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Lemma 52.1. Let \( X \) be a topological space. If \( f \) and \( g \) are two paths in \( X \) such that the endpoint of \( f \) agrees with the starting point of \( g \), then the singular 1-chains \( \Phi \left( f\right) + \Phi \left( g\right) \) and \( \Phi \left( {f * g}\right) \) are homologous. | Proof. The lemma follows quite easily from Lemma 41.3 We refer to Figure 820 for a sketch of the proof. We leave it to the reader to turn the sketch into a proper proof. | No |
Lemma 52.4. Let \( X \) be a set and let \( {x}_{1},\ldots ,{x}_{k} \) and \( {y}_{1},\ldots ,{y}_{l} \) be elements in \( X \) . If\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{k}{x}_{i} - \mathop{\sum }\limits_{{i = 1}}^{l}{y}_{i} = 0 \in {\mathbb{Z}}^{\left( X\right) }, \]\n\nthen \( k = l \) and there exists a permutatio... | Proof (*). Let \( X \) be a set and let \( {x}_{1},\ldots ,{x}_{k} \) and \( {y}_{1},\ldots ,{y}_{l} \) be elements in \( X \) such that\n\n\[ \underset{ = : a}{\underbrace{\mathop{\sum }\limits_{{i = 1}}^{k}{x}_{i}}} = \underset{ = : b}{\underbrace{\mathop{\sum }\limits_{{i = 1}}^{l}{y}_{i}}} \in {\mathbb{Z}}^{\left( ... | Yes |
Lemma 52.7. Let \( n \in \mathbb{N} \) . We denote by \( T = {\mathbb{R}}^{n}/{\mathbb{Z}}^{n} = {\left( {S}^{1}\right) }^{n} \) the \( n \) -dimensional torus.\n\n(1) We denote by \( \sigma \in {\mathrm{H}}_{1}\left( {S}^{1}\right) \) the standard generator introduced on page 1174. Given \( i \in \{ 1,\ldots, n\} \) w... | (1) This statement follows easily from the definitions, Theorem 16.16 and the Hurewicz Theorem 52.5.\n\n(2) Let \( \Theta : {\pi }_{1}\left( {T,0}\right) \rightarrow {\mathbb{Z}}^{n} \) be the isomorphism from Theorem 16.16. By Proposition 52.2 (4), Theorem 16.16 and Exercise 16.16 we obtain the following commutative d... | No |
Corollary 52.8. (*) Let \( p : Y \rightarrow X \) be a finite-index covering of path-connected topological spaces. Then the cokernel of the map\n\n\[ \n{p}_{ * } : {\mathrm{H}}_{1}\left( Y\right) \rightarrow {\mathrm{H}}_{1}\left( X\right)\n\]\n\nis finite. | Proof (*). We pick a base point \( {y}_{0} \) of \( Y \) and we write \( {x}_{0} = p\left( {y}_{0}\right) \) . We consider the following diagram of maps:\n\n\n\nHere it follows from the Hurewicz Theorem 52.5 that t... | No |
(1) Let \( X \) be a topological space, let \( {x}_{0} \in X \) and let \( n \geq 1 \) . The Hurewicz homomorphism\n\n\[{\Phi }_{\left( X,{x}_{0}\right) } : {\pi }_{n}\left( {X,{x}_{0}}\right) \rightarrow {\mathrm{H}}_{n}\left( X\right)\]\n\nis a homomorphism. | (1) We denote by \( \omega \in {\mathrm{H}}_{n}\left( {{I}^{n},\partial {I}^{n}}\right) \) the standard generator. We need to show that for any two maps \( f, g : \left( {{I}^{n},\partial {I}^{n}}\right) \rightarrow \left( {X,{x}_{0}}\right) \) we have\n\n\[ \left\lbrack {{f}_{ * }\left( \omega \right) }\right\rbrack +... | Yes |
For every \( n \in \mathbb{N} \) the Hurewicz homomorphism\n\n\[ \Phi : {\pi }_{n}\left( {{S}^{n}, * }\right) \rightarrow {\mathrm{H}}_{n}\left( {S}^{n}\right) \cong \mathbb{Z} \]\n\nsends \( \left\lbrack {{\operatorname{id}}_{{S}^{n}} : {S}^{n} \rightarrow {S}^{n}}\right\rbrack \in {\pi }_{n}\left( {S}^{n}\right) \) t... | Proof. We denote by \( \omega \in {\mathrm{H}}_{n}\left( {S}^{n}\right) \) the standard generator. By Lemma 53.1 (4) we have\n\n\[ \Phi \left( \underset{ \in {\pi }_{n}\left( {{S}^{n}, * }\right) }{\underbrace{\left\lbrack {\operatorname{id}}_{{S}^{n}} : {S}^{n} \rightarrow {S}^{n}\right\rbrack }}\right) = {\left( {\op... | Yes |
Lemma 53.3. For \( k < l \) the submanifold \( {\mathbb{{RP}}}^{k} \) is not a retract of \( {\mathbb{{RP}}}^{l} \) . | Proof. We have\n\n\[ \n{\pi }_{k}\left( {\mathbb{{RP}}}^{k}\right) \cong {\pi }_{k}\left( {S}^{k}\right) \neq 0\;\text{ and }\;{\pi }_{k}\left( {\mathbb{{RP}}}^{l}\right) \cong {\pi }_{k}\left( {S}^{l}\right) = 0.\n\] | Yes |
(1) For any \( n \in \mathbb{N} \) we have \( {\pi }_{n}\left( {S}^{n}\right) \cong \mathbb{Z} \), in fact \( {\pi }_{n}\left( {{S}^{n}, * }\right) = \mathbb{Z} \cdot \left\lbrack {\operatorname{id}}_{{S}^{n}}\right\rbrack \) . | (1) Let \( n \in \mathbb{N} \) . By Proposition 40.10 we know that \( {S}^{n} \) is \( \left( {n - 1}\right) \) -connected. Thus the statement follows immediately from the Hurewicz Theorems 52.5 and 53.5 and the fact that \( {\mathrm{H}}_{n}\left( {{S}^{n};\mathbb{Z}}\right) = \mathbb{Z} \cdot \left\lbrack {S}^{n}\righ... | Yes |
Corollary 53.7. Let \( \left( {X,{x}_{0}}\right) \) be a pointed topological space and let \( n \in \mathbb{N} \) . We suppose that \( X \) is simply connected. If given every \( k \in \{ 2,\ldots, n\} \) we have \( {\pi }_{k}\left( {X,{x}_{0}}\right) = 0 \) or \( {\mathrm{H}}_{k}\left( {X;\mathbb{Z}}\right) = 0 \), th... | Proof. We prove the corollary by induction on \( n \) . First consider the case \( n = 1 \) . Since \( X \) is simply connected we obtain from the Hurewicz Theorem 52.5 that \( {\mathrm{H}}_{1}\left( {X;\mathbb{Z}}\right) = 0 \) . Furthermore it follows from the Hurewicz Theorem 53.5 that \( {\Phi }_{\left( X,{x}_{0}\r... | Yes |
Lemma 53.8. Let \( X \) be a 1-dimensional CW-complex and let \( {x}_{0} \in X \) be base point. Then \( {\pi }_{n}\left( {X,{x}_{0}}\right) = 0 \) for every \( n \geq 2 \) . | Proof. Let \( X \) be a 1-dimensional CW-complex, let \( n \geq 2 \) and let \( {x}_{0} \in X \) be a base point. If necessary we can replace \( X \) by the path-component of \( {x}_{0} \) . In other words, without loss of generality we can assume that \( X \) is path-connected. We denote by \( \widetilde{X} \) the uni... | Yes |
Proposition 53.9. Let \( n \in \mathbb{N} \) and let \( f : {S}^{n} \rightarrow {S}^{n} \) be a homeomorphism with \( f\left( *\right) = * \) . Then the following equality holds in \( {\pi }_{n}\left( {{S}^{n}, * }\right) \) :\n\n\[ \left\lbrack f\right\rbrack = \left\{ \begin{matrix} \left\lbrack {\mathrm{{id}}}_{{S}^... | Proof. Let \( n \in \mathbb{N} \) and let \( f : {S}^{n} \rightarrow {S}^{n} \) be a homeomorphism with \( f\left( *\right) = * \) . By the naturality of the Hurewicz homomorphism, see Lemma 53.1 (3), we have the following commutative diagram\n\n\[ \begin{matrix} {\pi }_{n}\left( {{S}^{n}, * }\right) \xrightarrow[]{{f}... | Yes |
Lemma 53.10. Let \( n \in \mathbb{N} \) and let \( \left( {X,{x}_{0}}\right) \) be a pointed topological space.\n\n(1) Let \( f : \left( {{\bar{B}}^{n},{S}^{n - 1}}\right) \rightarrow \left( {X,{x}_{0}}\right) \) be a map. If \( \rho : {\bar{B}}^{n} \rightarrow {\bar{B}}^{n} \) is the reflection in some hyperplane of \... | (1) Let \( \varphi : \left( {{I}^{n},\partial {I}^{n}}\right) \rightarrow \left( {{\bar{B}}^{n},{S}^{n - 1}}\right) \) be the homeomorphism from page 126 that we use to identify the two different points of view regarding the homotopy group \( {\pi }_{n}\left( {X,{x}_{0}}\right) \) . It follows from Proposition 40.1 (2)... | Yes |
