left( X\right) \) is a free abelian group of rank \( m - 1 \) . | Proof. In view of the preceding corollary, Theorem 74.4 implies that \( {H}_{1}\left( X\right) \) is isomorphic to the quotient of the free abelian group \( {F}^{\prime } \) on the set \( {\alpha }_{1},\ldots ,{\alpha }_{m} \) by the subgroup generated by \( {\left( {\alpha }_{1}\right) }^{2}\cdots {\left( {\alpha }_{m... | Yes |
Lemma 77.1. Let \( w \) be a proper scheme of the form\n\n\[ w = \left\lbrack {y}_{0}\right\rbrack a\left\lbrack {y}_{1}\right\rbrack a\left\lbrack {y}_{2}\right\rbrack \]\n\nwhere some of the \( {y}_{i} \) may be empty. Then one has the equivalence\n\n\[ w \sim {aa}\left\lbrack {{y}_{0}{y}_{1}^{-1}{y}_{2}}\right\rbrac... | Proof. Step 1. We first consider the case where \( {y}_{0} \) is empty. We show that\n\n\[ a\left\lbrack {y}_{1}\right\rbrack a\left\lbrack {y}_{2}\right\rbrack \sim {aa}\left\lbrack {{y}_{1}^{-1}{y}_{2}}\right\rbrack \]\n\nIf \( {y}_{1} \) is empty, this result is immediate, while if \( {y}_{2} \) is empty, it follows... | No |
Corollary 77.2. If \( w \) is a scheme of projective type, then \( w \) is equivalent to a scheme of the same length having the form\n\n\[ \left( {{a}_{1}{a}_{1}}\right) \left( {{a}_{2}{a}_{2}}\right) \cdots \left( {{a}_{k}{a}_{k}}\right) {w}_{1} \]\n\nwhere \( k \geq 1 \) and \( {w}_{1} \) is either empty or of torus ... | Proof. The scheme \( w \) can be written in the form\n\n\[ w = \left\lbrack {y}_{0}\right\rbrack a\left\lbrack {y}_{1}\right\rbrack a\left\lbrack {y}_{2}\right\rbrack \]\n\nthen the preceding lemma implies that \( w \) is equivalent to a scheme of the form \( {w}^{\prime } = \) \( {aa}{w}_{1} \) that has the same lengt... | Yes |
Lemma 77.3. Let \( w \) be a proper scheme of the form \( w = {w}_{0}{w}_{1} \), where \( {w}_{1} \) is a scheme of torus type that does not contain two adjacent terms having the same label. Then \( w \) is equivalent to a scheme of the form \( {w}_{0}{w}_{2} \), where \( {w}_{2} \) has the same length as \( {w}_{1} \)... | Proof. This is the most elaborate proof of this section; three cuttings and pastings are involved. We show first that, switching labels and exponents if necessary, \( w \) can be written in the form\n\n(*)\n\n\[ \nw = {w}_{0}\left\lbrack {y}_{1}\right\rbrack a\left\lbrack {y}_{1}\right\rbrack b\left\lbrack {y}_{3}\righ... | Yes |
Lemma 77.4. Let \( w \) be a proper scheme of the form\n\n\[ w = {w}_{0}\left( {cc}\right) \left( {{ab}{a}^{-1}{b}^{-1}}\right) {w}_{1}. \]\n\nThen \( w \) is equivalent to the scheme\n\n\[ {w}^{\prime } = {w}_{0}\left( {aabbcc}\right) {w}_{1}. \] | Proof. Recall Lemma 77.1, which states that for proper schemes we have\n\n(*) \n\n\[ \left\lbrack {y}_{0}\right\rbrack a\left\lbrack {y}_{1}\right\rbrack a\left\lbrack {y}_{2}\right\rbrack \sim {aa}\left\lbrack {{y}_{0}{y}_{1}^{-1}{y}_{2}}\right\rbrack . \]\n\nWe proceed as follows:\n\n\[ w \sim \left( {cc}\right) \lef... | Yes |
Theorem 78.1. If \( X \) is a compact triangulable surface, then \( X \) is homeomorphic to the quotient space obtained from a collection of disjoint triangular regions in the plane by pasting their edges together in pairs. | Proof. Let \( {A}_{1},\ldots ,{A}_{n} \) be a triangulation of \( X \), with corresponding homeomorphisms \( {h}_{t} : {T}_{i} \rightarrow {A}_{i} \) . We assume the triangles \( {T}_{i} \) are disjoint; then the maps \( {h}_{i} \) combine to define a map \( h : E = {T}_{1} \cup \cdots \cup {T}_{n} \rightarrow X \) tha... | Yes |
