Finitely Generated Homology Groups

Question

Let $X$ be a topological space and let assume that it's cohomology groups $H^n(X; \mathbb{Z})$ (therefore considered as $\mathbb{Z}$ modules) are finitely generated. That means by classification theorem for finitely genenerated abelian groups that for every $n$ we have $$H^n(X; \mathbb{Z}) \cong \mathbb{Z}^{r_n} \oplus \bigoplus_{p \text{ prime }} \oplus _{k \ge 1} (\mathbb{Z}/(p^k\mathbb{Z}))^{r_n ^{p^k}}$$

with appropiate $r_n$ and $r_n ^{p^k} \in \mathbb{N}$.

My question is if and how to see that in this case the homology groups $H_n(X; \mathbb{Z})$ are also finitely generated?

I tried to use the Universal Efficient Theorem (correctly U Cocoefficient Thm :) ) to get following exact sequence

$${\displaystyle 0\to \operatorname {Ext} _{\mathbb{Z}}^{1}(\operatorname {H} _{i-1}(X;\mathbb{Z}),\mathbb{Z})\to H^{i}(X;\mathbb{Z})\,{\overset {h}{\to }}\,\operatorname {Hom} _{\mathbb{Z}}(H_{i}(X;\mathbb{Z}),\mathbb{Z})\to 0}$$

From this I can only conclude that $\operatorname {Hom} _{\mathbb{Z}}(H_{i}(X;\mathbb{Z}),\mathbb{Z})$ is also finitely generated, not more.

From here I can only conclude that the free part of $H_{i}(X;\mathbb{Z})$ finitely generated. But what about the torsion part $T$. Here I used the expression $H_{i}(X;\mathbb{Z})= \mathbb{Z}^d \oplus T$. By the way: Does there exits such splitting for non finitely generated abelian groups? If yes, what I know about $T$?

Or is there another way to show that $H_n(X; \mathbb{Z})$ are finitely generated?

Answer 1

This is true but not elementary. Proposition 3F.12 in Hatcher's algebraic topology book proves

If $A$ is not finitely generated, then either $\text{Hom}(A,\Bbb Z)$ or $\text{Ext}(A,\Bbb Z)$ is uncountable. Hence if $H_n(X;\Bbb Z)$ is not finitely generated, then either $H^n(X;\Bbb Z)$ or $H^{n+1}(X;\Bbb Z)$ is uncountable.

In particular, finite generation (even countability!) of cohomology implies the same of homology.