Lemma 53.12. (*) Let \( n \in \mathbb{N} \), let \( M \) be an \( n \) -dimensional smooth manifold and let \( Z \) be a subset of \( M \) . Furthermore suppose we are given a topological space \( X \), a base point \( {x}_{0} \in X \) and maps \( {f}_{1},\ldots ,{f}_{m} : \left( {{\bar{B}}^{n},{S}^{n - 1}}\right) \rig... | Proof (*). As before it follows from Lemmas 2.40 and 2.17, Lemma 6.28 and Lemma 3.10 that the map\n\n\[ \nH : M \times \left\lbrack {0,1}\right\rbrack \rightarrow X \n\]\n\n\[ \n\left( {P, t}\right) \mapsto \Phi \left( {{\Omega }_{1, t},\ldots ,{\Omega }_{m, t},{f}_{1},\ldots ,{f}_{m}}\right) \left( P\right) \n\]\n\nis... | Yes |
Lemma 53.13. Let \( \left( {X,{x}_{0}}\right) \) be a pointed topological space and let \( n \in {\mathbb{N}}_{0} \) . Furthermore let \( k \in \{ 0,\ldots, n\} \) . The following statements hold:\n\n(1) We have \( {\mathrm{C}}_{k}^{\left( n\right) }\left( {X,{x}_{0}}\right) = 0 \) ,\n\n(2) we have \( {\mathrm{H}}_{k}^... | Proof. Let \( k \in \{ 0,\ldots, n\} \) . Put differently, we have \( k \) with \( n \geq k \) . Note that this implies that \( {\Delta }^{k} \) is the \( n \) -skeleton of \( {\Delta }^{k} \) . In particular, by definition, \( {\mathrm{C}}_{k}^{\left( n\right) }\left( {X,{x}_{0}}\right) \) is the subgroup of \( {\math... | Yes |
Lemma 53.16. Let \( \left( {X,{x}_{0}}\right) \) be a pointed topological space and let \( n \in {\mathbb{N}}_{ \geq 2} \) . Furthermore let \( \sigma : \partial {\Delta }^{n + 1} \rightarrow X \) be a map that sends the \( \left( {n - 1}\right) \) -skeleton of \( \partial {\Delta }^{n + 1} \) to \( {x}_{0} \) . Then | \[ \left\lbrack {\sigma : \left( {\partial {\Delta }^{n + 1}, * }\right) \rightarrow \left( {X,{x}_{0}}\right) }\right\rbrack = \mathop{\sum }\limits_{{j = 0}}^{{n + 1}}{\left( -1\right) }^{j} \cdot \left\lbrack {\sigma \circ {i}_{j}^{n + 1} : \left( {{\Delta }^{n},\partial {\Delta }^{n}}\right) \rightarrow \left( {X,{... | No |
Corollary 54.3. Let \( k \in \mathbb{N} \) . The sphere \( {S}^{k} \) does not admit a social choice for any \( n \in {\mathbb{N}}_{ \geq 2} \) . | Proof. By Corollary 53.6 we know that \( {\pi }_{k}\left( {S}^{k}\right) \cong \mathbb{Z} \) . Since \( \mathbb{Z} \) is neither zero nor infinitely generated we obtain from Proposition 54.2 (1) that \( {S}^{k} \) does not admit a social choice for any \( n \in {\mathbb{N}}_{ \geq 2} \) . | Yes |
Lemma 54.4. Let \( X \) be a path-connected topological space. If \( X \) admits a social choice of some type \( n \geq 2 \), then for every \( k \in \mathbb{N} \) the homotopy group \( {\pi }_{k}\left( X\right) \) admits a social homomorphism of type \( n \) . | Proof. To simplify the notation a little bit we only consider the case \( n = 2 \) . Thus let \( f : X \times X \rightarrow X \) be a social choice of type 2 . Let \( k \in \mathbb{N} \) . We fix a base point \( {x}_{0} \in X \) . We consider the map\n\n\[ \Theta : {\pi }_{k}\left( {X,{x}_{0}}\right) \times {\pi }_{k}\... | Yes |
Lemma 54.5. Let \( G \) be a group and let \( n \in {\mathbb{N}}_{ \geq 2} \). The following two statements are equivalent:\n\n(1) The group \( G \) admits a social homomorphism of type \( n \).\n\n(2) The group \( G \) is abelian and it is strongly divisible by \( n \). | Proof. In the following let \( G \) be a group and let \( n \in {\mathbb{N}}_{ \geq 2} \). Since we will shortly see that the groups involved are abelian we will use additive notation throughout the argument.\n\nFirst we deal with the implication \ | No |
Lemma 54.6. Let \( n \in {\mathbb{N}}_{ \geq 2} \) .\n\n(1) The group \( \mathbb{Z} \) is not strongly divisible by \( n \) .\n\n(2) Let \( m \in \mathbb{N} \) . The cyclic group \( {\mathbb{Z}}_{m} \) is strongly divisible by \( n \) if and only if \( m \) and \( n \) are coprime.\n\n(3) If a finitely generated group ... | Proof. The first two statements are basically trivial. Finally the last statement follows from the previous two statements together with the classification of finitely generated abelian groups, see Theorem 19.4. | No |
Proposition 54.8. For any \( n \in \mathbb{N} \) the map\n\nset of homotopy equivalence classes of maps \( {S}^{n} \rightarrow {S}^{n} \)\n\n\[ \n\deg : \overbrace{\left\lbrack {S}^{n},{S}^{n}\right\rbrack } \rightarrow \mathbb{Z} \n\]\n\n\[ \n\left\lbrack {f : {S}^{n} \rightarrow {S}^{n}}\right\rbrack \rightarrow \deg... | Proof. First note that we know by Lemma 45.11 (3) and (4) that the degree map is well-defined and that it is a monoid morphism. It remains to show that the degree map is a bijection. Now let \( * \in {S}^{n} \) be the standard base point. We consider the following diagram\n\n![448f61af-e517-4f9c-831f-f6ce5868f6c0_1351_... | Yes |
Lemma 54.10. Let \( G \) be a group and let \( R \) be a commutative ring.\n\n(1) The group ring \( R\left\lbrack G\right\rbrack \) is an associative ring with a multiplicatively neutral element,\n\nnamely \( 1 \cdot e \) where \( e \) denotes the trivial element in \( G \) . | Proof. All the statements are verified easily. Perhaps the most interesting statement is (5). We will not rob the reader of the pleasure of proving this in Exercise 54.3 | No |
Lemma 54.11. Let \( \langle T\rangle \) be the free group on a generating set \( T \) . Given any ring \( S \) and given any map \( g : T \rightarrow S \) there exists a unique ring homomorphism \( \varphi : \mathbb{Z}\left\lbrack {\langle T\rangle }\right\rbrack \rightarrow S \) with \( \psi \left( t\right) = g\left( ... | Proof. It follows immediately from Lemma 19.14 that there exists a unique group homomorphism \( \varphi : \langle T\rangle \rightarrow {S}^{ * } \mathrel{\text{:=}} \{ \) units of \( S\} \) that extends the given map \( g \) . It is now straightforward to verify that the map\n\n\[ \mathbb{Z}\left\lbrack {\langle T\rang... | Yes |
Lemma 54.13. Let \( G \) be a group and let \( A \) be an abelian group (with additive notation for the group structure). Suppose we are given a \( G \) -action on the group \( A \) . Then the map \[ \mathbb{Z}\left\lbrack G\right\rbrack \times A \rightarrow A \] \[ \left( {\mathop{\sum }\limits_{{i = 1}}^{m}{r}_{i}{g}... | Proof. The lemma is indeed elementary and trivial. We leave it to the reader to ponder what \ | No |
Theorem 54.17. (Serre) If \( X \) is a simply-connected topological space such that all homology groups are finitely generated (for example \( X \) could be a compact topological manifold or a finite CW-complex), then all homotopy groups are also finitely generated abelian groups. | In fact this follows from the \ | No |
Proposition 54.18. The group \( {\pi }_{3}\left( {{S}^{1} \vee {S}^{2}}\right) \) is not finitely generated over the group ring \( \mathbb{Z}\left\lbrack {{\pi }_{1}\left( {{S}^{1} \vee {S}^{2}}\right) }\right\rbrack = \mathbb{Z}\left\lbrack {t}^{\pm 1}\right\rbrack \) | A proof for the proposition is for example sketched in Hat02, Chapter 4.2, Exercise 38]. | No |
