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Corollary 2.4.5. Let \( H \) be a subgroup of a free group \( G \) . Then \( H \) is a free group. | Proof. Consider a rose \( R \) whose edges are in 1-1 correspondence with the generators of \( G \) . By Theorem 2.2.19, there is a cover \( \widetilde{R} \) with \( {\pi }_{1}\left( {\widetilde{R},\widetilde{v}}\right) = H \) . But it is easy to see that, since a covering projection \( p : \widetilde{R} \rightarrow R ... | Yes |
Corollary 2.4.6. Let \( G \) be a free group on \( k \) elements and let \( H \) be a subgroup of \( G \) of index \( n \) . Then \( H \) is a free group on \( \left( {k - 1}\right) n + 1 \) elements. | Proof. We may consider \( G \) to be the fundamental group \( {\pi }_{1}\left( {R, v}\right) \), where \( R \) is a \( k \) - leafed rose. Then \( H \) is the fundamental group of an \( n \) -fold cover \( {\pi }_{1}\left( {\widetilde{R},\widetilde{v}}\right) \) . Now \( R \) has 1 vertex and \( k \) edges, so \( \wide... | Yes |
Theorem 2.5.1. Let \( X \) be a path connected space. Then there is a 1-1 correspondence between conjugacy classes of elements of \( {\pi }_{1}\left( {X,{x}_{0}}\right) \) and \( \pi \left( X\right) = \) \( \left\{ \right. \) homotopy classes of maps \( \left. {{S}^{1} \rightarrow X}\right\} \) . | Proof. Let \( \Phi : {\pi }_{1}\left( {X,{x}_{0}}\right) \rightarrow \pi \left( X\right) \) be the map given by \ | No |
Let \( X \) be the subspace of \( {\mathbb{R}}^{2} \) consisting of the closed line segments joining the points \( \left( {1/n,0}\right) \) to \( \left( {0,1}\right) \) for each positive integer \( n \), and also the closed line segment joining \( \left( {0,0}\right) \) to \( \left( {0,1}\right) \) . Give \( X \) the t... | Let \( A \) be the subspace of \( X \) consisting of the closed line segment joining \( \left( {0,0}\right) \) to \( \left( {0,1}\right) \) . Then \( X \) and \( A \) are both contractible to the point \( \left( {0,1}\right) \) by a homotopy leaving that point fixed. It then follows that \( A \) is a deformation retrac... | No |
Theorem 3.2.2. Let \( f : \\left( {X, A}\\right) \\rightarrow \\left( {Y, B}\\right) \) be a map of pairs and suppose that both \( f \) : \( X \\rightarrow Y \) and \( f \\mid A : A \\rightarrow B \) are homotopy equivalences. Then \( {f}_{i} : {H}_{i}\\left( {X, A}\\right) \\rightarrow {H}_{i}\\left( {Y, B}\\right) \)... | Proof. We have the commutative diagram of exact sequences:\n\n\n\nThe first, second, fourth, and fifth vertical arrows are isomorphisms. Hence, by Lemma A.1.8, so is the third. | Yes |
Theorem 3.2.4. Let \( X \) be a nonempty space and let \( {x}_{0} \) be an arbitrary point of \( X \) . Then for each \( i \) ,\n\n(i) \( {\widetilde{H}}_{i}\left( X\right) \cong {H}_{i}\left( {X,{x}_{0}}\right) \) ,\n\n(ii) \( {H}_{i}\left( X\right) \cong {H}_{i}\left( {x}_{0}\right) \oplus {\widetilde{H}}_{i}\left( X... | Proof. As \( {x}_{0} \) is a retract of \( X \), this is a special case of Lemma 3.2.1. | No |
Lemma 3.2.5. Let \( f : X \rightarrow Y \) be a map. Then \( f \) induces well-defined maps \( {\widetilde{f}}_{i} \) : \( {\widetilde{H}}_{i}\left( X\right) \rightarrow {\widetilde{H}}_{i}\left( Y\right) \) for every \( i \), where \( {\widetilde{f}}_{i} = {f}_{i} \mid {\widetilde{H}}_{i}\left( X\right) \) . | Proof. This follows immediately from the commutativity of the diagram\n\n | No |
Theorem 3.2.7. (1) Let \( A \) be a nonempty closed subset of \( X \). Suppose that \( \partial A \) has an open neighborhood \( C \) in \( A \) such that the inclusions \( \left( {X - A}\right) \cup C \rightarrow X - \operatorname{int}\left( A\right) \) and \( \partial A \rightarrow C \) are both homotopy equivalences... | Proof. (1) Let \( V = A - C \). Then \( V \) is a closed set in the interior of \( A \), so \( (X - V, A - V) \rightarrow \left( {X, A}\right) \) is excisive. But \( X - V = \left( {X - A}\right) \cup C \) and \( A - V = C \). By hypothesis the first of these is homotopy equivalent to \( X - \operatorname{int}\left( A\... | Yes |
Theorem 3.2.9. (1) Let \( A \) be a nonempty subset of \( X \) . Then for each \( i,{H}_{i}\left( {X, A}\right) \) is isomorphic to the reduced homology group \( {\widetilde{H}}_{i}\left( {X \cup {cA}}\right) \) . | Proof. (1) We follow the idea of the proof of Theorem 3.2.7. Let \( V = \{ \left( {a, s}\right) \mid s \geq \) \( \left. \frac{1}{2}\right\} \) so that \( V \) is a closed subset of \( X{ \cup }_{A}{cA} \) which is contained in the interior of \( \overline{cA} \) . Then the inclusion \( \left( {\left( {X{ \cup }_{A}{cA... | Yes |
Theorem 3.2.10 (Mayer-Vietoris). Let \( X = {X}_{1} \cup {X}_{2}, A = {X}_{1} \cap {X}_{2} \), and suppose that the inclusion \( \left( {{X}_{1}, A}\right) \rightarrow \left( {X,{X}_{2}}\right) \) is excisive. Then there is a long exact sequence in homology\n\n\[ \cdots \rightarrow {H}_{i}\left( A\right) \overset{\alph... | Proof. We have the long exact homology sequences\n\n\n\nwhere by assumption \( \varepsilon : {H}_{i}\left( {{X}_{1}, A}\right) \rightarrow {H}_{i}\left( {X,{X}_{2}}\right) \) is an isomorphism. Then the theorem follows... | Yes |
Theorem 3.2.11. Let \( \left( {X, A, B}\right) \) be a triad and suppose that the inclusion \( (A, A \cap B) \rightarrow \left( {A \cup B, B}\right) \) is excisive. Then there is an exact homology sequence\n\n\[ \cdots \rightarrow {H}_{i}\left( {X, A \cap B}\right) \rightarrow {H}_{i}\left( {X, A}\right) \oplus {H}_{i}... | Proof. Exactly the same as the proof of Theorem 3.2.10. | No |
Theorem 3.2.13. (1) For any space \( X \) there is an isomorphism, for any \( i \) ,\n\n\[ \sum : {\widetilde{H}}_{i + 1}\left( {\sum X}\right) \rightarrow {\widetilde{H}}_{i}\left( X\right) \] | Proof. We prove (1). We have the exact homology sequence of the pair \( \left( {{c}_{ + }X, X}\right) \) :\n\n\[ \cdots \rightarrow {H}_{i + 1}\left( {{c}_{ + }X, X}\right) \rightarrow {H}_{i}\left( X\right) \rightarrow {H}_{i}\left( {{c}_{ + }X}\right) \rightarrow \cdots \]\n\nNow \( {c}_{ + }X \) is contractible to t... | Yes |