Theorem 78.2. If \( X \) is a compact connected triangulable surface, then \( X \) is homeomorphic to a space obtained from a polygonal region in the plane by pasting the edges together in pairs. | Proof. It follows from the preceding theorem that there is a collection \( {T}_{1},\ldots ,{T}_{n} \) of triangular regions in the plane, and orientations and a labelling of the edges of these regions, where each label appears exactly twice in the total labelling scheme, such that \( X \) is homeomorphic to the quotien... | Yes |
Lemma 79.1 (The general lifting lemma). Let \( p : E \rightarrow B \) be a covering map; let \( p\left( {e}_{0}\right) = {b}_{0} \). Let \( f : Y \rightarrow B \) be a continuous map, with \( f\left( {y}_{0}\right) = {b}_{0} \). Suppose \( Y \) is path connected and locally path connected. The map \( f \) can be lifted... | Proof. If the lifting \( \widetilde{f} \) exists, then \[ {f}_{ * }\left( {{\pi }_{1}\left( {Y,{Y}_{0}}\right) }\right) = {p}_{ * }\left( {{\bar{f}}_{ * }\left( {{\pi }_{1}\left( {Y,{y}_{0}}\right) }\right) }\right) \subset {p}_{ * }\left( {{\pi }_{1}\left( {E,{e}_{0}}\right) }\right) . \] This proves the \ | No |
Theorem 79.2. Let \( p : E \rightarrow B \) and \( {p}^{\prime } : {E}^{\prime } \rightarrow B \) be covering maps; let \( p\left( {e}_{0}\right) = \) \( {p}^{\prime }\left( {e}_{0}^{\prime }\right) = {b}_{0} \) . There is an equivalence \( h : E \rightarrow {E}^{\prime } \) such that \( h\left( {e}_{0}\right) = {e}_{0... | Proof. We prove the \ | No |
Lemma 79.3. Let \( p : E \rightarrow B \) be a covering map. Let \( {e}_{0} \) and \( {e}_{1} \) be points of \( {p}^{-1}\left( {b}_{0}\right) \) , and let \( {H}_{i} = {p}_{ * }\left( {{\pi }_{1}\left( {E,{e}_{i}}\right) }\right) \). (a) If \( \gamma \) is a path in \( E \) from \( {e}_{0} \) to \( {e}_{1} \), and \( ... | Proof. (a) First, we show that \( \left\lbrack \alpha \right\rbrack * {H}_{1} * {\left\lbrack \alpha \right\rbrack }^{-1} \subset {H}_{0} \). Given an element \( \left\lbrack h\right\rbrack \) of \( {H}_{1} \), we have \( \left\lbrack h\right\rbrack = {p}_{ * }\left( \left\lbrack \bar{h}\right\rbrack \right) \) for som... | Yes |
Theorem 79.4. Let \( p : E \rightarrow B \) and \( {p}^{\prime } : {E}^{\prime } \rightarrow B \) be covering maps; let \( p\left( {e}_{0}\right) = \) \( {p}^{\prime }\left( {e}_{0}^{\prime }\right) = {b}_{0} \) . The covering maps \( p \) and \( {p}^{\prime } \) are equivalent if and only if the subgroups\n\n\[ \n{H}_... | Proof. If \( h : E \rightarrow {E}^{\prime } \) is an equivalence, let \( {e}_{1}^{\prime } = h\left( {e}_{0}\right) \), and let \( {H}_{1}^{\prime } = {p}_{ * }\left( {{\pi }_{1}\left( {{E}^{\prime },{e}_{1}^{\prime }}\right) }\right) \) Theorem 79.2 implies that \( {H}_{0} = {H}_{1}^{\prime } \), while the preceding ... | Yes |
Lemma 80.1. Let \( B \) be path connected and locally path connected. Let \( p : E \rightarrow B \) be a covering map in the former sense (so that \( E \) is not required to be path connected). If \( {E}_{0} \) is a path component of \( E \), then the map \( {p}_{0} : {E}_{0} \rightarrow B \) obtained by restricting \(... | Proof. We first show \( {p}_{0} \) is surjective. Since the space \( E \) is locally homeomorphic to \( B \), it is locally path connected. Therefore \( {E}_{0} \) is open in \( E \) . It follows that \( p\left( {E}_{0}\right) \) is open in \( B \) . We show that \( p\left( {E}_{0}\right) \) is also closed in \( B \), ... | Yes |