Lemma 55.3. Let\n\n\\[ \n{C}_{ * } \\mathrel{\\text{:=}} 0 \\rightarrow {\\mathrm{C}}_{k}\\overset{{\\partial }_{k}}{ \\rightarrow }{\\mathrm{C}}_{k - 1}\\overset{{\\partial }_{k - 1}}{ \\rightarrow }\\ldots {\\mathrm{C}}_{1}\\overset{{\\partial }_{1}}{ \\rightarrow }{\\mathrm{C}}_{0} \\rightarrow 0 \n\\]\n\nbe a chain... | Proof. For the above chain complex \\( \\left( {{C}_{n},{\\partial }_{n}}\\right) \\) we write for each \\( n \\in {\\mathbb{N}}_{0} \\)\n\n\\[ \n{Z}_{n} \\mathrel{\\text{:=}} \\ker \\left( {\\partial }_{n}\\right) ,\\;{B}_{n} \\mathrel{\\text{:=}} \\operatorname{im}\\left( {\\partial }_{n + 1}\\right) \\;\\text{ and }... | Yes |
Lemma 55.4. Let\n\n\\[ \n0 \rightarrow {A}_{k} \rightarrow {A}_{k - 1} \rightarrow \ldots \rightarrow {A}_{1} \rightarrow {A}_{0} \rightarrow 0 \n\\] \n\nbe an exact sequence of finitely generated abelian groups. Then\n\n\\[ \n\\mathop{\\sum }\\limits_{{n = 0}}^{k}{\\left( -1\\right) }^{n} \\cdot \\operatorname{rank}{A... | Proof. We can view this exact sequence of finitely generated abelian groups as a chain complex whose homology groups vanish. The lemma is therefore an immediate consequence of Lemma 55.3. | No |
Lemma 55.5. Let \( X = Y \cup Z \) be a decomposition of a finite CW-complex \( X \) into two subcomplexes \( Y \) and \( Z \) . Then the following equality holds\n\n\[ \chi \left( X\right) = \chi \left( Y\right) + \chi \left( Z\right) - \chi \left( {Y \cap Z}\right) . \] | Proof. We give two proofs for the lemma:\n\n(1) Given \( n \in {\mathbb{N}}_{0} \) and given a subset \( W \) we denote by \( n\left( W\right) \) the number of open \( n \) -cells that are contained in \( W \) . In our case we have\n\n\[ \begin{aligned} n\left( X\right) & = n\left( Y\right) + n\left( Z\right) - n\left(... | Yes |
(1) Let \( g \in {\mathbb{N}}_{0} \) and \( n \in {\mathbb{N}}_{0} \) . If we denote by \( {\sum }_{g, n} \) the surface \( {\sum }_{g} \) of genus \( g \) minus \( n \) open disks, then \( \chi \left( {\sum }_{g, n}\right) = 2 - {2g} - n \) . | (1) We proved this statement in the example preceding the lemma. | No |
Lemma 55.7. If \( M \) and \( N \) are two compact connected \( n \) -dimensional smooth manifolds. We have the following equality.\n\n\[ \chi \left( {M\# N}\right) = \chi \left( M\right) + \chi \left( N\right) + \left\{ \begin{matrix} - 2, & \text{ if }n\text{ is even,} \\ 0, & \text{ if }n\text{ is odd. } \end{matrix... | Proof \( \left( *\right) \) . We perform the following straightforward calculation: \n\nThe same way we also see that \( \chi \left( {N \smallsetminus {B}^{n}}\right) = \chi \left( N\right) + {\left( -1\right) }^{n... | Yes |
Lemma 55.8. Let \( X \) and \( Y \) be two finite CW-complexes, then\n\n\[ \chi \left( {X \times Y}\right) = \chi \left( X\right) \cdot \chi \left( Y\right) \] | Proof. Let \( X \) and \( Y \) be two finite CW-complexes. Given \( k \in {\mathbb{N}}_{0} \) we denote by \( {c}_{k} \) the number of \( k \) -cells of \( X \) and we denote by \( {d}_{k} \) the number of \( k \) -cells of \( Y \) .\n\nWe equip \( X \times Y \) with the product CW-structure from page [961]. By constru... | Yes |
Lemma 55.9. Let \( \sum \) be the surface of genus \( g \) and let \( p : \widetilde{\sum } \rightarrow \sum \) be a \( k \) -fold connected covering. Then \( \operatorname{genus}\left( \widetilde{\sum }\right) = k \cdot \left( {g - 1}\right) + 1. | Proof. We denote by \( \widetilde{g} \) the genus of \( \widetilde{\sum } \) . The lemma follows immediately from the following two facts:\n\n(1) By the remark on page 1363 we have \( \chi \left( \sum \right) = 2 - {2g} \) and \( \chi \left( \widetilde{\sum }\right) = 2 - 2\widetilde{g} \) .\n\n(2) By Proposition 37.4 ... | Yes |
Proposition 31.17. If \( g \geq 1 \), then the group\n\n\[ \n{\pi }_{1}\text{ (surface of genus }g\text{ ) } \cong \left\langle {{x}_{1},{y}_{1},\ldots ,{x}_{g},{y}_{g} \mid \left\lbrack {{x}_{1},{y}_{1}}\right\rbrack \cdots \cdots \left\lbrack {{x}_{g},{y}_{g}}\right\rbrack }\right\rangle \n\]\n\nis not isomorphic to ... | Proof. Let \( \sum \) be a surface of genus \( g \geq 1 \) . We write \( \pi = {\pi }_{1}\left( \sum \right) \) . Suppose that there exists an isomorphism \( \varphi : \pi \rightarrow F \) where \( F \) is the free group on \( {2g} \) generators. Let \( \alpha : \pi \rightarrow {\mathbb{Z}}_{k} \) be an epimorphism for... | No |
Proposition 55.10. If a finite group \( G \) acts freely and continuously on \( {S}^{2n} \), then \( G \) is either trivial or \( G \cong {\mathbb{Z}}_{2} \) . | Proof. Let \( G \) be a finite group that acts freely and continuously on \( {S}^{2n} \). By Lemma 16.5 the action is also discrete. By Proposition 16.9 the projection map \( {S}^{2n} \rightarrow {S}^{2n}/G \) is a covering of degree \( \left| G\right| \). We obtain the equalities\n\n\[ \begin{aligned} 2 & = \chi \left... | Yes |
Proposition 55.11. Let \( g \in {\mathbb{N}}_{0} \) . If a finite group \( G \) acts freely and continuously on the surface \( \sum \) of genus \( g \), then the order of \( G \) divides \( \chi \left( G\right) = 2 - {2g} \) . | Proof. The proof of Proposition [55.11] is verbatim the same as the proof of Proposition 55.10. The only change is that we need to use the fact, shown on page 1363, that \( \chi \left( \sum \right) = 2 - {2g} \) | Yes |
Lemma 55.12. Let \( G \) be a connected topological graph with \( v \) vertices and e edges. Then\n\n\[{\mathrm{H}}_{0}\left( G\right) \cong \mathbb{Z},\;{\mathrm{H}}_{1}\left( G\right) \cong {\mathbb{Z}}^{e - v + 1}\;\text{ and }\;{\mathrm{H}}_{i}\left( G\right) = 0\text{ for }i \geq 2.\] | Proof. Let \( G \) be a connected topological graph with \( v \) vertices and \( e \) edges. Recall that by the discussion on page 932 we can view \( G \) as a 1-dimensional CW-complex with \( {v0} \) -cells and \( e \) 1-cells.\n\nNow that we have developed so many techniques for computing homology groups it is perhap... | Yes |
Proposition 56.2. (Euler’s Formula) Let \( P \) be a \( k \) -dimensional convex polyhedron in \( {\mathbb{R}}^{k} \). (1) The interior \( \overset{ \circ }{P} \) of \( P \) is non-empty and all faces of \( P \) are of dimension \( \leq k - 1 \). (2) The boundary \( \partial P \) admits a CW-structure such that for eac... | Proof. We leave the pleasure of proving the first two statements to the reader as Exercise 56.1. Now we calculate that \[ \mathop{\sum }\limits_{{j = 0}}^{{k - 1}}{\left( -1\right) }^{j} \cdot \# j\text{-dimensional faces}\; = \;\chi \left( {\partial P}\right) \; = \;\chi \left( {S}^{k - 1}\right) \; = \;1 + {\left( -1... | No |