Theorem 3.2.14. Let \( A \) and \( B \) be subspaces of \( X \) with \( B \subseteq A \) . Then there is an exact homology sequence\n\n\[ \n\cdots \rightarrow {H}_{i}\left( {A, B}\right) \rightarrow {H}_{i}\left( {X, B}\right) \rightarrow {H}_{i}\left( {X, A}\right) \overset{\partial }{ \rightarrow }{H}_{i - 1}\left( {... | Proof. We merely remark here that the boundary map in the sequence is the composition\n\n\[ \n{H}_{i}\left( {X, A}\right) \rightarrow {H}_{i - 1}\left( A\right) \rightarrow {H}_{i - 1}\left( {A, B}\right)\n\]\n\nOtherwise, the result follows directly from Theorem A.2.12. | No |
Theorem 3.2.15. Let \( X = {X}_{1} \cup {X}_{2}, A = {X}_{1} \cap {X}_{2} \), and suppose that the inclusion \( \left( {{X}_{1}, A}\right) \rightarrow \left( {X,{X}_{2}}\right) \) is excisive. Let \( B \) be an arbitrary subspace of \( A \) . Then there is a long exact sequence in homology | \[ \cdots \rightarrow {H}_{i}\left( {A, B}\right) \rightarrow {H}_{i}\left( {{X}_{1}, B}\right) \oplus {H}_{i}\left( {{X}_{2}, B}\right) \rightarrow {H}_{i}\left( {X, B}\right) \rightarrow {H}_{i - 1}\left( {A, B}\right) \rightarrow \cdots . \] | Yes |
Theorem 3.3.4. Let \( X = {X}_{1} \cup {X}_{2}, A = {X}_{1} \cap {X}_{2} \), and suppose that the inclusion \( \left( {{X}_{1}, A}\right) \rightarrow \left( {X,{X}_{2}}\right) \) is excisive. Then there is a long exact sequence in cohomology | \[ \cdots \leftarrow {H}^{i}\left( A\right) \leftarrow {H}^{i}\left( {X}_{1}\right) \oplus {H}^{i}\left( {X}_{2}\right) \leftarrow {H}^{i}\left( X\right) \leftarrow {H}^{i - 1}\left( A\right) \leftarrow \cdots . \] | Yes |
Lemma 4.1.1. 1. \( {H}_{0}\left( {S}^{0}\right) \cong \mathbb{Z} \oplus \mathbb{Z} \) . More precisely, \( {H}_{0}\left( {S}^{0}\right) = \{ {mp} + {nq} \mid m, n \in \mathbb{Z}\} \) . | Proof. (1) Since \( \{ - 1\} \) and \( \{ 1\} \) are distinct components of \( {S}^{0} \), we have by Lemma 3.2.1 that \( {H}_{i}\left( {S}^{0}\right) \cong {H}_{i}\left( {\{ - 1\} }\right) \oplus {H}_{i}\left( {\{ 1\} }\right) \) . Since \( \{ - 1\} \) and \( \{ 1\} \) are both spaces consisting of a single point, the... | Yes |
Lemma 4.1.3. Fix a positive integer \( n \) .\n\n1. \( {H}_{n}\left( {S}^{n}\right) \cong \mathbb{Z} \) and \( {H}_{0}\left( {S}^{n}\right) \cong \mathbb{Z} \) .\n\n2. \( {\widetilde{H}}_{n}\left( {S}^{n}\right) \cong \mathbb{Z} \) .\n\n3. \( {H}_{i}\left( {S}^{n}\right) = 0 \) for \( i \neq 0, n \) and \( {\widetilde{... | Proof. Observe that for any \( k,\sum {S}^{k} \) is homeomorphic to \( {S}^{k + 1} \) . Then, if \( {\sum }^{i} \) denotes \( \sum \) applied \( i \) times, \( {\sum }^{i}{S}^{k} \) is homeomorphic to \( {S}^{k + i} \) . In particular \( {\sum }^{n}{S}^{0} \) is homeomorphic to \( {S}^{n} \) . Then, by repeated applica... | Yes |
Lemma 4.1.5. For any \( n \geq 1 \), there does not exist a retraction from \( {D}^{n} \) onto \( {S}^{n - 1} \) . | Proof. If there were such a retraction \( r : {D}^{n} \rightarrow {S}^{n - 1} \), then \( r \) would induce a surjection \( {r}_{i} : {H}_{i}\left( {D}^{n}\right) \rightarrow {H}_{i}\left( {S}^{n - 1}\right) \) for each \( i \), by Lemma 3.2.1(iii). But for \( i = n - 1,{H}_{n - 1}\left( {D}^{n}\right) = 0 \) and \( {H... | Yes |
Theorem 4.1.6 (Brouwer fixed-point theorem). Let \( f : {D}^{n} \rightarrow {D}^{n} \) be an arbitrary map. Then \( f \) has a fixed point, i.e. there is an \( {x}_{0} \in D \) with \( f\left( {x}_{0}\right) = {x}_{0} \) . | Proof. Suppose that \( f \) does not have a fixed point. Let \( r : {D}^{n} \rightarrow {S}^{n - 1} \) be the map defined as follows:\n\nFor \( x \in {D}^{n} \), take the line segment from \( f\left( x\right) \) to \( x \) and prolong it until it intersects \( {S}^{n - 1} \) at some point \( {x}^{\prime } \) . Then set... | Yes |
Theorem 4.1.7 (Invariance of domain). Let \( U \) be a nonempty open set in \( {\mathbb{R}}^{n} \) and \( V \) be a nonempty open set in \( {\mathbb{R}}^{m} \) and suppose there is a homeomorphism \( f : U \rightarrow V \) . Then \( m = n \) . | Proof. This is trivially true if \( m = 0 \) or \( n = 0 \), so we assume \( m \geq 1 \) and \( n \geq 1 \) . Although from a logical standpoint it is not necessary to begin with this special case, the basic idea of the proof comes through most clearly if we first consider the case \( U = {\mathbb{R}}^{m}, V = {\mathbb... | Yes |
Lemma 4.1.9. Let \( a : {S}^{n} \rightarrow {S}^{n} \) be the antipodal map, i.e., \( a\left( {{x}_{1},\ldots ,{x}_{n + 1}}\right) = \) \( \left( {-{x}_{1},\ldots , - {x}_{n + 1}}\right) \) . Then the degree of a is \( {\left( -1\right) }^{n + 1} \) . | Proof. We divide the proof into two cases.\n\nCase \( 1 \) ( \( n \) is odd, \( n = {2m} - 1 \) ). Then we may regard \( {S}^{n} \) as the unit sphere in \( {\mathbb{C}}^{m} \) , and \( a : {S}^{n} \rightarrow {S}^{n} \) is \( a\left( {{z}_{1},\ldots ,{z}_{m}}\right) = \left( {-{z}_{1},\ldots , - {z}_{m}}\right) \) . B... | Yes |
Lemma 4.2.7. A CW-complex \( X \) has the following properties:\n\n1. (Closure-finiteness) The closure of each cell in \( X \) intersects only finitely many other cells in \( X \) .\n\n2. (Weak topology) A subset A of \( X \) is closed if and only if the intersection of \( A \) with the closure of every cell in \( X \)... | Proof. (1) The closure of each cell is \( f\left( {D}_{\lambda }^{n}\right) \), the image of a compact set, and hence compact, and if (1) were false this set would have an infinite subset (one point from each other cell) without an accumulation point, which is impossible. | No |
Lemma 4.2.10. Let \( X \) be obtained from \( A \) by adjoining an \( n \) -cell. Then\n\n\[ \n{H}_{i}\left( {X, A}\right) = \left\{ \begin{array}{ll} \mathbb{Z} & i = n \\ 0 & i \neq n \end{array}\right.\n\] | Proof. Let \( C = \left\{ {x \in {D}^{n}\left| \right| x \mid \geq 1/2}\right\} \) . Then \( C \) is a \ | No |
