Singular chain complexes

Suppose we have a (finite or infinite) collection of objects that we call "cells". The collection $K$ is partitioned into sets of cells of various "dimensions" $Q_k,k=0,1,2,...$. They are understood as

• copies of (oriented) $k$-dimensional simplices $\Delta ^k$, or
• copies of (oriented) $k$-dimensional balls ${\bf B}^k$, or
• spaces homeomorphic to those.

However, the homeomorphisms are assumed but will remain unspecified.

Not every topological space can be directly built from cells. Instead we use continuous maps of these cells. A singular $k$-cell in topological space $X$ is a map $$\sigma :{\bf B}^k \rightarrow X.$$

One way to disassemble $X$ into singular cells is to take them all: $$X=\bigcup _k \bigcup _{\sigma \in Q_k} \text{im } \sigma.$$ There is a lot of redundancy.

We write $$S_k(X)= \{ \sigma :\Delta ^k \rightarrow X \}.$$ It is a chain complex we denote by $S(X)$ called the singular chain complex of $X$.

Any map $$f:X \rightarrow Y$$ induces a chain map (we use the same letter) $$f:S(X) \rightarrow S(Y),$$ defined simply as the composition: $$f_k(\sigma _k)=f\sigma _k,\forall \sigma _k \in S_k(X).$$

Theorem (Inclusion). (Rotman p. 71) Let $i:A\hookrightarrow X$ be the inclusion, then the induced chain map $i:S_k(A)\rightarrow S_k(X)$ is an injection for all $k\ge 0$.

Corollary (Exactness). (Rotman p. 90) Let $i:A\hookrightarrow X$ be the inclusion, then the following sequence is exact: $$0\rightarrow S(A)\rightarrow S(X) \rightarrow S(X)/i(S(A)) \rightarrow 0,$$ or simply $$0\rightarrow S(A)\rightarrow S(X) \rightarrow S(X,A) \rightarrow 0.$$

Corollary (Exactness Axiom of Homology). (Rotman p. 93) For each pair $(X, A)$ there are the connecting homomorphisms $$\partial _m: H_m (X,A) \rightarrow H_{m-1}(A),m=1,2,...,$$ with the following property: for any map of pairs $$f:(X,A) \rightarrow (Y,B)$$ the following commutative diagram has "long exact sequences" as rows: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccc} & ... & \ra{} & H_m(A) & \ra{i_m} & H_m(X) & \ra{j_m} & H_m(X,A) & \ra{\partial _m} & H_{m-1}(A) & \ra{} & ... \\ & & & \da{(f|_A)_m} & & \da{f_{m}} & & \da{f_{m}} & & \da{(f|_A)_{m-1}} &\\ & ... & \ra{} & H_m(B) & \ra{i_m} & H_m(Y) & \ra{j_m} & H_m(Y,B) & \ra{\partial _m} & H_{m-1}(B) & \ra{} & ... \\ \end{array} $$ where $i$ and $j$ are the inclusions.

Theorem (Intersection). (Rotman p. 90) Let $i:A\hookrightarrow X,j:B\hookrightarrow X$ be the inclusions, then $i(S(A)) < S(X), j(S(B)) < S(X)$ and $$i(S(A)) \cap j(S(B)) = S(k(A \cap B)),$$ where $k:A \cap B \hookrightarrow X$ is the inclusion, or simply $$S(A) \cap S(B) = S(A \cap B).$$

Theorem (Union). (Rotman p. 91) Suppose $X=\bigcup _{\alpha} X_{\alpha}$. Let $i^{\alpha}:X_{\alpha}\hookrightarrow X$ be the inclusions, then $i^{\alpha}(S(X_{\alpha})) < S(X)$ and $$\sum _{\alpha} i^{\alpha}(S(X_{\alpha})) = S(X),$$ or simply $$\sum _{\alpha} S(X_{\alpha})= S(X).$$

Corollary (Additivity). If a space is the disjoint union of a family of topological spaces $\{X_{\alpha}\}$: $$X = \coprod_{\alpha}{X_{\alpha}},$$ then its singular chain complex is the direct sum of their singular chain complexes: $$S(X) \cong \bigoplus_{\alpha} S(X_{\alpha}).$$

Corollary (Additivity Axiom of Homology). If space is the disjoint union of a family of topological spaces $\{X_{\alpha}\}$: $$X = \coprod_{\alpha}{X_{\alpha}},$$ then its homology is the direct sum of their homologies: $$H_m(X) \cong \bigoplus_{\alpha} H_m(X_{\alpha}).$$

Theorem (Dimension). If $P=\{ p\}$ is a one-point space, then its chain complex of the one-point space $P$ is acyclic. That is, the boundary operator $\partial$ of the chain complex $S(P)$ of $P$ satisfies $$\text{im } \partial _{k+1} = \ker \partial _{k},\forall k\ne 0.$$