Theorem 56.3. Up to rotation and stretching the only regular convex 3-dimensional polyhedra in \( {\mathbb{R}}^{3} \) are the platonic solids illustrated in Figure 860 870 They have the following types of faces and valences and they have the following rotational symmetry groups: | Sketch of PROOF. We will not provide a full proof for the theorem. We refer to Cox48, Section 6.7 for a full proof that the list of five platonic solids is indeed complete. Furthermore we refer to Arm88, p. 40] or alternatively to [Aa08, Chapter 5.5] for a proof of the statement regarding the rotational symmetry groups... | No |
Proposition 56.6. Let \( G \) be a graph with \( v \) vertices and e edges. We denote by \( g \) the girth of \( G \) . If \( G \) is planar and if \( g < \infty \), then\n\n\[ e \leq \frac{g}{g - 2} \cdot \left( {v - 2}\right) \] | Sketch of PROOF. Let \( G \) be a graph with \( v \) vertices and \( e \) edges. We suppose that the girth \( g \) of \( G \) is finite. Furthermore we suppose that \( G \) is planar, i.e. we suppose that there exists an embedding \( \varphi : \left| G\right| \rightarrow {\mathbb{R}}^{2} \) . We write \( X = \varphi \l... | Yes |
Proposition 56.7. If a graph \( G \) contains the complete graph \( {K}_{5} \) or the complete bipartite graph \( {K}_{3,3} \) as a subgraph, then \( G \) itself cannot be planar. | Proof.\n\n(1) Let \( m \in {\mathbb{N}}_{ \geq 3} \) . It is straightforward to see that the complete graph \( {K}_{m} \) has \( m \) vertices and \( \frac{1}{2}m\left( {m - 1}\right) \) edges. Furthermore the girth is three. It is straightforward to verify that the inequality of Proposition 56.6 is only satisfied for ... | Yes |
Lemma 56.8. Let \( G \) be a graph. If \( {G}^{\prime } \) is a subdivision of \( G \), then the topological realizations of \( {G}^{\prime } \) and \( G \) are homeomorphic. | Proof. We leave it to the reader to provide the elementary proof of this lemma. | No |
Lemma 57.1. Let \( A \) and \( B \) be two abelian groups.\n\n(1) For any \( a,{a}^{\prime } \in A \) and \( b,{b}^{\prime } \in B \) we have\n\n\[ \left( {a + {a}^{\prime }}\right) \otimes b = a \otimes b + {a}^{\prime } \otimes b \in A \otimes B \]\n\n\[ a \otimes \left( {b + {b}^{\prime }}\right) = a \otimes b + a \... | Proof. Let \( A \) and \( B \) be two abelian groups.\n\n(1) This statement is an immediate consequence of the definition of the tensor product. | No |
Lemma 57.3. Let \( A, B, C \) and \( {A}_{i}, i \in I \) be abelian groups.\n\n(1) The maps\n\n\[ A \otimes B \rightarrow B \otimes A \]\n\n\[ \left( {{\bigoplus }_{i \in I}{A}_{i}}\right) \otimes B \rightarrow {\bigoplus }_{i \in I}{A}_{i} \otimes B \]\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i} \otimes {b}_{i} \... | Proof. We refer to [Lan93, Corollary XVI.2.2] for a proof for the distributivity in (1). The other statements in (1) to (3) follow fairly easily from the definitions and Lemma 57.1. We refer to [Mun84, Chapter 50] for details. A proof of the last statement is given in Bou07, Chapter II.6.3] or alternatively in [Mats89,... | No |
Lemma 57.4. Let \( f : A \rightarrow {A}^{\prime } \) and \( {f}^{\prime } : {A}^{\prime } \rightarrow {A}^{\prime \prime } \) be homomorphisms and let \( g : B \rightarrow {B}^{\prime } \) and \( {g}^{\prime } : {B}^{\prime } \rightarrow {B}^{\prime \prime } \) be homomorphisms between abelian groups. Then we have\n\n... | Proof. The statement follows immediately from the definitions. | No |
Lemma 57.5. Let \( R \) be a commutative ring.\n\n(1) Given any abelian group \( A \) the map\n\n\[ \left( {A \otimes R}\right) \times R \rightarrow A \otimes R \]\n\n\[ \left( {\mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i} \otimes {b}_{i}, r}\right) \; \mapsto \;\mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i} \otimes {b}_{i}r... | Proof. Once again the statements follow immediately from the definitions. | No |
Lemma 57.6. Let\n\n\[ 0 \rightarrow A\overset{i}{ \rightarrow }B\overset{p}{ \rightarrow }C \rightarrow 0 \]\n\nbe a short exact sequence and let \( G \) be an abelian group. If the short exact sequence splits, then\n\n\[ 0 \rightarrow A \otimes G\xrightarrow[]{i \otimes \mathrm{{id}}}B \otimes G\xrightarrow[]{p \otime... | Proof. According to Splitting Lemma 46.2 we can without loss of generality suppose that the short exact sequence is of the form\n\n\[ 0 \rightarrow A\xrightarrow[]{a \mapsto \left( {a,0}\right) }A \oplus C\xrightarrow[]{\left( {a, c}\right) \mapsto c}C \rightarrow 0. \]\n\nIf we tensor this short exact sequence with \(... | Yes |
Lemma 57.9. Let \( G \) be an abelian group. If \( G \) is free abelian, then tensoring with \( G \) is exact. | Proof. Let\n\n\[ 0 \rightarrow A\overset{\varphi }{ \rightarrow }B\overset{\psi }{ \rightarrow }C \rightarrow 0 \]\n\nbe a short exact sequence and let \( G \cong {\mathbb{Z}}^{\left( S\right) } \) be a free abelian group. It follows from Lemma 57.3 (1) and (3) that tensoring the above short exact sequence with \( G \c... | Yes |
Lemma 57.13. Let \( \alpha : H \rightarrow {H}^{\prime } \) be a homomorphism between abelian groups. Furthermore let \( {F}_{ * } \) be a free resolution of \( H \) and let \( {F}_{ * }^{\prime } \) be a free resolution of \( {H}^{\prime } \) .\n\n(1) There exists an extension of \( \alpha \) to the free resolutions \... | Proof \( \left( *\right) \) . Let \( {F}_{ * } \) be a free resolution of \( H \) and let \( {F}_{ * }^{\prime } \) be a free resolution of \( {H}^{\prime } \) . Statements (1) and (2) are both a straightforward consequence of some mild diagram chasing and the flexibility which Lemma 19.1 provides for defining homomorp... | Yes |
Lemma 57.14. Let\n\n\\[ \ldots \rightarrow {F}_{2}\overset{{f}_{2}}{ \rightarrow }{F}_{1}\overset{{f}_{1}}{ \rightarrow }{F}_{0}\overset{{f}_{0}}{ \rightarrow }H \rightarrow 0 \\]\n\nand\n\n\\[ \ldots \rightarrow {F}_{2}^{\prime }\overset{{f}_{2}^{\prime }}{ \rightarrow }{F}_{1}^{\prime }\overset{{f}_{1}^{\prime }}{ \r... | Proof. We pick an extension \( {\alpha }_{i}, i \in \mathbb{N} \) of the identity map id: \( H \rightarrow H \) .\n\n(1) We apply Lemma 57.13 to the free resolutions \( {F}_{ * }^{\prime } \) and \( {F}_{ * } \) (i.e. we swap the roles of the free resolutions) and we obtain maps \( {\left\{ {\beta }_{i}\right\} }_{i \i... | Yes |
Lemma 57.15. Let \( G \) and \( H \) be abelian groups, then there exists a natural isomorphism 883\n\n\[{\operatorname{Tor}}_{0}\left( {H, G}\right) \overset{ \cong }{ \rightarrow }H \otimes G.\] | Proof. Let \( G \) and \( H \) be abelian groups. Let\n\n\[ \cdots \rightarrow {F}_{2}\overset{{f}_{2}}{ \rightarrow }{F}_{1}\overset{{f}_{1}}{ \rightarrow }{F}_{0}\overset{{f}_{0}}{ \rightarrow }H \rightarrow 0 \]\n\nbe the canonical free resolution of \( H \). We know from Lemma 57.10 that the sequence\n\n\( \left( *... | Yes |
(1) Every abelian group \( H \) admits a free resolution of length 1 . | Let \( H \) be an abelian group. If \( H \) is finitely generated, then we already saw on page 1408 that \( H \) admits a free resolution of length 1 . Now suppose that \( H \) is any abelian group. We have the exact sequence\n\n\[ 0 \rightarrow \ker \left( {\alpha \left( H\right) : {\mathbb{Z}}^{\left( H\right) } \rig... | Yes |
