If $X$ is a CW complex with finite dimension, $H_k(X)$ is finitely generated for all $k$.

Question

I found the following result in a document without proof:

Then I found another document with a proof that I modified to get a proof for the first document's result:

But I think this can be used to prove that $H_k(X)$ is finitely generated for $k>n$ (since $0$ is finitely generated) and for $k=n$, since $H_n(X)=\bigoplus_{c_n} \Bbb Z$ (I used $c_n$ to denote the number of $n$ cells), but not for $k<n$ (is that right?) So I tried doing this to find $H_{n-1}(X)$:

Consider the long exact sequence for $(X^n,X^{n-1})$:

$H_n(X^n, X^{n-1}) \longrightarrow H_{n-1}(X^{n-1}) \longrightarrow^g H_{n-1}(X^n) \longrightarrow H_{n-1}(X^n,X^{n-1})$

The first group is $\bigoplus_{c_n} \Bbb Z$. The second group is $\bigoplus_{c_{n-1}} \Bbb Z$ by (a) and the last gruoup is $0$ because $n-1 \neq n$. So, $g$ is surjective and by the first isomorphism theorem, $H_{n-1}(X)=H_{n-1} (X^n) \simeq \bigoplus_{c_{n-1}} \Bbb Z / \ker g$.

I got stuck here because I don't know that $\ker g$ is. I got the feeling that I'm taking the wrong way...

Answer 1

You don't need to know what $g$ is or what $\ker g$ is. A $\mathbb{Z}$-module $M$ is finitely generated exactly when there is a surjective morphism $\mathbb{Z}^{\oplus n} \to M$. Then $M$ will be generated by the images of $(0,\dots,0,1,0,\dots,0)$.

The idea here is quite simple: the $k$-cells of $X$ generate $H_k(X)$. There may be relations among those cells, some of those cells may be homotopic to $0$, but certainly every $k$-homology class is a sum of $k$-cells and if there are finitely many $k$-cells then it is finitely generated.

Answer 2

It is not true. Let $X$ be an infinite discrete space. This is a $0$-dimensional CW-complex such that $H_0(X)$ is not finitely generated.

However, your question differs from its title. What you want to prove is that if $X$ is dimension-wise finite, then all $H_k(X)$ are finitely generated. A dimension-wise finite CW-complex may be infinite-dimensional, but it has only finitely many cells in each dimension.

Your proof is essentially correct, but has a little gap.

You show correctly that $H_{n-1}(X^n)$ is a quotient of the free abelian group $H_{n-1}(X^{n-1})$ with $c_{n-1}$ generators which follows from (a) and (b). By the way, you do not need the first group for this conclusion. This quotient is then trivially finitely generated. See the examples in https://en.wikipedia.org/wiki/Finitely_generated_group.

The gap is that this not show immediately that $H_{n-1}(X)$ is finitely generated. To fill the gap, we observe that $H_{n-1}(X^n) = H_{n-1}(X^{n+m})$ for each $m \ge 0$. This follows by induction on $m$ by considering the exact sequence $$H_n(X^{n+m+1}, X^{n+m}) \to H_{n-1}(X^{n+m}) \to H_{n-1}(X^{n+m+1}) \to H_{n-1}(X^{n+m+1}, X^{n+m}) $$ in which the first and the last group vanish.

But now each element of $H_{n-1}(X)$ is the homology class $[c]$ of a singular $(n-1)$-cycle $c$. Since singular simplices are defined on a compact space, their images are compact and thus contained in some finite skeleton of $X$. We conclude that the whole cycle $c$ is contained in some $X^{n+m}$, i.e $[c]$ is in the image of $H_{n-1}(X^{n+m}) \to H_{n-1}(X)$. Since $H_{n-1}(X^n) = H_{n-1}(X^{n+m})$, we conclude that $[c]$ is in the image of $H_{n-1}(X^n) \to H_{n-1}(X)$. Thus this homomorphism is surjective and we conclude that $H_{n-1}(X)$ is finitely generated.

Although it is irrelevant here, let us observe that $H_{n-1}(X^n) \to H_{n-1}(X)$ is an isomorphism. In fact, if an $(n-1)$-cycle $d$ in $X^n$ bounds an $n$-chain in $X$, this chain lies in some $X^{n+m}$ and we see that $[d] = 0$ via $H_{n-1}(X^n) = H_{n-1}(X^{n+m})$.