(a) If \( p \) and \( r \) are covering maps, so is \( q \) . | Proof. By our convention, \( X, Y \), and \( Z \) are path connected and locally path connected. Let \( {x}_{0} \in X \) ; set \( {y}_{0} = q\left( {x}_{0}\right) \) and \( {z}_{0} = p\left( {x}_{0}\right) \) .\n\n(a) Assume that \( p \) and \( r \) are covering maps. We show first that \( q \) is surjective. Given \( ... | Yes |
Theorem 80.3. Let \( p : E \rightarrow B \) be a covering map, with \( E \) simply connected. Given any covering map \( r : Y \rightarrow B \), there is a covering map \( q : E \rightarrow Y \) such that \( r \circ q = p \) . | Proof. Let \( {b}_{0} \in B \) ; choose \( {e}_{0} \) and \( {y}_{0} \) so that \( p\left( {e}_{0}\right) = {b}_{0} \) and \( r\left( {y}_{0}\right) = {b}_{0} \) . We apply Lemma 79.1 to construct \( q \) . The map \( r \) is a covering map, and the condition\n\n\[ \n{p}_{ * }\left( {{\pi }_{1}\left( {E,{e}_{0}}\right)... | Yes |
Lemma 80.4. Let \( p : E \rightarrow B \) be a covering map; let \( p\left( {e}_{0}\right) = {b}_{0} \). If \( E \) is simply connected, then \( {b}_{0} \) has a neighborhood \( U \) such that inclusion \( i : U \rightarrow B \) induces the trivial homomorphism\n\n\[ \n{i}_{ * } : {\pi }_{1}\left( {U,{b}_{0}}\right) \r... | Proof. Let \( U \) be a neighborhood of \( {b}_{0} \) that is evenly covered by \( p \) ; break \( {p}^{-1}\left( U\right) \) up into slices; let \( {U}_{\alpha } \) be the slice containing \( {e}_{0} \). Let \( f \) be a loop in \( U \) based at \( {b}_{0} \). Because \( p \) defines a homeomorphism of \( {U}_{\alpha ... | Yes |
Lemma 81.1. The image of the map \( \Psi \) equals the image under \( \Phi \) of the subgroup \( N\left( {H}_{0}\right) /{H}_{0} \) of \( {\pi }_{1}\left( {B,{b}_{0}}\right) /{H}_{0} \) . | Proof. Recall that the lifting correspondence \( \phi : {\pi }_{1}\left( {B,{b}_{0}}\right) \rightarrow F \) is defined as follows: Given a loop \( \alpha \) in \( B \) at \( {b}_{0} \), let \( \gamma \) be its lift to \( E \) beginning at \( {e}_{0} \) ; let \( {e}_{1} = \gamma \left( 1\right) \) ; and define \( \phi ... | Yes |
Theorem 81.2. The bijection\n\n\[ \n{\Phi }^{-1} \circ \Psi : \mathcal{C}\left( {E, p, B}\right) \rightarrow N\left( {H}_{0}\right) /{H}_{0} \n\]\n\nis an isomorphism of groups. | Proof. We need only show that \( {\Phi }^{-1} \circ \Psi \) is a homomorphism. Let \( h, k : E \rightarrow E \) be covering transformations. Let \( h\left( {e}_{0}\right) = {e}_{1} \) and \( k\left( {e}_{0}\right) = {e}_{2} \) ; then\n\n\[ \n\Psi \left( h\right) = {e}_{1}\;\text{ and }\;\Psi \left( k\right) = {e}_{2}, ... | No |
Corollary 81.3. The group \( {H}_{0} \) is a normal subgroup of \( {\pi }_{1}\left( {B,{b}_{0}}\right) \) if and only if for every pair of points \( {e}_{1} \) and \( {e}_{2} \) of \( {p}^{-1}\left( {b}_{0}\right) \), there is a covering transformation \( h : E \rightarrow \) \( E \) with \( h\left( {e}_{1}\right) = {e... | \[ {\Phi }^{-1} \circ \Psi : \mathcal{C}\left( {E, p, B}\right) \rightarrow {\pi }_{1}\left( {B,{b}_{0}}\right) /{H}_{0}. \] | No |
Corollary 81.4. Let \( p : E \rightarrow B \) be a covering map. If \( E \) is simply connected, then | \[ \mathcal{C}\left( {E, p, B}\right) \cong {\pi }_{1}\left( {B,{b}_{0}}\right) \] | No |