Lemma 57.17. Let \( G, H,{\left\{ {\mathrm{H}}_{i}\right\} }_{i \in I} \) and \( {\left\{ {G}_{j}\right\} }_{j \in J} \) be abelian groups. Then the following holds:\n\n(1) There exists a natural isomorphism \( \operatorname{Tor}\left( {\oplus {\mathrm{H}}_{i}, G}\right) \cong \oplus \operatorname{Tor}\left( {{\mathrm{... | Proof.\n\n(1) For each \( i \in I \) we choose a free resolution \( {F}_{ * }^{i} \) of \( {\mathrm{H}}_{i} \) . Then \( \oplus {F}_{ * }^{i} \) is a free resolution\nof \( \mathop{\bigoplus }\limits_{i}{\mathrm{H}}_{i} \) . The statement follows easily from the natural isomorphisms\n\n\[ \left( {{\bigoplus }_{i}{F}_{j... | Yes |
Theorem 57.18. (The Algebraic Universal Coefficient Theorem) Let \( \\left( {{C}_{n},{\partial }_{n}}\\right) \) be a chain complex of free abelian groups and let \( G \) be an abelian group. Then for each \( n \\in {\\mathbb{N}}_{0} \) there exists a natural homomorphism \( p : {\\mathrm{H}}_{n}\\left( {C;G}\\right) \... | Proof. Let \( \\left( {{C}_{n},{\partial }_{n}}\\right) \) be a chain complex of free abelian groups. For each \( n \\in {\\mathbb{N}}_{0} \) we write as always \( {Z}_{n} \\mathrel{\\text{:=}} \\ker \\left( {\\partial }_{n}\\right) \) and \( {B}_{n} \\mathrel{\\text{:=}} \\operatorname{im}\\left( {\\partial }_{n + 1}\... | Yes |
Theorem 57.19. (Universal Coefficient Theorem) Let \( \left( {X, A}\right) \) be a pair of topological spaces and let \( G \) be an abelian group. Then for each \( n \in {\mathbb{N}}_{0} \) there exists a natural homomorphism \( {\mathrm{H}}_{n}\left( {X, A;G}\right) \rightarrow \operatorname{Tor}\left( {{\mathrm{H}}_{... | Proof. Let \( \left( {X, A}\right) \) be a pair of topological spaces and let \( G \) be an abelian group. As we pointed out on page 1120, the chain groups \( {\mathrm{C}}_{n}\left( {X, A}\right), n \in {\mathbb{N}}_{0} \), are free abelian groups. Thus we can apply the Universal Coefficient Theorem 57.18 and we immedi... | No |
Let \( f : \left( {X, A}\right) \rightarrow \left( {Y, B}\right) \) be a map between pairs of topological spaces and let \( G \) be an abelian group. Then\n\n\[ \n\text{the induced map}\n\]\n\n\[ \n{f}_{ * } : {\mathrm{H}}_{n}\left( {X, A}\right) \rightarrow {\mathrm{H}}_{n}\left( {Y, B}\right) \text{is}\; \Rightarrow ... | Proof. The corollary follows immediately from the naturality of the short exact sequence of the Universal Coefficient 57.19 together with the Five-Lemma 43.12. | No |
(1) Let \( \\left( {X, A}\\right) \) be a pair of topological spaces and let \( G \) be a subgroup of \( \\left( {\\mathbb{C}, + }\\right) \) . Then\n\nfor each \( n \\in {\\mathbb{N}}_{0} \) the map\n\n\[ \n\\mu : {\\mathrm{H}}_{n}\\left( {X, A;\\mathbb{Z}}\\right) \\otimes G \\rightarrow {\\mathrm{H}}_{n}\\left( {X, ... | (1) This statement is an immediate consequence of the Universal Coefficient Theorem 57.19 and the fact, obtained in Lemma 57.17 (4), that the torsion-groups are zero for any subgroup of \( \\left( {\\mathbb{C}, + }\\right) \) . | Yes |
Lemma 57.22. Let \( f : {S}^{n} \rightarrow {S}^{n} \) be a map and let \( k \in \mathbb{N} \) . Then the induced map\n\n\[ \n{f}_{ * } : {\mathrm{H}}_{n}\left( {{S}^{n};{\mathbb{Z}}_{k}}\right) \rightarrow {\mathrm{H}}_{n}\left( {{S}^{n};{\mathbb{Z}}_{k}}\right)\n\]\n\nis given by multiplication by \( \deg \left( f\ri... | Proof (*). Let \( f : {S}^{n} \rightarrow {S}^{n} \) be a map and let \( k \in \mathbb{N} \) . Since the short exact sequence of the Universal Coefficient Theorem 57.19 is natural we obtain the following commutative diagram\n\n\[ \n\begin{array}{l} 0\xrightarrow[]{\;}{\mathrm{H}}_{n}\left( {S}^{n}\right) \otimes {\math... | Yes |
Let \( \\left( {X, A}\\right) \) and \( \\left( {Y, B}\\right) \) be topological spaces and let \( G \) be an abelian group. Then\n\n\[ \n\\begin{matrix} {\\mathrm{H}}_{n}\\left( {X, A}\\right) \\text{ and }{\\mathrm{H}}_{n}\\left( {Y, B}\\right) \\text{ are } \\\\ \\text{ isomorphic for all }n \\in {\\mathbb{N}}_{0} \... | ## Remark.\n\n(1) Corollary 57.25 is in some sense a disappointment, it shows that homology groups with coefficients are not better at distinguishing topological spaces than ordinary homology groups.\n\n(2) Corollary 57.25 is also an almost immediate consequence of Propositions 49.1 and 49.2 | No |
Proposition 57.26. It is not possible assign to each topology space \( X \) an isomorphism\n\n\[ \n{\Phi }_{X} : {\mathrm{H}}_{n}\left( {X;{\mathbb{F}}_{2}}\right) \rightarrow {\mathrm{H}}_{n}\left( X\right) \otimes {\mathbb{F}}_{2} \oplus \operatorname{Tor}\left( {{\mathrm{H}}_{n - 1}\left( X\right) ,{\mathbb{F}}_{2}}... | Proof. So suppose instead that such a map \( {\Phi }_{X} \) exists for every topological space \( X \) . We consider the map\n\n\[ \nf : {\mathbb{{RP}}}^{2} = {\bar{B}}^{2}/z \sim - z \rightarrow {S}^{2} = {\bar{B}}^{2}/{S}^{1} \n\]\n\nfrom page 1420. We obtain the following diagram:\n\n\[ \n\overset{ \cong {\mathbb{F}... | Yes |
Theorem 57.27. Let \( \mathcal{H} \) and \( {\mathcal{H}}^{\prime } \) be two homology theories and let \( \varphi : {\mathcal{H}}_{0}\left( \star \right) \rightarrow {\mathcal{H}}_{0}^{\prime }\left( \star \right) \) be an isomorphism. Then there exists a natural isomorphism between \( \mathcal{H} \) and \( {\mathcal{... | Sketch of A PROOF. Let \( G \) be an abelian group. We start our proof with the following observation: Given a pair \( \left( {X, A}\right) \) of CW-complexes we saw in Corollary 36.35 (3) that we can equip the mapping cone \( \operatorname{Cone}\left( {i : A \rightarrow X}\right) \) of the inclusion map with a natural... | No |
Lemma 58.2. Let \( \mathcal{C} = \left( {{\mathrm{C}}_{ * },{\partial }_{ * }}\right) \) and \( {\mathcal{C}}^{\prime } = \left( {{\mathrm{C}}_{ * }^{\prime },{\partial }_{ * }^{\prime }}\right) \) be two chain complexes. Given \( n \in {\mathbb{N}}_{0} \) we define\n\n\[ \n{\left( \mathcal{C} \otimes {\mathcal{C}}^{\p... | Proof. We still have to show that the maps \( d \) are indeed boundary maps of a chain complex, i.e. we have to verify that \( d \circ d = 0 \). This is the case since for any \( {c}_{p} \in {\mathrm{C}}_{p} \) and \( {c}_{q}^{\prime} \in {\mathrm{C}}_{q}^{\prime} \) we have\n\n\[ \nd\left( {d\left( {{c}_{p} \otimes {c... | Yes |
Lemma 58.3. Let \( \mathcal{C} \) and \( \mathcal{D} \) be two chain complexes.\n\n(1) Given two chain maps \( f : \mathcal{C} \rightarrow \mathcal{D} \) and \( {f}^{\prime } : {\mathcal{C}}^{\prime } \rightarrow {\mathcal{D}}^{\prime } \) the induced map\n\n\[ f \otimes {f}^{\prime } : \mathcal{C} \otimes {\mathcal{C}... | Proof. The proof of the lemma is elementary. For completeness' sake and for a good conscience we provide the proof of the first statement. So let \( \mathcal{C} \) and \( \mathcal{D} \) be two chain complexes and let \( f : \mathcal{C} \rightarrow \mathcal{D} \) and \( {f}^{\prime } : {\mathcal{C}}^{\prime } \rightarro... | Yes |