Lemma 4.2.11. 1. \( {H}_{i}\left( {{X}^{n},{X}^{n - 1}}\right) = 0 \) for \( i \neq n \) . | Proof. This is just an elaboration of Lemma 4.2.10.\n\nLet \( \left( {{D}^{n}\left( \frac{1}{2}\right) ,{S}^{n - 1}\left( \frac{1}{2}\right) }\right) \) be the pair consisting of the disk of radius \( \frac{1}{2} \) and its boundary. Then the inclusions induce isomorphisms on homology\n\n\[ {H}_{ * }\left( {{D}^{n}\lef... | Yes |
Lemma 4.2.14. \( {C}_{ * }^{\text{cell }}\left( X\right) \) is a chain complex. | Proof. We need only check that \( {\partial }_{n - 1}{\partial }_{n} = 0 \) . But this is the composition\n\n\[ \n{H}_{n}\left( {{X}^{n},{X}^{n - 1}}\right) \overset{\partial }{ \rightarrow }{H}_{n - 1}\left( {X}^{n - 1}\right) \rightarrow {H}_{n - 1}\left( {{X}^{n - 1},{X}^{n - 2}}\right) \]\n\n\[ \n\overset{\partial ... | Yes |
Lemma 4.2.16. The group \( {C}_{n}^{\text{cell }}\left( X\right) \) is the free abelian group on the n-cells of \( X \) . If \( {\alpha }_{\lambda }^{n} \) is the generator corresponding to the n-cell \( {D}_{\lambda }^{n},\lambda \in {\Lambda }_{n} \), then \( \partial \left( {\alpha }_{\lambda }^{n}\right) \) is give... | Proof. This follows directly from Lemma 4.2.11 and its proof. | No |
Theorem 4.2.20. Let \( X \) be a finite CW-complex. Then\n\n\[ \chi \left( X\right) = \mathop{\sum }\limits_{i}{\left( -1\right) }^{i} \cdot \text{ number of }i\text{-cells of }X. \] | Proof. Let \( X \) have \( {d}_{i}i \) -cells and suppose \( {d}_{i} = 0 \) for \( i > n \) . We have the cellular chain complex of \( X \)\n\n\[ 0 \rightarrow {C}_{n}^{\mathrm{{cell}}}\left( X\right) \rightarrow {C}_{n - 1}^{\mathrm{{cell}}}\left( X\right) \rightarrow \cdots \rightarrow {C}_{1}^{\mathrm{{cell}}}\left(... | Yes |
Theorem 4.2.23. Let \( X \) be a finite \( {CW} \) -complex. Let \( \widetilde{X} \) be an \( n \) -fold cover of \( X \) . Then \( \chi \left( \widetilde{X}\right) = {n\chi }\left( X\right) \) . | Proof. Given any cell decomposition of \( X \), we may refine it to obtain a cell decomposition so that every cell is evenly covered by the covering projection. Then the inverse image of every cell is \( n \) cells, so the theorem immediately follows from Theorem 4.2.20. | Yes |
Let \( R \) be a \( k \) -leafed rose, and let \( \widetilde{R} \) be any \( n \) -fold cover of \( R \) . Then \( R \) has one 0 -cell and \( {k1} \) -cells, so \( \chi \left( R\right) = 1 - k \) (which of course agrees with \( {H}_{0}\left( R\right) = \mathbb{Z} \) and \( {H}_{1}\left( \mathbb{R}\right) = {\mathbb{Z}... | Now \( {H}_{0}\left( \widetilde{R}\right) = \mathbb{Z} \) (as by definition, a cover is connected), so we must have\n\n\[ 1 - \operatorname{rank}{H}_{1}\left( \widetilde{R}\right) = n\left( {1 - k}\right) \]\n\nand hence \( {H}_{1}\left( \widetilde{R}\right) = {\mathbb{Z}}^{n\left( {k - 1}\right) + 1} \) . (Compare Cor... | Yes |
Lemma 4.2.26. Let \( f : X \rightarrow Y \) be a cellular map. Then for each \( i, f \) induces a map \( {f}_{i}^{\text{cell }} : {H}_{i}^{\text{cell }}\left( X\right) \rightarrow {H}_{i}^{\text{cell }}\left( Y\right) . | Proof. By hypothesis, \( f \) induces a map \( {H}_{i}\left( {{X}^{n},{X}^{n - 1}}\right) \rightarrow {H}_{i}\left( {{Y}^{n},{Y}^{n - 1}}\right) \) for each \( i \) and \( n \), and then it is easy to check this induces a map on cellular homology. | No |
Theorem 4.2.27. Let \( X \) and \( Y \) be \( {CW} \) -complexes and let \( f : X \rightarrow Y \) be a cellular map. Then the following diagram commutes: | Proof. This follows easily from the commutativity of the diagram where the vertical maps are all induced by \( f \) . | No |
Theorem 4.2.30. Let \( \\left( {X, A}\\right) \) be a CW-pair and let \( U \\subseteq A \) be such that \( \\left( {Y, B}\\right) = \) \( \\left( {X - U, A - U}\\right) \) is a CW-pair. Then the inclusion \( \\left( {Y, B}\\right) \\rightarrow \\left( {X, A}\\right) \) is excisive for cellular homology. | Proof. First observe that the hypothesis on \( U \) implies that \( U \) is a union of open cells of \( A \) . Let \( {F}_{n} \) be the free abelian group on the \( n \) -cells of \( X \) that are not contained in \( A \), which are exactly the \( n \) -cells of \( Y \) that are not contained in \( B \) . Then we have ... | Yes |
Theorem 4.2.33. Let \( X \) be a CW-complex with only finitely many cells in each dimension. Then for each \( n,{H}_{n}^{\text{cell }}\left( X\right) \) and \( {H}_{\text{cell }}^{n}\left( X\right) \) are finitely generated abelian groups. | Proof. \( {H}_{n}^{\text{cell }}\left( X\right) \) is a quotient of \( {Z}_{n}^{\text{cell }}\left( X\right) \), which is a subgroup of a finitely generated free abelian group, and hence itself is a finitely generated free abelian group, and similarly for \( {H}_{\text{cell }}^{n}\left( X\right) \) . | Yes |
Theorem 4.3.2. Let \( d = {\dim }_{\mathbb{R}}\mathbb{F} \) (so that \( d = 1 \) if \( \mathbb{F} = \mathbb{R} \) and \( d = 2 \) if \( \mathbb{F} = \mathbb{C} \) ). Then \( \mathbb{F}{P}^{n} \) has a CW-structure with one cell in dimension di for each \( i = 0,\ldots, n \) . | Proof. By induction on \( n \) .\n\nFor \( n = 0,\mathbb{F}{P}^{0} \) is just a point.\n\nAssume now the theorem is true for \( n - 1 \) . We shall show that \( \mathbb{F}{P}^{n} - \mathbb{F}{P}^{n - 1} \) is a single cell of dimension \( {dn} \), which, by induction, completes the proof.\n\nNow \( \mathbb{F}{P}^{n} - ... | Yes |
Theorem 4.3.3. The homology of \( \mathbb{C}{P}^{n} \) is as follows:\n\n\[ {H}_{i}\left( {\mathbb{C}{P}^{n}}\right) = \left\{ \begin{array}{ll} 0 & i > {2n} \\ \mathbb{Z} & 0 \leq i \leq {2n}\text{ even } \\ 0 & 0 < i < {2n}\text{ odd. } \end{array}\right. \] | Proof. The cellular chain complex of \( \mathbb{C}{P}^{n} \) is\n\n\[ 0 \rightarrow \mathbb{Z} \rightarrow 0 \rightarrow \mathbb{Z} \rightarrow \cdots \rightarrow \mathbb{Z} \rightarrow 0 \rightarrow \mathbb{Z} \rightarrow 0 \]\n\nwith \( \mathbb{Z} \) in every even dimension between 0 and \( {2n} \), and 0 otherwise. | Yes |