Theorem 81.5. Let \( X \) be path connected and locally path connected; let \( G \) be a group of homeomorphisms of \( X \) . The quotient map \( \pi .X \rightarrow X/G \) is a covering map if and only if the action of \( G \) is properly discontinuous. In this cases, the covering map \( \pi \) is regular and \( G \) i... | Proof. We show \( \pi \) is an open map. If \( U \) is open in \( X \), then \( {\pi }^{-1}\pi \left( U\right) \) is the union of the open sets \( g\left( U\right) \) of \( X \), for \( g \in G \) . Hence \( {\pi }^{-1}\pi \left( U\right) \) is open in \( X \), so that \( \pi \left( U\right) \) is open in \( X/G \) by ... | Yes |
Theorem 81.6. If \( p : X \rightarrow B \) is a regular covering map and \( G \) is its group of covering transformations, then there is a homeomorphism \( k : X/G \rightarrow B \) such that \( p = k \circ \pi \), where \( \pi : X \rightarrow X/G \) is the projection. | Proof. If \( g \) is a covering transformation, then \( p\left( {g\left( x\right) }\right) = p\left( x\right) \) by definition. Hence \( p \) is constant on each orbit, so it induces a continuous map \( k \) of the quotient space \( X/G \) into \( B \) . On the other hand, \( p \) is a quotient map because it is contin... | Yes |
Lemma 83.2. Let \( X \) be a linear graph. If \( C \) is a compact subspace of \( X \), there exists a finite subgraph \( Y \) of \( X \) that contains \( C \) . If \( C \) is connected, \( Y \) can be chosen to be connected. | Proof. First, note that \( C \) contains only finitely many vertices of \( X \) . For \( C \cap {X}^{0} \) is a closed discrete subspace of the compact space \( C \) ; since it has no limit point, it must be finite. Similarly, there are only finitely many values of \( \alpha \) for which \( C \) contains an interior po... | Yes |
Lemma 84.1. A graph \( X \) is connected if and only if every pair of vertices of \( X \) can be joined by an edge path in \( X \) . | Proof. Suppose \( X \) is connected. Define \( x \sim y \) if there is an edge path in \( X \) from \( x \) to \( y \) . For any edge of \( X \), its end points belong to the same equivalence class; let \( {Y}_{x} \) denote the union of all edges whose end points are equivalent to \( x \) . Then \( {Y}_{x} \) is a subg... | Yes |
Lemma 84.2. If \( T \) is a tree in \( X \), and if \( A \) is an edge of \( X \) that intersects \( T \) in a single vertex, then \( T \cup A \) is a tree in \( X \) . Conversely, if \( T \) is a finite tree in \( X \) that consists of more than one edge, then there is a tree \( {T}_{0} \) in \( X \) and an edge \( A ... | Proof. Suppose \( T \) is a tree in \( X \) and \( A \) is an edge that intersects \( T \) in a single vertex. Clearly \( T \cup A \) is connected; we show it contains no closed reduced edge paths. Let \( a \) and \( b \) be the end points of \( A \), with \( \{ a\} = T \cap A \) . See Figure 84.3. Suppose \( {x}_{0},\... | Yes |
Theorem 84.4. Let \( X \) be a connected graph. A tree \( T \) in \( X \) is maximal if and only if it contains all the vertices of \( X \) . | Proof. Suppose \( T \) is a tree in \( X \) that contains all the vertices of \( X \) . If \( Y \) is a subgraph of \( X \) that properly contains \( T \), we show that \( Y \) contains a closed reduced edge path; it follows that \( T \) is maximal. Let \( A \) be an edge of \( Y \) that is not in \( T \) ; by hypothes... | Yes |