Theorem 58.4. (Eilenberg-Zilber) Let \( X \) and \( Y \) be topological spaces. There exist natural maps \[ \Upsilon : {\mathrm{C}}_{ * }\left( X\right) \otimes {\mathrm{C}}_{ * }\left( Y\right) \rightarrow {\mathrm{C}}_{ * }\left( {X \times Y}\right) \] and \[ \Theta : {\mathrm{C}}_{ * }\left( {X \times Y}\right) \rig... | Proof. We postpone this technically intricate proof to Section 80.2 In that section the above statement follows from Theorem 80.2 together with the proof of the Eilenberg-Zilber Theorem provided on page 1962. | No |
Proposition 58.5. Let \( X \) and \( Y \) be two CW-complexes with finitely many cells in each dimension. Then there exists a natura 900 isomorphism \[ {C}_{ * }^{\mathrm{{CW}}}\left( {X \times Y}\right) \overset{ \cong }{ \rightarrow }{C}_{ * }^{\mathrm{{CW}}}\left( X\right) \otimes {C}_{ * }^{\mathrm{{CW}}}\left( Y\r... | Proof. Let \( X \) and \( Y \) be two CW-complexes with finitely many cells in each dimension. On page [960] we introduced a CW-structure on \( X \times Y \) where the \( n \) -cells of \( X \times Y \) are precisely of the form \( e \times f \) where \( e \) is a \( p \) -cell of \( X \) and \( f \) is an \( \left( {n... | No |
Lemma 58.6. Let \( \mathcal{C} \) and \( {\mathcal{C}}^{\prime } \) be two chain complexes. The map\n\n\[ \Omega : {\mathrm{H}}_{p}\left( \mathcal{C}\right) \otimes {\mathrm{H}}_{q}\left( {\mathcal{C}}^{\prime }\right) \rightarrow {\mathrm{H}}_{p + q}\left( {\mathcal{C} \otimes {\mathcal{C}}^{\prime }}\right) \]\n\n\[ ... | Proof. Let \( \mathcal{C} = \left( {{C}_{n},{\partial }_{n}}\right) \) and \( {\mathcal{C}}^{\prime } = \left( {{\mathrm{C}}_{n}^{\prime },{\partial }_{n}^{\prime }}\right) \) be two chain complexes. First note that if \( {c}_{i} \in {\mathrm{C}}_{p} \) and \( {c}_{i}^{\prime } \in {\mathrm{C}}_{q} \) are cycles, then\... | Yes |
Theorem 58.7. (Künneth Theorem for Chain Complexes) \( {}^{902} \) Let \( \mathcal{C} \) and \( {\mathcal{C}}^{\prime } \) be two chain complexes. If \( \mathcal{C} \) is free, i.e. if all chain groups of the chain complex \( \mathcal{C} \) are free abelian groups, then there exists a natural short exact sequence\n\n\[... | Proof. Let \( \mathcal{C} = \left( {{C}_{n},{\partial }_{n}}\right) \) and \( {\mathcal{C}}^{\prime } = \left( {{\mathrm{C}}_{n}^{\prime },{\partial }_{n}^{\prime }}\right) \) be two chain complexes such that all \( {C}_{n} \) are free abelian groups. As we will see, the proof of the theorem is quite similar to the pro... | No |
Proposition 59.1. Let \( p : \widetilde{X} \rightarrow X \) be a covering of degree \( k \in \mathbb{N} \) of a topological space \( X \) and let \( n \in {\mathbb{N}}_{0} \) . Then the following hold:\n\n(1) For any commutative ring \( R \) the transfer map \( {p}^{ * } : {\mathrm{C}}_{n}\left( {X;R}\right) \rightarro... | Proof. Let \( p : \widetilde{X} \rightarrow X \) be a \( k \) -fold covering of a topological space \( X \) .\n\n(1) Let \( c \in {\mathrm{C}}_{n}\left( {X;R}\right) \) be non-trivial. We write \( c = \mathop{\sum }\limits_{{i = 1}}^{n}{\sigma }_{i} \otimes {r}_{i} \) where \( {\sigma }_{1},\ldots ,{\sigma }_{n} \) are... | Yes |
Theorem 59.5. (Lusternik-Schnirelmann Theorem) Let \( {A}_{1},\ldots ,{A}_{n + 1} \) be closed subsets of \( {S}^{n} \) with \( {A}_{1} \cup \cdots \cup {A}_{n + 1} = {S}^{n} \) . Then there exists an \( i \in \{ 1,\ldots, n + 1\} \) such that \( {A}_{i} \) contains antipodal points, i.e. there exists a point \( x \) o... | Proof. Let \( {A}_{1},\ldots ,{A}_{n + 1} \) be closed subsets of \( {S}^{n} \) with \( {A}_{1} \cup \cdots \cup {A}_{n + 1} = {S}^{n} \) . Without loss of generality we can assume that all \( {A}_{i} \) are non-empty. (Indeed, at least one of the \( {A}_{i} \) is evidently non-empty, if one of them is empty we just re... | Yes |
Proposition 59.7. Suppose that for each \( t \in \mathbb{R} \) we have a measurable subset \( {X}_{t} \) of \( {\mathbb{R}}^{n} \) such that for each \( s \leq t \in \mathbb{R} \) we have \( {X}_{s} \subset {X}_{t} \) . Then \( \mathop{\bigcup }\limits_{{t \in \mathbb{R}}}{X}_{t} \) is also measurable and\n\n\[ \mathop... | Proof of Proposition 59.7. The proposition can be deduced easily from basic facts of the Lebesgue measure, see [Caro00, Chapter 16] or [Frie16b, Proposition 6.2 (6)]. | No |
Theorem 59.8. Given any \( n \in {\mathbb{N}}_{0} \) there exists a commutative ring \( R \) with \( s\left( R\right) = n \) . | Proof. We consider the ring \( R = \mathbb{R}\left\lbrack {{x}_{1}^{2},\ldots ,{x}_{n}^{2}}\right\rbrack /\left( {1 + {x}_{1}^{2} + \cdots + {x}_{n}^{2}}\right) \) . Evidently \( s\left( R\right) \leq n \) . We need to show that \( s\left( R\right) \geq n \) . So suppose that \( s\left( R\right) < n \) . This implies t... | Yes |
The statements of Lemma 60.1 (1) to (7) also hold for octonions. For completeness' sake we spell out the precise statements:\n\n(1) The octonions are a non-commutative algebra over \( \\mathbb{R} \) with identity.\n\n(2) For any \( z \\in \\mathbb{O} \) we have \( \\bar{z} + z \\in \\mathbb{R} \).\n\n(3) For all \( z, ... | Proof. As in the case of Lemma 60.1 the verification of (1) to (3) is totally elementary. The task of filling in the details is once again left to the reader. For later on it is helpful to consider the elementary prof of (4) in the notes:\n\n(4) Given \( \\left( {z, w}\\right) \\in \\mathbb{O} \) we calculate that\n\n\... | No |
Lemma 60.5. Let \( A \) be a finite-dimensional algebra over a field \( \mathbb{F} \) . The following two statements are equivalent:\n\n(1) \( A \) is a division algebra,\n\n(2) A has no zero-divisors, i.e. for any \( a, b \in A \) with \( a \cdot b = 0 \) we have \( a = 0 \) or \( b = 0 \) . | Proof. Let \( A \) be a finite-dimensional algebra over a field \( \mathbb{F} \) . Let \( a \in A \) be non-zero. Then the following holds:\n\nfor any \( y \in A \) there exists \( x \in A \) with \( {ax} = y \Leftrightarrow \) the map \( A \rightarrow A, x \mapsto {ax} \) is surjective\n\n\( \Leftrightarrow \) the map... | Yes |
Theorem 60.6. Any finite-dimensional division algebra over \( \mathbb{R} \) which is associative and which has an identity is isomorphic to \( \mathbb{R},\mathbb{C} \) or to \( \mathbb{H} \) . | Proof. The statement was first proved by Frobenius in 1878. A short self-contained proof is for example given in [Pal68]. | No |
Lemma 60.8. Given \( v, w \in {\mathbb{H}}^{n + 1} \smallsetminus \{ \left( {0,\ldots ,0}\right) \} \) we write \( v \sim w \) if there exists a non-zero \( h \in \mathbb{H} \) with \( h \cdot v = w \) . This defines an equivalence relation on \( {\mathbb{H}}^{n + 1} \smallsetminus \{ \left( {0,\ldots ,0}\right) \} \) ... | Proof.\n\n(1) It is clear that \( \sim \) is reflexive.\n\n(2) Since \( \mathbb{H} \) is a skew field we see that \( \sim \) is symmetric.\n\n(3) Finally suppose that \( u \sim v \) and \( v \sim w \) . Thus there exist \( g, h \in \mathbb{H} \smallsetminus \{ 0\} \) with \( {gu} = v \) and \( {hv} = w \) . Then\n\n\[ ... | Yes |