Lemma 5.1.3. For any \( n,\partial \left( {\partial {I}^{n}}\right) = 0 \) . | Proof. For \( n \leq 1 \) this is clear.\n\nFor \( n \geq 2,\partial \left( {\partial {I}^{n}}\right) \) is an element in the free abelian group generated by the \( \left( {n - 2}\right) \) - faces of \( {I}^{n} \), i.e., by the subsets, for each \( i \neq j \) and each \( {\varepsilon }_{i} = 0 \) or \( 1,{\varepsilon... | No |
Lemma 5.1.12. Let \( f : X \rightarrow Y \) be a map. Then finduces a chain map \( \left\{ {{f}_{n} : {C}_{n}\left( X\right) \rightarrow }\right. \) \( \left. {{C}_{n}\left( Y\right) }\right\} \) where \( {f}_{n} : {C}_{n}\left( X\right) \rightarrow {C}_{n}\left( Y\right) \) as follows. Let \( \Phi : {I}^{n} \rightarro... | Proof. This would be immediate if we were dealing with \( {Q}_{n}\left( X\right) \) and \( {Q}_{n}\left( Y\right) \) . But since \( {f\Phi } \) is degenerate wherever \( \Phi \) is, it is just about immediate for \( {C}_{n}\left( X\right) \) and \( {C}_{n}\left( Y\right) \) . Then the fact that we have maps on homology... | No |
Theorem 5.1.14. Singular homology satisfies Axioms 1 and 2. | Proof. Immediate from the definition of the induced map on singular cubes as composition. | No |
Theorem 5.1.15. Singular homology satisfies Axiom 3. | Proof. Immediate from the definition of the boundary map on singular cubes and from the definition of the induced map on singular cubes as composition. | No |
Theorem 5.1.16. Singular homology satisfies Axiom 4. | Proof. We have defined \( {C}_{n}\left( {X, A}\right) = {C}_{n}\left( X\right) /{C}_{n}\left( A\right) \) . Thus for every \( n \), we have a short exact sequence\n\n\[ 0 \rightarrow {C}_{n}\left( A\right) \rightarrow {C}_{n}\left( X\right) \rightarrow {C}_{n}\left( {X, A}\right) \rightarrow 0.\]\n\nIn other words, we ... | Yes |
Theorem 5.1.17. Singular homology satisfies Axiom 5. | Proof. For simplicity we consider the case of homotopic maps of spaces \( f : X \rightarrow Y \) and \( g : X \rightarrow Y \) (rather than maps of pairs). Then by definition, setting \( {f}_{0} = f \) and \( {f}_{1} = g \), there is a map \( F : X \times I \rightarrow Y \) with \( F\left( {x,0}\right) = {f}_{0}\left( ... | Yes |
Theorem 5.1.19. Let \( X \) be the space consisting of a single point. Then \( {H}_{0}\left( X\right) \cong \mathbb{Z} \) and \( {H}_{i}\left( X\right) = 0 \) for \( i \neq 0 \) . Thus singular homology satisfies the dimension axiom, Axiom 7, and has coefficient group \( \mathbb{Z} \) . | Proof. Let \( \Phi : {I}^{0} \rightarrow X \) be the unique map. Then \( {C}_{0}\left( X\right) \) is the free abelian group generated by \( \Phi \) . On the other hand, for any \( i > 0,\Phi : {I}^{i} \rightarrow X \) is a degenerate \( i \) -cube. Hence \( {C}_{i}\left( X\right) = \{ 0\} \) for \( i > 0 \) . Thus \( ... | Yes |
Theorem 5.1.24. For any singular chain \( c \) , \( \operatorname{supp}\left( c\right) \) is a compact subset of \( X \) . | Proof. For any \( \Phi : {I}^{n} \rightarrow X,\Phi \left( {I}^{n}\right) \) is a compact subset of \( X \) as it is the continuous image of a compact set. Then for any singular chain \( c \), supp \( \left( c\right) \) is a finite union of compact sets and hence is compact. | Yes |
Corollary 5.1.25. Let \( X \) be a union of components, \( X = \mathop{\bigcup }\limits_{{i \in I}}{X}_{i} \) . Then for any \( n \) , \( {H}_{n}\left( X\right) = {\bigoplus }_{i \in I}{H}_{n}\left( {X}_{i}\right) \) | Proof. This follows for any generalized homology theory from Lemma 3.2.1 if there are only finitely many components. But for singular homology theory, if \( c \in {C}_{n}\left( X\right) \) is any chain, then \( \operatorname{supp}\left( c\right) \) is compact, by Theorem 5.1.24, so is contained in \( \mathop{\bigcup }\... | Yes |
Lemma 5.1.26. (1) For any space \( X \), the group of singular \( n \) -chains \( {C}_{n}\left( X\right) \) is isomorphic to the free abelian group with basis the non-degenerate n-cubes. | Proof. This follows easily once we recall that \( {C}_{n}\left( X\right) = {Q}_{n}\left( X\right) /{D}_{n}\left( X\right) \) where \( {Q}_{n}\left( X\right) \) is the free abelian group on all \( n \) -cubes and \( {D}_{n}\left( X\right) \) is the free abelian group on the degenerate \( n \) -cubes | Yes |
Theorem 5.2.1. Let \( X \) be a space. Then \( {H}_{0}\left( X\right) \) is isomorphic to the free abelian group on the path components of \( X \) . | Proof. We assume \( X \) nonempty. We have already seen in Corollary 5.1.25 that if \( X = {X}_{1} \cup {X}_{2} \cup \cdots \) is a union of path components, then \( {H}_{i}\left( X\right) = {\bigoplus }_{k}{H}_{i}\left( {X}_{k}\right) \) . Thus it satisfies to prove the theorem in case \( X \) is path connected, so we... | Yes |
Lemma 5.2.2. Let \( f : I \rightarrow X \) and \( g : I \rightarrow X \) with \( f\left( 1\right) = g\left( 0\right) \) . Define \( h : I \rightarrow X \) by \( h\left( t\right) = f\left( {2t}\right) \) for \( 0 \leq t \leq \frac{1}{2} \), and \( h\left( t\right) = g\left( {{2t} - 1}\right) \) for \( \frac{1}{2} \leq t... | Proof. We exhibit a 2-cell \( C \) with \( \partial C = f + g - h.C : I \rightarrow I \rightarrow X \) is given by following \( f \) and then \( g \) along each of the heavy solid lines as indicated:\n\n\n\nThen \( \pa... | Yes |
Corollary 5.2.5. Let \( X \) be a path-connected space. The map \( \theta \) induces a bijection (of sets)\n\n\[ \left\{ \right. \text{free homotopy classes of maps:}\left. {{S}^{1} \rightarrow X}\right\} \rightarrow {H}_{1}\left( X\right) \text{.} \] | Proof. Immediate from Theorems 5.2.4 and 2.5.1. | No |
Lemma 5.2.6. Let \( f : \left( {X,{x}_{0}}\right) \rightarrow \left( {Y,{y}_{0}}\right) \) . Then the following diagram commutes: |  | No |