Proof. (Rotman p. 68) The only singular $k$-simplex for $k>0$ is the constant $\sigma _k: \Delta ^k \rightarrow P, \sigma _k(x)=p,\forall x$. Therefore, $$S_k(P)=< \sigma _k > \cong {\bf Z }.$$

Consider the boundary operator: $$\partial _k \sigma _k= \Sigma _{i=0}^k\sigma _k \epsilon _i,$$ where $$\epsilon _i = \epsilon _i^k: \Delta ^{k-1} \rightarrow \Delta ^k$$ is the affine map taking the vertices $\{e_0,...,e_{k-1}\}$ of the first simplex to the vertices $\{e_0,...,\widehat{e_i},...,e_{k}\}$ of the second while preserving the ordering.

Since $\sigma _k \epsilon _i$ is a $(k-1)$-simplex in $P$, which has to be $\sigma _{k-1}$, we have $$\partial _k \sigma _k= \big[ \Sigma _{i=0}^k (-1)^i \big] \sigma _{k-1}.$$ Therefore, for $k>1$, we have

• if $k$ is odd, $\partial _k \sigma _k= 0$, so that $\partial _k$ is the zero homomorphism; and
• if $k$ is even, $\partial _k \sigma _k= \sigma _{k-1}$, so that $\partial _k$ is an isomorphism.

Then, in the diagram $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccc} S_{k+1}(P) & \ra{\partial _{k+1}} & S_k(P) & \ra{\partial _{k}} & S_{k-1}(P),\\ \end{array} $$ we have

• if $k$ is odd, then
  • $\partial _k$ is zero, so that $\ker \partial _k =S_k(P)$, and, since $k+1$ is even,
  • $\partial _{k+1}$ is an isomorphism, so that $\text{im } \partial _{k+1} =S_k(P)$ as well;
• if $k$ is even, then
  • $\partial _k$ is an isomorphism, so that $\ker \partial _k =0$, and, since $k+1$ is even,
  • $\partial _{k+1}$ is zero, so that $\text{im } \partial _{k+1} =0$ as well.

So our identity is satisfied. $\blacksquare$

Corollary (Dimension Axiom of Homology). For the one-point space $P$, we have $$H_n(P) = 0,n \neq 0.$$

Theorem (Homotopy of inclusions). The chain maps $$ i^1, i^0:S(X)\rightarrow S(X\times [0,1])$$ are induced by the (inclusion/embedding) maps: $$ i^1, i^0:X\rightarrow X\times [0,1],$$ given by $$i^q(x)=(x,q),q=0,1,$$ are chain homotopic.

Proof. Cone construction, Rotman p. 75. $\blacksquare$

Corollary (Homotopy). Homotopic maps induce chain homotopic chain maps. That is, if two maps are homotopic: $$f \sim g:(X, A) \rightarrow (Y,B),$$ then their induced chain maps are chain homotopic: $$S(f) \simeq S(g):S(X, A) \rightarrow S(Y,B).$$

Proof. From the previous theorem, Rotman p. 76. Or via Acyclic models, Rotman p. 244. $\blacksquare$

Corollary (Homotopy Axiom of Homology). Homotopic maps induce the same homomorphisms in homology. That is, if two maps are homotopic: $$f \sim g:(X, A) \rightarrow (Y,B),$$ then their induced homology maps are equal: $$H(f) = H(g):H(X, A) \rightarrow H(Y,B).$$

Theorem (Eilenberg-Zilber). There is a chain equivalence $$\zeta : S(X \times Y) \rightarrow S(X) \otimes S(Y)$$ that is natural in the sense that maps $f:X \rightarrow X',g:Y \rightarrow Y'$, induce chain maps that satisfy: $$\zeta (f \times g)(a \times b)=f(a) \otimes g(b).$$

Proof. Via acyclic models, Bredon p. 227. $\blacksquare$

Theorem (Refinement). If $(X,A)$ is a pair and $\gamma$ is an open cover of $X$, then the chain map generated by the inclusion of the sum of subcomplexes $$C(i) : \sum _{U \in \gamma} C(X \cap U, A \cap U) \to C(X, A)$$ is a chain equivalence.

Proof. Via subdivision, Bredon p. 227. Or via acyclic models, Schon, R. Acyclic models and excision. Proc. Amer. Math. Soc. 59 (1) (1976) 167--168 $\blacksquare$

Corollary (Excision Axiom of Homology). If $(X,A)$ is a pair and $U$ is a subset of $X$ such that the closure of $U$ is contained in the interior of $A$, then the inclusion map $$i : (X-U, A-U) \to (X, A)$$ induces an isomorphism in homology: $$H(i) : H(X-U, A-U) \to H(X, A)$$

Proof. Bredon p. 228. $\blacksquare$