More (co)homological constructions

 Preliminary exam prep

Recall a previous post (2016-09-16, "Complexes and their homology") that focused on constructing topological spaces in different ways and recovering the homology. Here we complete that task, introducing cellular homology. Recall a cell complex (or CW complex) $X$ was a sequence of skeleta $X_k$ for $k=0,\dots,\dim(X)$ consisting of $k$-cells $e^k_i$ and their attaching maps to the $(k-1)$-skeleton.

Cellular homology

Definition: The long exact sequence in relative homology for the pair $X_k,X_{k-1}$ shares terms with the long exact sequence for the pair $X_{k+1},X_k$, as well as $X_{k-1},X_{k-2}$. By letting $d_k$ be the composition of maps in different long exact sequences, for $k>1$, that make the diagram

commute, we get a complex of equivalence classes of chains
$$\cdots \to H_{k+1}(X_{k+1},X_k) \tov{d_{k+1}} H_k(X_k,X_{k-1})\tov{d_k} H_{k-1}(X_{k-1},X_{k-2})\to \cdots \to H_1(X_1,X_0)\tov{d_1} H_0(X_0) \tov{d_0} 0,$$
whose homology $H_k^{CW}(X) = \ker(d_k)/\text{im}(d_{k-1})$ is called the cellular homology of $X$. The map $d_1$ is the connecting map in the long exact sequence of the pair $X_1,X_0$, and $d_0=0$.

This seems quite a roundabout way of defining homology groups, but it turns out to be very useful. Note that for $k=1$, the map $d_1$ is the same as for a simplicial complex, hence

Theorem: In the context above,

1. for $k\>0$, $H^{CW}_k(X)\cong H_k(X)$;
2. for $k\>1$, $H_k(X_k,X_{k-1})=\Z^\ell$, where $\ell$ is the number of $k$-cells in $X$; and
3. for $k\>2$, $d_k(e^k_i) = \displaystyle\sum_j\deg(\underbrace{\dy e^k_i}_{S^{k-1}}\tov{f_{k,i}} X_{k-1}\tov{\pi} \underbrace{X_{k-1}/X_{k-1}-e^{k-1}_j}_{S^{k-1}})e^{k-1}_j$.

Example: Real projective space $\R\P^n$ has a cell decomposition with one cell in each dimension, and 2-to-1 attaching maps $\dy(e_k) =2X_{k-1}$ for $k>1$. This gives us a construction
$$X_0 = e_0, \hspace{1cm} X_1 = e_1 \bigsqcup_{\dy(e_1)=e_0} X_0, \hspace{1cm} X_2 = e_2 \bigsqcup_{\dy(e_2)=2e_1} X_1, \hspace{1cm} X_3 = e_3 \bigsqcup_{\dy(e_3)=2e_2} X_2, \dots$$ It is immediate that $d_0=d_1=0$, and for higher degrees, we have
$$d_k(e^k) = \deg(S^{k-1}\to \R\P^{k-1}\to S^{k-1})e^{k-1}.$$
Since this is a map between spheres, we may apply local degree calculations. The first part is the 2-to-1 cover, where every point in $\R\P^{k-1}$ is covered by two points from $S^{k-1}$, one in each hemisphere. One covers it via the identity, the other via the antipodal map. As long as we choose a point not in $\R\P^{k-2}\subset \R\P^{k-1}$, the second step doesn't affect these degree calculations. The antipodal map $S^{k-1}\to S^{k-1}$ has degree $(-1)^k$, hence for $a$ the antipodal map, the composition has degree
$$\deg(S^{k-1}\to \R\P^{k-1}\to S^{k-1}) = \deg(\id_{S^{k-1}}) + \deg(a_{S^{k-1}}) = 1+(-1)^k = \begin{cases} 2 & k\text{ even}, \\ 0 & k \text{ odd.}\end{cases}$$

Products in (co)homology

Recall that an $n$-chain on $X$ is a map $\sigma:\Delta^n\to X$, where $\Delta^n=[v_0,\dots,v_n]$ is an $n$-simplex. These form the group $C_n$ of $n$-chains. An $n$-cochain is an element of $C^n = \Hom(C_n,\Z)$, though the coefficient group does not need to be $\Z$ necessarily.

Definition: The diagonal map $X\to X\times X$ induces a map on cohomology $H^*(X\times X)\to H^*(X)$, and by Kunneth, this gives a map $H^*(X)\otimes H^*(X)\to H^*(X)$, and is called the cup product.

For $a\in H^p(X)$ and $b\in H^q(X)$, representatives of the class $a$ are in $\Hom(C_p,\Z)$ and representatives of the class $b$ are in $\Hom(C_q,\Z)$, though we will conflate the notation for the class with that of a representative. Hence for a $(p+q)$-chain $\sigma$ the cup product of $a$ and $b$ acts as
$$(a\smile b)\sigma = a\left(\sigma|_{[v_0,\dots,v_p]}\right)\cdot b\left(\sigma|_{[v_p,\dots,v_{p+q}]}\right).$$
Definition: The cap product combines $p$-cochains with $q$-chains to give $(q-p)$-chains, by
$$\begin{array}{r c l} \frown\ :\ H^p(X) \times H_q(X) & \to & H_{q-p}(X), \\\ (a, \sigma) & \mapsto & a\left(\sigma|_{[v_0,\dots,v_p]}\right)\cdot \sigma|_{[v_p,\dots,v_q]}. \end{array}$$
The cap product with the orientation form of an orientable manifold $X$ gives the isomorphism of Poincare duality.

Remark: Given a map $f:X\to Y$, the cup and cap products satisfy certain identities via the induced map on cohomology groups. Let $a,b\in H^*(Y)$ and $c\in H_*(X)$ be cochain and chain classes, for which
$$f^*(a\smile b) = f^*(a)\smile f^*(b), \hspace{1cm} a\frown f_*c = f_*(f^*a\frown c).$$
The first identity asserts that $f^*$ is a ring homomorphism and the second describes the commutativity of an appropriate diagram. The cup and cap products are related by the equation
$$a(b\frown \sigma) = (a\smile b)\sigma,$$ for $a\in H^p$, $b\in H^q$ and $\sigma\in C_{p+q}$.

References: Hatcher (Algebraic topology, Chapter 2.2), Prasolov (Elements of homology theory, Chapter 2)