Theorem 5.2.7. Let \( d \) be any integer. Then for any integer \( n \geq 1 \), there exists a map \( f : {S}^{n} \rightarrow {S}^{n} \) of degree \( d \) . | Proof. Again the key step is the \( n = 1 \) case, and we provide an alternate proof of that (with the remainder of the proof being the same as in the previous proof).\n\nAgain we claim that \( f : {S}^{1} \rightarrow {S}^{1} \) by \( f\left( z\right) = {z}^{d} \) has degree \( d \) .\n\nTo prove that, we consider the ... | Yes |
Here is a pair of examples to show that the condition closure \( \left( U\right) \subseteq \) interior \( \left( A\right) \) cannot in general be relaxed to \( U \subseteq \operatorname{interior}\left( A\right) \) for the inclusion \( \left( {X - U, A - U}\right) \rightarrow \left( {X, A}\right) \) to be excisive. | (a) Let \( X = {\mathbb{R}}^{2} \) and let \( A \) be the subset of \( {\mathbb{R}}^{2} \) that is on or below the graph of the function\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} \sin \left( \frac{1}{x}\right) & x > 0 \\ 1 & x \leq 0. \end{array}\right. \]\n\nNote that \( \partial A \) consists of the union of... | Yes |
Lemma 5.3.2. \( {C}_{n}\left( {X;G}\right) \) is isomorphic to \( {C}_{n}\left( X\right) \otimes G \) (and similarly for \( A \) ). Also, \( {C}_{n}\left( {X, A;G}\right) \) is isomorphic to \( {C}_{n}\left( {X, A}\right) \otimes G \) . | Proof. Clear from Definition 5.3.1 and the fact that for any two abelian groups \( A \) and \( B,\left( {A \oplus B}\right) \otimes G \approx \left( {A \otimes G}\right) \oplus \left( {B \otimes G}\right) \), and hence, in this situation, \( \left( {A \otimes G}\right) \approx \) \( \left( {\left( {A \oplus B}\right) \... | No |
Lemma 5.3.3. With the above identifications, \( {C}_{n}\left( {X;G}\right) \) is a chain complex with boundary map \( \partial \otimes 1 : {C}_{n}\left( {X;G}\right) \rightarrow {C}_{n - 1}\left( {X;G}\right) \), and similarly for \( {C}_{n}\left( {A;G}\right) \) and \( {C}_{n}\left( {X, A;G}\right) \) . | Proof. The only thing to check is that \( {\left( \partial \otimes 1\right) }^{2} = 0 \) . But \( {\left( \partial \otimes 1\right) }^{2} = {\partial }^{2} \otimes 1 = 0 \) . | Yes |
Lemma 5.3.4. There is a split short exact sequence\n\n\[ 0 \rightarrow {C}_{n}\left( {A;G}\right) \rightarrow {C}_{n}\left( {X;G}\right) \rightarrow {C}_{n}\left( {X, A;G}\right) \rightarrow 0, \]\n\nand hence \( {C}_{n}\left( {X;G}\right) \) is isomorphic to \( {C}_{n}\left( {A;G}\right) \oplus {C}_{n}\left( {X, A;G}\... | Proof. We have the short exact sequence\n\n\[ 0 \rightarrow {C}_{n}\left( A\right) \rightarrow {C}_{n}\left( X\right) \rightarrow {C}_{n}\left( {X, A}\right) \rightarrow 0. \]\n\nTensoring such a sequence with \( G \) does not in general produce an exact sequence. But if this sequence is split short exact, tensoring wi... | Yes |
Lemma 5.3.5. With the identification in Lemma 5.3.2, \( f : X \rightarrow Y \) induces \( {f}_{ * } \otimes 1 \) : \( {C}_{ * }\left( {X;G}\right) \rightarrow {C}_{ * }\left( {Y;G}\right) \), and similarly for \( f : \left( {X, A}\right) \rightarrow \left( {Y, B}\right) \) . | In concrete terms, if \( \left\{ {{\Phi }_{i} : {I}^{n} \rightarrow X}\right\} \) are singular \( n \) -cubes, \( \left( {{f}_{ * } \otimes 1}\right) \left( {\mathop{\sum }\limits_{i}{g}_{i}{\Phi }_{i}}\right) = \) \( \sum {g}_{i}\left( {f{\Phi }_{i}}\right) \) | Yes |
Theorem 5.3.7. Singular homology with coefficients in \( G \) is an ordinary homology theory with coefficient group \( G \) . | Proof. First we check Axiom 7, the dimension axiom. If \( X \) consists of a single point, then \( {C}_{ * }\left( {X;G}\right) \) is isomorphic to\n\n\[ \cdots \rightarrow 0 \rightarrow 0 \rightarrow G \rightarrow 0 \rightarrow 0 \rightarrow \cdots \]\n\nwith homology as claimed.\n\nThe proof that this theory satisfie... | No |
Lemma 5.3.8. The map \( \tau : {C}_{n}\left( X\right) \rightarrow {C}_{n}\left( {X;G}\right) \) given by \( \tau \left( \Phi \right) = \Phi \otimes 1 \) where \( \Phi \) is a singular \( n \) -cube induces a map\n\n\[ \tau : {H}_{n}\left( X\right) \otimes G \rightarrow {H}_{n}\left( {X;G}\right) \] | Proof. Lemma 5.3.3 implies that \( \tau : {Z}_{n}\left( X\right) \rightarrow {Z}_{n}\left( {X;G}\right) \) and \( \tau : {B}_{n}\left( X\right) \rightarrow {B}_{n}\left( {X;G}\right) \) , where, as usual, \( {Z}_{n}\left( X\right) = \operatorname{Ker}\left( {\partial }_{n}\right) \) and \( {B}_{n}\left( X\right) = \ope... | Yes |
Theorem 5.3.9 (Universal coefficient theorem). (1) For any space \( X \) and abelian group \( G \), there is a split short exact sequence\n\n\[ 0 \rightarrow {H}_{n}\left( X\right) \otimes G\overset{\tau }{ \rightarrow }{H}_{n}\left( {X;G}\right) \rightarrow \operatorname{Tor}\left( {{H}_{n - 1}\left( X\right), G}\righ... | Proof. This is a purely algebraic fact about the homology of chain complexes, and we omit the proof. | No |
Example 5.3.12. By Lemma A.3.8, \( \operatorname{Tor}\left( {{\mathbb{Z}}_{2},{\mathbb{Z}}_{m}}\right) \approx {Z}_{2} \) for \( m \) even. Thus, from Theorem 4.3.4, for real projective spaces we have, for \( m \) even, | \[ {H}_{i}\left( {\mathbb{R}{P}^{n};{\mathbb{Z}}_{m}}\right) = \left\{ \begin{array}{ll} 0 & i > n \\ {\mathbb{Z}}_{m} & i = n\text{ odd } \\ {\mathbb{Z}}_{2} & i = n\text{ even } \\ {\mathbb{Z}}_{2} & 1 \leq i \leq n - 1 \\ {\mathbb{Z}}_{m} & i = 0. \end{array}\right. \] | Yes |
Corollary 5.3.13. Let \( f : X \rightarrow Y \) and suppose that \( {f}_{ * } : {H}_{n}\left( X\right) \rightarrow {H}_{n}\left( Y\right) \) is an isomorphism for all \( n \) . Then \( {f}_{ * } : {H}_{n}\left( {X;G}\right) \rightarrow {H}_{n}\left( {Y;G}\right) \) is an isomorphism for all \( n \) . | Proof. This follows directly from the universal coefficient theorem and the short five lemma. | Yes |
Lemma 5.4.2. The cross product induces a map\n\n\[ \n{C}_{j}\left( X\right) \otimes {C}_{k}\left( Y\right) \rightarrow {C}_{j + k}\left( {X \times Y}\right) \n\] | Proof. If either \( \Phi \) or \( \Psi \) is degenerate, so is \( \Phi \times \Psi \) . | No |