Theorem 84.5. If \( X \) is a linear graph, every tree \( {T}_{0} \) in \( X \) is contained in a maximal tree in \( X \) . | Proof. We apply Zom’s lemma to the collection \( \mathcal{T} \) of all trees in \( X \) that contain \( {T}_{0} \) , strictly partially ordered by proper inclusion. To show this collection has a maximal element, we need only prove the following:\n\nIf \( {\mathcal{T}}^{\prime } \) is a subcollection of \( \mathcal{T} \... | Yes |
Lemma 84.6. Suppose \( X = U \cup V \), where \( U \) and \( V \) are open sets of \( X \). Suppose that \( U \cap V \) is the union of two disjoint open path-connected sets \( A \) and \( B \), that \( \alpha \) is a path in \( U \) from the point \( a \) of \( A \) to the point \( b \) of \( B \), and that \( \beta \... | Proof. The situation is similar to that of Theorem 59.1, except that \( U \cap V \) has two path components instead of one. The proof is also similar.\n\nLet \( f \) be a loop in \( X \) based at \( a \). Choose a subdivision \( 0 = {a}_{0} < {a}_{1} < \cdots < {a}_{n} = \) 1 of \( \left\lbrack {0,1}\right\rbrack \) su... | Yes |
Theorem 85.1. If \( H \) is a subgroup of a free group \( F \), then \( H \) is free. | Proof. Let \( \{ \alpha \mid \alpha \in J\} \) be a system of free generators for \( F \) . Let \( X \) be a wedge of circles \( {S}_{\alpha } \), one for each \( \alpha \in J \) ; let \( {x}_{0} \) be their common point. We can give \( X \) the structure of a linear graph by breaking each circle \( {S}_{\alpha } \) in... | Yes |
Lemma 85.2. If \( X \) is a finite, connected linear graph, then the cardinality of a system of free generators for the fundamental group of \( X \) is \( 1 - \chi \left( X\right) \) . | Proof. Step I. We first show that for any finite tree \( T \), we have \( \chi \left( T\right) = 1 \) . We proceed by induction on the number \( n \) of edges in \( T \) . If \( n = 1 \), then \( T \) has one edge and two vertices, so \( \chi \left( T\right) = 1 \) . If \( n > 1 \), we can write \( T = {T}_{0} \cup A \... | Yes |
Theorem 85.3. Let \( F \) be a free group with \( n + 1 \) free generators; let \( H \) be a subgroup of \( F \). If \( H \) has index \( k \) in \( F \), then \( H \) has \( {kn} + 1 \) free generators. | Proof. We apply the construction given in the proof of Theorem 85.1. We can assume that \( F = {\pi }_{1}\left( {X,{x}_{0}}\right) \), where \( X \) is a linear graph whose underlying space is a wedge of \( n + 1 \) circles. Given \( H \), we choose a path-connected covering space \( p : E \rightarrow X \) such that \(... | Yes |
Proposition 1.2.2. The union of two closed sets is closed. | Proof: Let the two closed sets be \( E \) and \( F \) . Then \( X \smallsetminus E \) and \( X \smallsetminus F \) are open. So\n\n\[ S \equiv \left( {X \smallsetminus E}\right) \cap \left( {X \smallsetminus F}\right) \]\n\nis open. But then\n\n\[ {}^{c}S \equiv X \smallsetminus S = E \cup F \]\n\nis closed. | Yes |
Proposition 1.2.3. Let \( {\left\{ {E}_{\beta }\right\} }_{\beta \in B} \) be closed sets. Then \( { \cap }_{\beta }{E}_{\beta } \) is also closed. | Proof: Exercise for the reader. | No |