Lemma 61.1. Let \( K = \left( {V, S}\right) \) be an abstract simplicial complex. Furthermore let \( s \) and \( t \) be two simplices. Then either the intersection \( s \cap t \) is empty, or the intersection \( s \cap t \) is again a simplex and it is a face of \( s \) and \( t \) . | Proof (*). Let \( s \) and \( t \) be two simplices of an abstract simplicial complex \( K = \left( {V, S}\right) \) . If \( s \cap t = \varnothing \) then there is nothing to show. Now suppose that \( s \cap t \neq \varnothing \) . Evidently \( s \cap t \) is a non-empty subset of \( s \) and \( t \) . By definition o... | Yes |
Lemma 61.2. Let \( {\left\{ {L}_{i} = \left( {W}_{i},{T}_{i}\right) \right\} }_{i \in I} \) be a family of abstract simplicial complexes. The following statements hold:\n\n(1)\n\n\[ \mathop{\bigcap }\limits_{{i \in I}}{L}_{i} \mathrel{\text{:=}} \left( {\mathop{\bigcap }\limits_{{i \in I}}{W}_{i},\mathop{\bigcap }\limi... | Proof. The lemma is close to being a tautology. | Yes |
Lemma 61.3. Let \( K \) and \( L \) be two abstract simplicial complexes.\n\n(1) If \( K \) and \( L \) are finite, then there exist only finitely many simplicial maps from \( K \) to \( L \) . | Proof.\n\n(1) This statement is obvious. | No |
Lemma 61.4. Given any abstract simplicial complex \( K = \left( {V, S}\right) \) we have\n\n\[ \left| K\right| = \left\{ {\alpha \in {\mathbb{R}}^{\left( V\right) } \mid }\right. \]\n\n\[ \left. \begin{array}{ll} \text{ (1) } & \text{ the set }\{ v \in V \mid \alpha \left( v\right) \neq 0\} \text{ is a simplex of }K \\... | Proof. We leave the elementary task of proving the equality to the reader. Note though that one actually needs to use that \( K = \left( {V, S}\right) \) is an abstract simplicial complex. | No |
Lemma 61.5. Let \( K \) be an abstract simplicial complex.\n\n(1) A subset \( A \subset \left| K\right| \) is closed if and only if for every simplex \( s \) of \( K \) the preimage \( {\Phi }_{s}^{-1}\left( A\right) \) under a characteristic map \( {\Phi }_{s} \) is closed. | Proof. Let \( K = \left( {V, S}\right) \) be an abstract simplicial complex.\n\n(1) This statement follows almost immediately from the definition of the topology on\n\n\( \left| K\right| \) together with the elementary Lemma \( \left| \overline{1.3}\right| \left( 7\right) \) . | No |
Lemma 61.6. Let \( K = \left( {V, S}\right) \) be an abstract simplicial complex and let \( f : \left| K\right| \rightarrow X \) be a map to some topological space \( X \) .\n\n(1) If for each \( k \) -simplex \( s \in S \) with characteristic map \( {\Phi }_{s} : {\Delta }^{k} \rightarrow \left| K\right| \) the compos... | Proof.\n\n(1) This statement follows immediately from the definition of the topology on \( \left| K\right| \) .\n\n(2) This statement follows immediately from (1) and the observation that characteristic maps are continuous. | Yes |
Lemma 61.8. Let \( K = \left( {V, S}\right) \) be a subcomplex of some given abstract simplicial complex \( L = \left( {W, T}\right) \) . (1) The inclusion \( {\mathbb{R}}^{\left( V\right) } \rightarrow {\mathbb{R}}^{\left( W\right) } \) restricts to an inclusion \( \left| K\right| \rightarrow \left| L\right| \) . (2) ... | Proof. Let \( L = \left( {W, T}\right) \) be a simplicial complex and let \( K = \left( {V, S}\right) \) be a subcomplex. We denote by \( i : K \rightarrow L \) the inclusion map. (1) It follows immediately from the definitions that the inclusion \( {\mathbb{R}}^{\left( V\right) } \rightarrow {\mathbb{R}}^{\left( W\rig... | Yes |
Lemma 61.9. Let \( K = \left( {V, S}\right) \) be an abstract simplicial complex. If \( K \) is finite, then the following two conclusions hold:\n\n(1) The topological realization \( \left| K\right| \) is compact.\n\n(2) The topology of the topological realization \( \left| K\right| \), defined on page 1487, agrees wit... | Proof. Let \( K = \left( {V, S}\right) \) be a finite abstract simplicial complex.\n\n(1) By Lemma 61.5 we know in particular that for each \( k \) -simplex \( s \) the characteristic map \( {\Phi }_{s} : {\Delta }^{k} \rightarrow \left| K\right| \) is continuous. Since \( {\Delta }^{k} \) is compact we obtain from Lem... | Yes |
Lemma 61.10. Let \( K = \left( {V, S}\right) \) be a finite abstract simplicial complex and let \( \varphi : {\mathbb{R}}^{V} \rightarrow {\mathbb{R}}^{n} \) be a linear map such that the restriction of \( \varphi \) to \( \left| K\right| \) is an injection. Then \( \varphi : \left| K\right| \rightarrow {\mathbb{R}}^{n... | Proof. By Lemma 61.9 (1) we know that \( \left| K\right| \) is compact. Furthermore, note that it follows from Lemma 61.9 (2) that the linear map \( \varphi : {\mathbb{R}}^{V} \rightarrow {\mathbb{R}}^{n} \) restricts to a continuous map \( \varphi : \left| K\right| \rightarrow {\mathbb{R}}^{n} \) . Since \( \overline{... | Yes |
Lemma 61.11. Let \( K = \left( {V, S}\right) \) be an abstract simplicial complex.\n\n(1) Let \( s \in S \) be a \( k \) -simplex.\n\n(a) The closure of the open simplex \( \langle s\rangle \) is given by \( \left| s\right| \).\n\n(b) Every characteristic map \( {\Phi }_{s} : {\Delta }^{k} \rightarrow \left| s\right| \... | Proof (*).\n\n(1) Let \( s \in S \) be a \( k \) -simplex and let \( {\Phi }_{s} : {\Delta }^{k} \rightarrow \left| s\right| \) be a characteristic map. By Lemma 61.5 (3) we know that the map \( {\Phi }_{s} : {\Delta }^{k} \rightarrow \left| s\right| \) is a homeomorphism and that \( \left| s\right| \) is a closed subs... | Yes |
Proposition 61.12. Let \( L \) be an abstract simplicial complex.\n\n(1) If \( X \subset \left| L\right| \) is a compact subset, then there exists a finite subcomplex \( K \) of \( L \) such that \( X \subset \left| K\right| \) . | Proof. Let \( L = \left( {W, T}\right) \) be an abstract simplicial complex. Clearly (2) is just a special case of (1). Thus it suffices to prove (1). Now let \( X \subset \left| L\right| \) be a compact subset. For each simplex \( t \) of \( L \) with \( \langle t\rangle \cap X \neq \varnothing \) we pick a point \( {... | No |
Lemma 61.13. (*) Let \( L = \left( {W, T}\right) \) be an abstract simplicial complex and let \( X \subset \left| L\right| \) be a closed subset. If for every simplex \( t \in T \) we have either \( \langle t\rangle \cap X = \varnothing \) or \( \langle t\rangle \subset X \), then there exists a subcomplex \( K \) of \... | Proof \( \left( *\right) \) . We consider\n\n\[ \nV \mathrel{\text{:=}} W \cap X\;\text{ and }\;S \mathrel{\text{:=}} \{ t \in T \mid \langle t\rangle \subset X\} .\n\]\n\nClaim. We claim that \( \left( {V, S}\right) \) is a subcomplex of \( L = \left( {W, T}\right) \) .\n\nLet \( s \in S \) be a simplex and let \( r \... | Yes |