Lemma 5.4.3. In this situation, \n\n\[ \n\\partial \\left( {\\Phi \\times \\Psi }\\right) = \\left( {\\partial \\Phi }\\right) \\times \\Psi + {\\left( -1\\right) }^{j}\\Phi \\times \\left( {\\partial \\Psi }\\right) .\n\] | Proof. Direct calculation, with careful attention to signs. | No |
Lemma 5.4.5. The cross product induces a map\n\n\\[ \n{H}_{j}\\left( X\\right) \\otimes {H}_{k}\\left( Y\\right) \\rightarrow {H}_{j + k}\\left( {X \\times Y}\\right) \n\\] | Proof. First we show that we obtain a map\n\n\\[ \n{Z}_{j}\\left( X\\right) \\otimes {Z}_{k}\\left( Y\\right) \\rightarrow {Z}_{j + k}\\left( {X \\times Y}\\right) \n\\]\n\nLet \\( c \\in {Z}_{j}\\left( X\\right) \\) and \\( d \\in {Z}_{k}\\left( Y\\right) \\) be singular cycles, so that \\( \\partial c = 0 \\) and \\(... | Yes |
Theorem 5.4.6 (Künneth formula). (1) For any spaces \( X \) and \( Y \), there is a split short exact sequence\n\n\[ 0 \rightarrow {\left( {H}_{ * }\left( X\right) \otimes {H}_{ * }\left( Y\right) \right) }_{n} \rightarrow {H}_{n}\left( {X \times Y}\right) \rightarrow {\left( \mathrm{{Tor}}\left( {H}_{ * }\left( X\righ... | Proof. This is a purely algebraic result, whose proof we omit, but we again remark that it crucially uses the fact that \( {C}_{ * }\left( X\right) \) and \( {C}_{ * }\left( Y\right) \) are chain complexes of free abelian groups. | No |
Lemma 5.4.8. Let \( Y \) be a path connected space and let \( \pi : X \times Y \rightarrow X \) be projection on the first factor. For any element \( \alpha \) of \( {H}_{n}\left( X\right) \) , \[ {\pi }_{ * }\left( {\alpha \otimes {1}_{Y}}\right) = \alpha \] | Proof. Clear from the construction in Lemma 5.4.5. | No |
Lemma 5.5.3. The map \( {f}^{ * } : {C}^{ * }\left( Y\right) \rightarrow {C}^{ * }\left( X\right) \) induces a map \( {f}^{ * } : {H}^{ * }\left( Y\right) \rightarrow {H}^{ * }\left( X\right) \) . | Proof. It is routine to check that \( {f}^{ * }\left( {{Z}^{n}\left( Y\right) }\right) \subseteq {Z}^{n}\left( X\right) \) and \( {f}^{ * }\left( {{B}^{n}\left( Y\right) }\right) \subseteq {B}^{n}\left( X\right) \) . | No |
Theorem 5.5.4. Singular cohomology is an ordinary cohomology theory with \( \mathbb{Z} \) coefficients. | Proof. This proof entirely mimics the proof that singular homology is an ordinary homology theory with \( \mathbb{Z} \) coefficients. There is just one subtlety, Axiom 4, the exactness axiom. Exactness for homology followed from the short exactness of the sequence of singular chain complexes \[ 0 \rightarrow {C}_{ * }\... | Yes |
Theorem 5.5.7. Singular cohomology with coefficients in \( G \) is an ordinary cohomology theory with coefficient group \( G = {H}^{0}\left( {X;G}\right) \) . | Proof. Again this mirrors the proof for singular homology in Sect. 5.1. Again Axiom 4 uses the fact that, for every \( n \), the sequence \( 0 \rightarrow {C}^{n}\left( {X, A}\right) \rightarrow {C}^{n}\left( X\right) \rightarrow \) \( {C}^{n}\left( A\right) \rightarrow 0 \) is split exact. | No |
Theorem 5.5.8 (Universal coefficient theorem). (1) Let \( X \) be a space and let \( G \) be an abelian group. Suppose that \( X \) is of finite type or that \( G \) is of finitely generated. Then there is a split short exact sequence\n\n\[ 0 \rightarrow {H}^{n}\left( X\right) \otimes G \rightarrow {H}^{n}\left( {X;G}\... | Proof. Again we omit the purely algebraic argument, but we note that, while the cochain groups \( {C}^{ * }\left( X\right) \) are not in general free, they are torsion-free, and that fact, together with our additional hypotheses, suffices to be able to apply that argument. | No |
Lemma 5.5.11. The evaluation map e induces a map\n\n\[ e : {H}^{n}\left( X\right) \otimes {H}_{n}\left( X\right) \rightarrow \mathbb{Z} \]\n\nby \( e\left( {\left\lbrack \gamma \right\rbrack ,\left\lbrack c\right\rbrack }\right) = \gamma \left( c\right) \), where \( \gamma \) (resp. \( c \) ) is a representative of the... | Proof. We can restrict \( e \) to evaluate cocycles on cycles,\n\n\[ e : {Z}^{n}\left( X\right) \otimes {Z}_{n}\left( X\right) \rightarrow \mathbb{Z} \]\n\nby \( e\left( {\gamma, c}\right) = \gamma \left( c\right) \) . But then if \( c \) is a boundary, \( c = \partial d, e\left( {\gamma, c}\right) = e\left( {\gamma ,\... | Yes |
Theorem 5.5.12 (Universal coefficient theorem). (1) For any space \( X \) and abelian group \( G \), there is a split short exact sequence\n\n\[ 0 \rightarrow \operatorname{Ext}\left( {{H}_{n - 1}\left( X\right), G}\right) \rightarrow {H}^{n}\left( {X;G}\right) \overset{e}{ \rightarrow }\operatorname{Hom}\left( {{H}_{n... | Proof. Again this is a purely algebraic argument which we omit. | No |
Corollary 5.5.15. Let \( X \) be a space of finite type and suppose that \( {H}_{n}\left( X\right) \approx {F}_{n} \oplus {T}_{n} \) , where \( {F}_{n} \) is a free abelian group and \( {T}_{n} \) is a torsion group, for each \( n \) . Then \[ {H}^{n}\left( X\right) \approx {F}_{n} \oplus {T}_{n - 1} \] for each \( n \... | Proof. This follows from the computation of Ext in Lemma A.3.12. | No |
The integral singular cohomology of \( \mathbb{R}{P}^{n} \) is as follows:\n\n\[ \n{H}^{k}\left( {\mathbb{R}{P}^{n}}\right) = \left\{ \begin{array}{ll} 0 & k > n \\ \mathbb{Z} & k = n\text{ odd } \\ 0 & k = n\text{ even } \\ {\mathbb{Z}}_{2} & 1 \leq k \leq n - 1\text{ even } \\ 0 & 1 \leq k \leq n - 1\text{ odd } \\ \... | as we see from Corollary 5.5.15 and Theorem 4.3.4. | No |
Theorem 5.5.19 (Universal coefficient theorem). (1) Let \( X \) be a space of finite type. For any abelian group \( G \) there is a split short exact sequence\n\n\[ 0 \rightarrow \operatorname{Ext}\left( {{H}^{n + 1}\left( X\right), G}\right) \rightarrow {H}_{n}\left( {X;G}\right) \overset{e}{ \rightarrow }\operatornam... | Proof. Again we omit this purely algebraic proof. | No |