Lemma 1.2.4. Let \( S \) be a set in a topological space \( X \) . If each point \( s \in S \) has a neighborhood that lies in \( S \), then \( S \) is open. | Proof: Let \( s \in S \) and let \( {U}_{s} \) be the neighborhood of \( s \) that lies in \( S \) . We have\n\n\[ S = \mathop{\bigcup }\limits_{{s \in S}}\{ s\} \subseteq \mathop{\bigcup }\limits_{{s \in S}}{U}_{s} \subseteq S \]\n\nWe conclude that\n\n\[ \mathop{\bigcup }\limits_{{s \in S}}{U}_{s} = S \]\n\nBut the s... | Yes |
Proposition 1.2.5. The interior of any set \( S \) is open. | Proof: If \( p \in S \) and \( U \) is a neighborhood of \( p \) that lies inside \( S \) then any point \( x \) of \( U \) is also in the interior of \( S \) . For \( U \) will be the required neighborhood of \( x \) that lies in \( S \) . | Yes |
Proposition 1.2.6. The boundary of any set \( S \) is closed. | Proof: Let \( x \) be a point that is not in \( \partial S \), the boundary of \( S \) . Then there is some neighborhood \( U \) of \( x \) that does not intersect both \( S \) and \( {}^{c}S \) . It follows that any point \( t \in U \) also has such a neighborhood, namely \( U \) itself. So \( U \) lies in the complem... | Yes |
Proposition 1.2.7. The set \( \bar{S} \) equals the union of \( S \) and \( \partial S \) . | Proof: Suppose that \( x \) is a point that is not in \( S \cup \partial S \) . Since \( x \) is not in \( \partial S \) there is a neighborhood \( U \) of \( x \) that either does not intersect \( S \) or does not intersect \( {}^{c}S \) . We know that \( x \notin S \), so it must be that \( U \subseteq {}^{c}S \) . S... | Yes |
Proposition 1.2.8. Let \( \left( {X,\mathcal{U}}\right) \) be a topological space and \( S \subseteq X \) . Then the closure of \( S \) equals the union of \( S \) and its accumulation points. | Proof: Exercise for the reader. Imitate the method of Proposition 1.2.7. | No |
Proposition 1.3.3. On the real line, the definition of continuity in Remark 1.3.2 is equivalent to Definition 1.3.1 (formulated in the language of inverse images of open sets). | Proof: Suppose that \( I \) is an interval in \( \mathbb{R} \) and \( f : I \rightarrow \mathbb{R} \) . Assume that \( f \) is continuous according to the classical definition in 1.3.2. Let \( V \) be an open subset of \( \mathbb{R} \) ; we must show that \( {f}^{-1}\left( V\right) \) is open. Consider the set \( {f}^{... | Yes |
Theorem 1.4.1. Let \( \left( {X,\mathcal{U}}\right) \) be a normal space and let \( E \) and \( F \) be disjoint, closed sets in \( X \) . Then there is a continuous function \( f : X \rightarrow \left\lbrack {0,1}\right\rbrack \) such that \( f\left( E\right) = \{ 0\} \) and \( f\left( F\right) = \{ 1\} \) (that is, t... | Proof of the Theorem: By normality, there are disjoint open sets \( U \) and \( V \) such that \( E \subseteq U \) and \( F \subseteq V \) . For technical (and also traditional) reasons we shall denote this set \( U \) by \( {U}_{1/2} \) . Now we see that \( E \) and \( X \smallsetminus {U}_{1/2} \) are closed and disj... | Yes |
Proposition 1.4.3. A regular, Lindelöf space is normal. | Proof: Let \( X \) be as in the hypothesis, and let \( E, F \) be disjoint, closed sets in \( X \) . For each point \( e \in E \), let \( {U}_{e} \) be an open set containing \( e \) such that \( {\bar{U}}_{e} \cap F = \varnothing \) . The set exists by the regularity hypothesis. Similarly, for each \( f \in F \), we f... | Yes |