Lemma 61.14. Let \( Y \) be a simplicial complex.\n\n(1) Any \( k \) -simplex of \( Y \) is homeomorphic to \( {\Delta }^{k} \) . | (1) This statement is a reformulation of Lemma 61.5. | No |
Lemma 61.15. Let \( \\left( {X,\\left( {K = \\left( {V, S}\\right) ,\\Theta : \\left| K\\right| \\rightarrow X}\\right) }\\right) \) and \( \\left( {Y,\\left( {L = \\left( {W, T}\\right) ,\\Omega : \\left| L\\right| \\rightarrow Y}\\right) }\\right) \) be two simplicial complexes. Given a simplicial map \( \\varphi : K... | Proof. The map \( \\Phi \) is given by \( \\Omega \\circ \\left| \\varphi \\right| \\circ {\\Theta }^{-1} \) . Since \( \\Omega \) and \( \\Theta \) are homeomorphisms we see that \( \\Phi \) is also unique. | Yes |
Lemma 61.17. Let \( n \in \mathbb{N} \) . We denote by \( \left( {K = \left( {V, S}\right) ,\Theta : \left| K\right| \rightarrow {\left\lbrack 0,1\right\rbrack }^{n}}\right) \) the canonical simplicial structure of \( {\left\lbrack 0,1\right\rbrack }^{n} \) . We define a new simplicial complex \( \widetilde{K} = \left(... | Proof. It is elementary to see that the map \( \widetilde{\Theta } : \left| \widetilde{K}\right| \rightarrow {\mathbb{R}}^{n} \), once it is written down properly, is continuous and a bijection. It follows from Proposition 2.45 that the map is in fact a homeomorphism. | No |
Lemma 61.20. Let \( K \) and \( L \) be two abstract simplicial complexes. If one of \( K \) or \( L \) is empty, then by definition we have a natural homeomorphism \( \left| K\right| * \left| L\right| \cong \left| {K * L}\right| \). If \( K \) and \( L \) are non-empty, then the map\n\n\[ \Theta : \left( {\left| K\rig... | Proof. Let \( K = \left( {V, S}\right) \) and \( L = \left( {W, T}\right) \) be two abstract simplicial complexes. If one of \( K \) or \( L \) is empty, then the statement is true basically by definition. Thus let us now assume that \( K \) and \( L \) are both non-empty. We consider the map\n\n\[ \Theta : \overset{ =... | No |
Corollary 61.21. If \( X = \left( {K,\Theta : \left| K\right| \rightarrow X}\right) \) is a finite simplicial complex, then the suspension \( \sum \left( X\right) \) and the cone \( \operatorname{Cone}\left( X\right) \) admit natural simplicial structures where the corresponding abstract simplicial complexes are given ... | Proof. The corollary follows immediately from Lemma 61.20 together with the two examples on page 1505. | Yes |
Lemma 61.22. Every finite simplicial complex is simplicially isomorphic to a finite linear simplicial complex. | Proof. By definition a finite simplicial complex is homeomorphic to the topological realization \( \left| K\right| \) of a finite abstract simplicial complex \( K = \left( {V, S}\right) \) . In Lemma 61.9 we showed that we can view \( \left| K\right| \) as a linear simplicial complex in \( {\mathbb{R}}^{V} = {\mathbb{R... | Yes |
(1) Let \( K \) be a linear simplicial complex in \( {\mathbb{R}}^{n} \). (a) Every \( k \)-simplex \( s \) of \( K \) is of the form \( s = \left| \left\{ {{v}_{0},\ldots ,{v}_{k}}\right\} \right| \) for some distinct vertices \( {v}_{0},\ldots ,{v}_{k} \). (b) Every simplex of \( K \) is a convex subset of \( {\mathb... | Proof. (1) Almost all of these statements follow easily from the definitions. The only statement which needs a little bit of thought is the fact that (iii) implies (i). Since we will not make use of this statement we feel comfortable with the thought of leaving the details to the reader. | No |
Lemma 61.24. Every (ordered) simplicial complex admits a (natural) CW-complex structure where given any \( n \in \mathbb{N} \) the \( n \) -simplices of the simplicial structure are precisely the n-cells \( {}^{952} \) of the CW-structure. In particular the following statements hold:\n\n(1) The simplicial subcomplexes ... | SKETCH OF PROOF. Basically by definition of a simplicial complex it suffices to prove the statement for (ordered) abstract simplicial complexes and their topological realizations. The lemma now follows from fleshing out the following steps for a given (ordered) abstract simplicial complex \( K = \left( {V, S}\right) \)... | No |
Corollary 61.26. Let \( Y \) be a simplicial complex and let \( X \) be a subcomplex. If \( Y \) is finite, then there exists a finite simplicial complex, namely \( Y \cup \operatorname{Cone}\left( X\right) \), that is homotopy equivalent to \( Y/X \) . | Proof. Basically by definition we only need to deal with abstract simplicial complexes and their topological realizations. Thus let \( L = \left( {W, T}\right) \) be a finite abstract simplicial complex and let \( K = \left( {V, S}\right) \) be a subcomplex. We denote by \( i : K \rightarrow L \) the inclusion map. Rec... | Yes |
Proposition 61.27. Every (countable) regular CW-complex admits the structure of a (countable) simplicial complex. | Sketch of PROOF. We will not make use of this theorem, thus we refer to Geo08, Corol- \n\nHere we only provide a sketch of a sketch of the proof. Thus let \( X \) be a regular CW-complex. We turn inductively the sk... | No |
Proposition 61.28. There exists a finite CW-complex that does not admit a simplicial structure. | Proof. We consider the function\n\n\[\n\begin{aligned} f : \left\lbrack {0,1}\right\rbrack & \rightarrow \mathbb{R} & \text{ and we consider the map } & g : {\left\lbrack 0,1\right\rbrack }^{2} \rightarrow {\mathbb{R}}^{3} \\ t & \mapsto \left\{ \begin{array}{ll} 0, & \text{ if }t = 0, \\ t \cdot \cos \left( \frac{2\pi... | Yes |
Lemma 62.1. (*) Let \( K = \left( {V, S}\right) \) be an abstract simplicial complex.\n\n(1) The map \( S \mapsto {\mathbb{R}}^{\left( V\right) } \) given by \( s \mapsto \underline{s} \) is natural.\n\n(2) For simplices \( s, t \in S \) with \( s \neq t \) we have \( \underline{s} \neq \underline{t} \).\n\n(3) For eac... | Proof (*).\n\n(1) The most difficult part is to figure out what \ | No |
Lemma 62.4. For every \( k \in {\mathbb{N}}_{0} \) the barycentric subdivision of a linear simplicial complex in \( {\mathbb{R}}^{n} \) is again a linear simplicial complex in \( {\mathbb{R}}^{n} \) . | Proof. This statement follows almost immediately from the definition of the barycentric subdivision. | No |
Proposition 62.6. Let \( K = \left( {V, S}\right) \) be an abstract simplicial complex. The following three statements are equivalent:\n\n(1) The abstract simplicial complex \( K \) is locally finite.\n\n(2) For every vertex \( v \in V \) the star \( \operatorname{St}\left( {K,\{ v\} }\right) \) is a finite simplicial ... | Proof \( \left( *\right) \) . Let \( K = \left( {V, S}\right) \) be an abstract simplicial complex.\n\nWe start out with \( \left( 1\right) \Leftrightarrow \left( 2\right) \Leftrightarrow \left( 3\right) \) . This equivalence of statements follows almost immediately from the definitions and the observation that an abst... | Yes |
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