Tools of (co)homology

 Preliminary exam prep

Let $X,Y$ be topological spaces, $G$ a group, and $R$ a unital commutative ring.

Defining homology groups

Theorem: If $(X,A)$ is a good pair (there exists a neighborhood $U\subset X$ of $A$ such that $U$ deformation retracts onto $A$), then for $i:A\hookrightarrow X$ the inclusion and $q:X\twoheadrightarrow X/A$ the quotient maps, there exists a long exact sequence of reduced homology groups
$$\cdots \to \widetilde H_n(A) \tov{i_*} \widetilde H_n(X) \tov{q_*} \widetilde H_n(X/A) \to \cdots.$$

Theorem: For any pair $(X,A)$, there exists a long exact sequence of homology groups
$$\cdots \to H_n(A) \to H_n(X) \to H_n(X,A) \to \cdots,$$
where the last is called a relative homology group. Hence $H_n(X,A)\cong \widetilde H_n(X/A)$ for a good pair $(X,A)$.

Theorem (Excision): For any triple of spaces $(Z,A,X)$ with $\text{cl}(Z)\subset \text{int}(A)$, there is an isomorphism $H_n(X-Z,A-Z)\cong H_n(X,A)$.

For any $x\in X$, the local homology of $X$ at $x$ is the relative homology groups $H_n(X,X-\{x\})$. By excision, these are isomorphic to $H_n(U,U-\{x\})$ for $U$ any neighborhood of $x$. If $X$ is nice enough around $x$ (that is, if $U\cong \R^k$), then these groups are isomorphic to $H_n(\R^k,\R^k-\{x\})\cong H_n(D^k,\dy D^k) = H_n(S^k)$.

Theorem (Mayer-Vietoris): For $X=A\cup B$, there is a long exact sequence of homology groups
$$\cdots \to H_n(A\cap B) \to H_n(A)\oplus H_n(B) \to H_n(X) \to \cdots,$$

and if $A\cap B$ is non-empty, there is an analogous sequence for reduced homology groups.

Extending with coefficients

Recall the $\Tor$ and $\Ext$ groups, which were, respectively, the left and right derived functors of, respectively, $\otimes$ and $\Hom$ (see post "Exactness and derived functors," 2016-03-20). Here we only need $\Tor_1$ and $\Ext^1$, which are given by, for any groups (that is, $\Z$-modules) $A$, $B$,
$$\begin{array}{r c c c l} \Tor(A,B) & = & H_1(\text{projres}(A)\otimes B) & = & H_1(A\otimes \text{projres}(B)), \\ \Ext(A,B) & = & H^1(\Hom(A,\text{injres}(B))) & = & H^1(\Hom(\text{projres}(A),B)). \end{array}$$
Note that $\Tor$ is symmetric in its arguments, while $\Ext$ is not. Recall that $\Tor_0(A,B)=A\otimes B$ and $\Ext^0(A,B) = \Hom(A,B)$.

Theorem (Universal coefficient theorem): There exist isomorphisms
$$\begin{array}{r c c c l} H_n(X;G) & \cong & \Hom(H^n(X),G)\oplus \Ext(H^{n+1}(X),G) & \cong & H_n(X)\otimes G\ \oplus\ \Tor(H_{n-1}(X),G),  \\ H^n(X;G) & \cong & \Hom(H_n(X),G)\oplus \Ext(H_{n-1}(X),G) & \cong & H^n(X)\otimes G\ \oplus\ \Tor(H^{n+1}(X),G). \end{array}$$

Here are some common $\Hom$, $\Tor$, and $\Ext$ groups:
$$\begin{align*} \Hom(\Z,G) & = G & \Tor(\Z,G) & = 0 & \Ext(\Z,G) & = 0 \\ \Hom(\Z_m,\Z) & = 0 & \Tor(G,\Z) & = 0 & \Ext(\Z_m,\Z) & = \Z_m \\ \Hom(\Z_m,\Z_n) & = \Z_{\gcd(m,n)} & \Tor(\Z_m,\Z_n) & = \Z_{\gcd(m,n)} & \Ext(\Z_m,\Z_n) & = \Z_{\gcd(m,n)} \\ \Hom(\Q,\Z_n) & = 0 & & & \Ext(\Q,\Z_n) & = 0 \\ \Hom(\Q,\Q) & = \Q & & & \Ext(G,\Q) & = 0 \end{align*}$$

Theorem (Künneth formula): For $X,Y$ CW-complexes, $F$ a field, and $H^k(Y;G)$ or $H^k(X;G)$ finitely generated for all $k$, there are isomorphisms, for all $k$,

$$H_k(X\times Y;F) \cong \bigoplus_{i+j=k} H_i(X;F)\otimes_FH_j(Y;F), \hspace{1cm} H^k(X\times Y;G) \cong \bigoplus_{i+j=k} H^i(X;G)\otimes_GH^j(Y;G)$$

Dualities

Theorem (Poincaré duality): For $X$ a closed $n$-manifold (compact, without boundary) that is $R$-orientable (consistent choice of $R$-generator for each local homology group), for $k=0,\dots,n$ there are isomorphisms

$$H^k(X;R)\cong H_{n-k}(X;R).$$

Note that a simply orientable manifold means $\Z$-orientable. A manifold that is not $\Z$-orientable is always $\Z_2$-orientable (in fact all manifolds are $\Z_2$-orientable).

Theorem (Alexander duality): For $X\subsetneq S^n$ a non-empty closed locally contractible subset, for $k=0,\dots,n-1$ there are isomorphisms
$$\widetilde H^k(X) \cong \widetilde H_{n-k-1}(S^n-X).$$

References: Hatcher (Algebraic topology, Chapters 2, 3), Aguilar, Gitler, and Prieto (Algebraic Topology from a Homotopical Viewpoint, Chapter 7)