Proposition 1.4.5 (Tychanoff). Every Tychanoffspace \( X \) can be embedded as a subspace of a (possibly infinite-dimensional) cube. | Proof: Let \( C \) denote the family of all continuous real functions from \( X \) to \( I \), the closed unit interval. The complete regularity tells us that \( C \) can distinguish points from closed sets in \( X \) . The \( {\mathbf{T}}_{\mathbf{1}} \) property tells us that every singleton set is closed, hence \( C... | Yes |
Proposition 1.5.1. Let \( K \) be a compact set in a Hausdorff (i.e., \( {\mathbf{T}}_{\mathbf{2}} \) ) space and let \( x \) be a point that is not in \( K \) . Then there are disjoint open sets \( U \) and \( V \) such that \( U \supseteq K \) and \( V \ni x \) . | Proof: This proof is a nice illustration of how compactness works. By the Hausdorff property, for each point \( k \) in \( K \), there is a neighborhood \( {U}_{k} \) of \( k \) and a neighborhood \( {V}_{k} \) of \( x \) such that \( {U}_{k} \cap {V}_{x} = \varnothing \) . The sets \( \left\{ {U}_{k}\right\} \) form a... | Yes |
Proposition 1.5.3. A compact set in a Hausdorff space is closed. | Proof: Let \( K \) be a compact set and \( x \) a point that is not in \( K \) . By the preceding proposition, there is a neighborhood \( U \) of \( x \) that is disjoint from \( K \) . That shows that the complement of \( K \) is open. So \( K \) is closed. | Yes |
Proposition 1.5.4. Let \( f : X \rightarrow Y \) be a continuous mapping of topological spaces. If \( K \subseteq X \) is compact then \( f\left( K\right) \equiv \{ f\left( k\right) : k \in K\} \) is compact. | Proof: Let \( W = {\left\{ {W}_{\alpha }\right\} }_{\alpha \in A} \) be an open covering of \( f\left( K\right) \) . Then, since \( f \) is continuous, \( {\left\{ {f}^{-1}\left( {W}_{\alpha }\right) \right\} }_{\alpha \in A} \) is an open covering of \( K \) . Therefore there is a finite subcovering \( {f}^{-1}\left( ... | Yes |
Theorem 1.5.5 (Heine-Borel). A set \( E \subseteq \mathbb{R} \) is compact if and only if it is closed and bounded. | Proof: If the set is closed and bounded then compactness follows precisely as in the proof of Example 1.5.5. We leave the details to the reader. Now suppose that \( E \subseteq \mathbb{R} \) is compact. Since \( \mathbb{R} \) is Hausdorff, we can be sure by Proposition 1.5.3 that \( E \) is closed. It remains to show t... | No |
Proposition 1.5.6. Let \( X \) be a topological space. A closed subset of a compact set in \( X \) is compact. | Proof: Let \( K \) be a compact set and \( E \subseteq K \) a closed subset. Let \( W = \) \( {\left\{ {W}_{\alpha }\right\} }_{\alpha \in A} \) be an open covering of \( E \) . Let \( {W}^{\prime } = X \smallsetminus E \) . Then \( {W}^{\prime } \) is open and \( \mathcal{W} \cup \left\{ {W}^{\prime }\right\} \) is an... | Yes |
Proposition 1.5.7. A one-to-one, continuous map from a compact space \( X \) onto a Hausdorff space \( Y \) is bicontinuous. | Proof: Let \( f \) be such a map. Let \( U \) be an open subset of \( X \) . Then \( E = X \smallsetminus \) \( U \) is closed. Hence, by 1.5.6, it is compact. It follows from 1.5.4 that \( f\left( E\right) \) is compact in \( Y \) . So, by 1.5.3, \( f\left( E\right) \) is closed. But then \( f\left( U\right) = Y \smal... | Yes |