Theorem 5.5.20. Let \( X \) be a space with finitely generated homology. Let \( \mathbb{F} \) be an arbitrary field. Then \( \chi \left( X\right) \) is given by\n\n\[ \chi \left( X\right) = \left\{ \begin{array}{l} \mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( -1\right) }^{n}\operatorname{rank}{H}_{n}\left( {X;\math... | Proof. This follows directly from the universal coefficient theorems. (If \( \mathbb{F} \) is a field of characteristic zero, then all of these ranks are equal for every integer \( n \) . If \( \mathbb{F} \) does not have characteristic 0, that may not be the case, depending on the space \( X \), but nevertheless the a... | Yes |
Lemma 5.5.23. Let \( \pi : X \times Y \rightarrow X \) be projection on the first factor. For any element \( \alpha \) of \( {H}^{n}\left( X\right) \) , | \[ {\pi }^{ * }\left( \alpha \right) = \alpha \otimes {1}^{Y} \] | Yes |
Theorem 5.6.1. There are natural maps of chain complexes\n\n\[ E : {C}_{ * }\left( X\right) \otimes {C}_{ * }\left( Y\right) \rightarrow {C}_{ * }\left( {X \times Y}\right) \]\n\nand\n\n\[ F : {C}_{ * }\left( {X \times Y}\right) \rightarrow {C}_{ * }\left( X\right) \otimes {C}_{ * }\left( Y\right) \]\n\nthat are invers... | Here \( E \) is the map of Lemma 5.4.2, and we do not define \( F \) . | No |
Lemma 5.6.3. In this situation, \n\n\[ \n\delta \left( {f \times g}\right) = \left( {\delta f}\right) \times g + {\left( -1\right) }^{j}f \times \left( {\delta g}\right) . \n\] | Proof. Entirely analogous to the proof of Lemma 5.4.3. | No |
Lemma 5.6.4. The cross product induces a map\n\n\[ \times : {H}^{j}\left( X\right) \otimes {H}^{k}\left( Y\right) \rightarrow {H}^{j + k}\left( {X \times Y}\right) . \] | Proof. Entirely analogous to the proof of Lemma 5.4.5. | No |
Theorem 5.6.5. Let \( \alpha : {X}_{1} \rightarrow {X}_{2} \) and \( \beta : {Y}_{1} \rightarrow {Y}_{2} \) be maps. Then there are commutative diagrams\n\n\[ \n{H}_{j}\left( {X}_{1}\right) \otimes {H}_{k}\left( {Y}_{1}\right) \rightarrow {H}_{j + k}\left( {{X}_{1} \times {Y}_{1}}\right) \]\n\n\[ \n{\alpha }_{ * } \oti... | Proof. This follows directly from the covariance/contravariance of the maps on homology/cohomology and the naturality of the Eilenberg-Zilber maps. | Yes |
Let \( \alpha \in {H}^{j}\left( X\right) \) and \( \beta \in {H}^{k}\left( Y\right) \). Then\n\n\[ \alpha \times \beta = {\pi }_{1}{}^{ * }\left( \alpha \right) \cup {\pi }_{2}{}^{ * }\left( \beta \right) = \left( {\alpha \times {1}^{Y}}\right) \cup \left( {{1}^{X} \times \beta }\right) . \] | Proof. We prove the first of these. The last equality is just Lemma 5.5.23. To prove the first, let \( \bigtriangleup : X \times Y \rightarrow \left( {X \times Y}\right) \times \left( {X \times Y}\right) \) be the diagonal. Then, by definition,\n\n\[ {\pi }_{1}{}^{ * }\left( \alpha \right) \cup {\pi }_{2}{}^{ * }\left(... | Yes |
Theorem 5.6.14. (1) Let \( \alpha \in {H}^{j}\left( X\right) ,\beta \in {H}^{k}\left( X\right) ,\gamma \in {H}^{l}\left( Y\right) \), and \( \delta \in {H}^{m}\left( Y\right) \). Then, if \( n = j + k + l + m \), \[ \left( {\alpha \cup \beta }\right) \times \left( {\gamma \cap \delta }\right) = {\left( -1\right) }^{kl}... | This follows from the previous properties we have obtained with enough careful attention to detail (including signs). | No |
Theorem 5.6.17. Let \( X \) be a space and let \( C \) and \( D \) be subspaces of \( X \) . Assume that \( \{ X \times C, D \times X\} \) and \( \{ C, D\} \) are both excisive couples. Then there is a cup product\n\n\[ \cup : {H}^{j}\left( {X, C}\right) \otimes {H}^{k}\left( {X, D}\right) \rightarrow {H}^{j + k}\left(... | Proof. The condition that \( \{ X \times C, D \times X\} \) be excisive is necessary in order to apply the Eilenberg-Zilber theorem, and the condition that \( \{ C, D\} \) be excisive is necessary to obtain the analog of Lemma 5.6.12. Otherwise, the constructions are entirely analogous (though more complicated). | No |
Corollary 5.6.18. Let \( \left( {X, A}\right) \) be a pair. In part (1), assume that \( \{ X \times A, A \times X\} \) is an excisive couple.\n\n(1) There is a cup product\n\n\[ \cup : {H}^{j}\left( {X, A}\right) \otimes {H}^{k}\left( {X, A}\right) \rightarrow {H}^{j + k}\left( {X, A}\right) \]\n\nand a cap product\n\n... | Proof. (1) This is the special case \( C = D = A \) of Theorem 5.6.17. | No |
We take \( \mathbb{Z} \) coefficients. Let \( p, q \geq 1 \) . Then \( {H}^{p}\left( {S}^{p}\right) \cong \mathbb{Z} \) and we choose a generator \( \alpha \) . Also, \( {H}^{q}\left( {S}^{q}\right) \cong \mathbb{Z} \) and we choose a generator \( \beta \) . Now consider \( {H}^{ * }\left( {{S}^{p} \times {S}^{q}}\righ... | But by Lemma 5.6.12 this gives\n\n\[ \n\widetilde{\gamma } = \widetilde{\alpha } \cup \widetilde{\beta }\n\] | Yes |
Example 5.7.2. Again we take \( \\mathbb{Z} \) coefficients. Let \( p, q \\geq 1 \) . Let \( Y = {S}^{p} \\vee {S}^{q} \\vee {S}^{p + q} \) , i.e., the union of \( {S}^{p},{S}^{q} \), and \( {S}^{p + q} \) with all three spaces identified at one point. | Let \( Z = {S}^{p} \\vee {S}^{q} \\subset Y \) and note that we have a retraction \( f : Y \\rightarrow Z \) given by collapsing \( {S}^{p + q} \) to the identification point. Then \( {f}^{ * } : {H}^{n}\\left( Z\\right) \\rightarrow {H}^{n}\\left( Y\\right) \) is an isomorphism for \( n = p, q \) . Let \( \\alpha \) b... | Yes |
Theorem 5.7.3. Let \( X = {S}^{p} \times {S}^{q} \) and \( Y = {S}^{p} \vee {S}^{q} \vee {S}^{p + q} \) . If \( p \) and \( q \) are not both 0, then \( X \) and \( Y \) are not homotopy equivalent. | Proof. If \( p = 0 \) or \( q = 0 \) this is trivial.\n\nSuppose \( p, q \geq 1 \) . Then by Examples 5.7.1 and 5.7.2 \( X \) and \( Y \) have nonisomorphic cohomology rings, so by Corollary 5.6.16 they are not homotopy equivalent. | Yes |
Corollary 5.7.5. Let \( n > m \geq 1 \) . If \( f : \mathbb{R}{P}^{n} \rightarrow \mathbb{R}{P}^{m} \) is any map, then \( {f}_{ * } : {H}_{1}\left( {\mathbb{R}{P}^{n}}\right) \rightarrow {H}_{1}\left( {\mathbb{R}{P}^{m}}\right) \) is the zero map. | Proof. If \( m = 1 \), then \( {H}_{1}\left( {\mathbb{R}{P}^{n}}\right) = {\mathbb{Z}}_{2} \) and \( {H}_{1}\left( {\mathbb{R}{P}^{m}}\right) = \mathbb{Z} \) and the only map from \( {\mathbb{Z}}_{2} \) to \( \mathbb{Z} \) is the zero map. Suppose \( m > 1 \) . Then \( {H}_{1}\left( {\mathbb{R}{P}^{n}}\right) = {\mathb... | Yes |
Theorem 6.1.8. If \( M \) is an \( n \) -manifold with nonempty boundary then \( \operatorname{int}\left( M\right) \) is an \( n \) -manifold and \( \partial M \) is an \( \left( {n - 1}\right) \) -manifold. | Proof. The first statement is clear. As for the second, if \( x \in \partial M \) and \( {\varphi }_{x} : {\mathbb{R}}_{ + }^{n} \rightarrow U \) is a homeomorphism with \( x \in {\varphi }_{x}\left( {\partial {\mathbb{R}}_{ + }^{n}}\right) \), then \( {\varphi }_{x} \mid \partial {\mathbb{R}}_{ + }^{n} \) is a homeomo... | Yes |
Lemma 6.2.1. Let \( M \) be an n-manifold and let \( x \in M \) be arbitrary. Then \( {H}_{n}\left( {M, M - x;G}\right) \) is isomorphic to \( G \) . | Proof. Let \( \left( {{U}_{\alpha },{\varphi }_{\alpha }}\right) \) be a coordinate patch with \( x \in {\varphi }_{\alpha } \) . Let \( p = {\varphi }_{\alpha }^{-1}\left( x\right), p \in {\mathbb{R}}^{n} \) . Then we have maps\n\n\[ \left( {{\mathbb{R}}^{n},{\mathbb{R}}^{n}-\{ p\} }\right) \rightarrow \left( {{U}_{\a... | Yes |
Theorem 6.2.6. Every manifold is \( \mathbb{Z}/2\mathbb{Z} \) -orientable. | Proof. Following the diagram in Definition 6.1.3 all the way around from \( G \) to \( G \) gives an isomorphism from \( G \) to \( G \), and \( {\bar{\varphi }}_{x} \) and \( {\bar{\varphi }}_{y} \) are compatible if and only if this isomorphism is the identity. But the only isomorphism \( \bar{\varphi } : \mathbb{Z}/... | Yes |
Theorem 6.2.7. (1) Let \( M \) be the union of components \( M = {M}_{1} \cup {M}_{2} \cup \cdots \) . Then \( M \) is \( \mathbb{Z} \) -orientable if and only if each \( {M}_{i} \) is \( \mathbb{Z} \) -orientable. | Proof. The important thing to note is that if \( {\bar{\varphi }}_{x} : \mathbb{Z} \rightarrow {H}_{n}\left( {M, M - x;\mathbb{Z}}\right) \) is a local \( \mathbb{Z} \) -orientation, there is exactly one other local \( \mathbb{Z} \) -orientation at \( x \), namely \( - {\bar{\varphi }}_{x} \), where \( - {\bar{\varphi ... | No |
Theorem 6.2.10. Let \( M \) be a connected \( n \) -manifold. Then a system of local orientations \( \left\{ {\bar{\varphi }}_{x}\right\} \) is an orientation of \( M \) if and only if for every \( x, y \in M \) and every path \( f : I \rightarrow M \) with \( f\left( 0\right) = x \) and \( f\left( 1\right) = y,{\bar{\... | Proof. If \( M \) is orientable, let \( \left\{ {\bar{\varphi }}_{x}\right\} \) be an orientation, i.e., a compatible system of local orientations. Then for any path \( f,{f}_{y}\left( {\bar{\varphi }}_{x}\right) = {\bar{\varphi }}_{y} \) is independent of the choice of \( f \) .\n\nConversely, if \( {f}_{y}\left( {\ba... | Yes |
Lemma 6.2.11. (1) Let \( x \) and \( y \) be two points in \( M \) that are both contained in some coordinate patch \( {U}_{\alpha } \) . Then for any two paths \( f \) and \( g \) from \( x \) to \( y \) with \( f\left( I\right) \subset {U}_{\alpha } \) and \( g\left( I\right) \subset {U}_{\alpha },{f}_{y}\left( {\bar... | Proof. (1) Since \( I \) is compact, \( f\left( I\right) \) is a compact subset of \( {U}_{\alpha } \), and hence \( {\varphi }_{\alpha }^{-1}\left( {f\left( I\right) }\right) \) is a compact subset of \( {\mathbb{R}}^{n} \), as is \( {\varphi }_{\alpha }^{-1}\left( {g\left( I\right) }\right) \) . But then we may choos... | Yes |
Theorem 6.2.13. Let \( M \) be a connected \( n \) -manifold. If \( M \) is simply connected, then \( M \) is orientable. | Proof. By Theorem 6.2.10, we must show that \( \left\{ {{f}_{y}\left( {\bar{\varphi }}_{x}\right) }\right\} \) is independent of the choice of \( f \) . By Lemma 6.2.12(2), that will be the case if \( {h}_{x}\left( {\bar{\varphi }}_{x}\right) = {\bar{\varphi }}_{x} \) for any loop \( {h}_{x} \) based at \( x \) . But b... | Yes |
Theorem 6.2.15. A connected manifold \( M \) is orientable if and only if its orientation character \( w\left( f\right) = 0 \) for every loop \( f \) in \( M \). | Proof. In light of Lemma 6.2.12, this is just a restatement of Theorem 6.2.10. | No |
Lemma 6.2.16. Let \( M \) be a connected manifold. The orientation character gives a homomorphism \[ w : {\pi }_{1}\left( {M, x}\right) \rightarrow \mathbb{Z}/2\mathbb{Z} \] defined by \( w\left( \alpha \right) = w\left( f\right) \) where \( f \) is a loop in \( M \) representing \( \alpha \in {\pi }_{1}\left( {M, x}\r... | Proof. By Lemma 6.2.12(2), \( w \) depends only on the homotopy class of \( f \), and by Lemma 6.2.12(1), \( w \) is a homomorphism. | Yes |
Corollary 6.2.17. Let \( M \) be a connected nonorientable manifold. Then \( M \) has a unique 2-fold cover \( N \) that is orientable. | Proof. \( N \) is the cover of \( M \) corresponding to the subgroup \( \operatorname{Ker}\left( w\right) \subset {\pi }_{1}\left( {M, x}\right) \) of index 2 as in Theorem 2.2.19. | No |
Lemma 6.2.18. Let \( M \) be a manifold. The orientation character gives a homomorphism\n\n\[ w : {H}_{1}\left( {M;\mathbb{Z}}\right) \rightarrow \mathbb{Z}/2\mathbb{Z} \] | Proof. The map \( w : {\pi }_{1}\left( {M, x}\right) \rightarrow \mathbb{Z}/2\mathbb{Z} \) is a map to an abelian group, so factors through the abelianization of \( {\pi }_{1}\left( {M, x}\right) \) . In case \( M \) is connected that is just \( {H}_{1}\left( {M;\mathbb{Z}}\right) \) by Theorem 5.2.4. In the general ca... | Yes |
Theorem 6.2.19. Let \( M \) be a manifold. The orientation character gives a homomorphism \[ w : {H}_{1}\left( {M;\mathbb{Z}/2\mathbb{Z}}\right) \rightarrow \mathbb{Z}/2\mathbb{Z} \] | Proof. The map \( w : {H}_{1}\left( {M;\mathbb{Z}}\right) \rightarrow \mathbb{Z}/2\mathbb{Z} \) factors through \( {H}_{1}\left( {M;\mathbb{Z}}\right) /2{H}_{1}\left( {M;\mathbb{Z}}\right) \) (i.e. \( w\left( {2\alpha }\right) = {2w}\left( \alpha \right) = 0 \) for any \( \alpha \in {H}_{1}\left( {M;\mathbb{Z}}\right) ... | No |
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