Theorem 1.5.8. A topological space \( \left( {X,\mathcal{U}}\right) \) is compact if and only if any family \( \mathcal{F} = {\left\{ {F}_{\alpha }\right\} }_{\alpha \in A} \) of closed sets in \( X \) with the finite intersection property actually satisfies \( { \cap }_{\alpha \in A}{F}_{\alpha } \neq \varnothing \) . | Proof: First suppose that \( X \) is compact. Let \( \mathcal{F} = {\left\{ {F}_{\alpha }\right\} }_{\alpha \in A} \) be a family of closed sets in \( X \) and suppose that \( { \cap }_{\alpha \in A}{F}_{\alpha } = \varnothing \) . Now look at \( \left\{ {X \smallsetminus {F}_{\alpha }}\right\} \) . This must (by De Mo... | Yes |
Proposition 1.7.1. Let \( f : X \rightarrow Y \) be a continuous mapping. Let \( E \subseteq X \) be connected. Then \( f\left( E\right) \) is connected. | Proof: Suppose to the contrary that \( f\left( E\right) \) is disconnected. Write \( f\left( E\right) = \) \( A \cup B \) (both nonempty) with disjoint open sets \( U \) and \( V \) so that \( A \subseteq U \) and \( B \subseteq V \) . Then \( {f}^{-1}\left( U\right) \) and \( {f}^{-1}\left( V\right) \) are disjoint, n... | Yes |
Proposition 1.7.2. Let \( \left( {X,\mathcal{U}}\right) \) be a topological space. If \( A \) and \( B \) in \( X \) are connected sets with a common point \( p \) then \( A \cup B \) is connected. | Proof: Suppose not. Say that the disjoint open sets \( U \) and \( V \) disconnect \( A \cup B \) . Then \( p \) must lie in one of these two open sets. Say that it lies in \( U \) . Since \( A \) cannot be disconnected, it follows that \( A \subseteq U \) . A similar argument shows that \( B \subseteq U \) . Thus \( A... | Yes |
Proposition 1.8.1. Let \( \left( {X,\mathcal{U}}\right) \) be a topological space. If \( X \) is path-connected then \( X \) is connected. | Proof: Suppose to the contrary that \( X \) is disconnected. So there are disjoint open sets \( U, V \) that disconnect \( X \) . Let \( P \) be a point of \( U \cap X \) and \( Q \) be a point of \( V \cap X \) and \( \gamma : \left\lbrack {0,1}\right\rbrack \rightarrow X \) a path that connects them. Then \( {\gamma ... | Yes |
Proposition 1.8.2. Every connected open set \( U \) in Euclidean space is path-connected. | Proof: Fix a point \( P \in U \) . Define\n\n\( S = \{ u \in U : \) the point \( u \) can be connected to \( P \) by a path \( \} . \)\n\n\n\nFIGURE 1.14. Connected open sets in Euclidean space are path-connected.\n\nT... | Yes |
Lemma 1.9.1. If \( K \) is a metric space that is a continuum, and if \( K \) has exactly two noncut points, then \( K \) is homeomorphic to the unit interval. | Proof: Although this result is intuitively appealing, it is remarkably tricky to prove (relying as it does on the construction of the real numbers and other subtle ideas). We refer the reader to [WIL, pp. 206-207] for the details. | No |
Proposition 1.10.1. The product of totally disconnected spaces is totally disconnected. Also every subspace of a totally disconnected space is totally disconnected. | Proof: Let \( {X}_{1} \) and \( {X}_{2} \) be totally disconnected and let \( {\pi }_{j} \) be the projection from \( X = {X}_{1} \times {X}_{2} \) to \( {X}_{j} : {\pi }_{j}\left( {{x}_{1},{x}_{2}}\right) = {x}_{j} \) . Of course the projection is a continuous mapping. If \( S \) is a nonempty, connected subset of \( ... | Yes |
Lemma 1.10.2. A nonempty subset \( S \) of a 0-dimensional space \( X \) is 0- dimensional. | Proof: The set \( S \) clearly has dimension at most 0 . It is not empty, so does not have dimension -1 . So the dimension must be 0 . | No |
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