Homology – Is $H^n(X,G)$ Isomorphic to $\operatorname{Hom}(H_n(X), G)$?

Question

Here is the question I am trying to understand it:

Prove or give a counter-example.

$(a)$ $H^1(X,G)$ is isomorphic to $\operatorname{Hom}(H_1(X), G),$ for any space $X.$

$(b)$ $H^n(X,G)$ is isomorphic to $\operatorname{Hom}(H_n(X), G),$ for any space $X.$

$(c)$ $H^1(\mathcal{C},G)$ is isomorphic to $\operatorname{Hom}(H_1(\mathcal{C}), G),$ for any chain complex $\mathcal{C}$ of free abelian groups.

How can I think about these statements? Should I start thinking using the universal coefficient theorem and when the ext functor vanishes or what(say for letter (c))? How about if I am speaking for spaces not chain complexes what should I do?

Any help will be greatly appreciated!

EDIT:

I think the only UCT that we have in Hatcher's for cohomology is for chain complexes of abelian groups …. But then we also know that the first cohomology group is just the Ext functor if we have a free resolution for the space….. But then what?

Answer 1

The universal coefficient theorem for cohomology (UCT) says that there is a short exact sequence of the form

$$0\to \text{Ext}^1(H_{n-1}(X),G)\to H^n(X,G)\to \text{Hom}(H_n(X), G)\to 0$$

for any abelian group $G$.

$(a)$ $H^1(X,G)$ is isomorphic to $\operatorname{Hom}(H_1(X), G),$ for any space $X.$

It is true. Because $H_0(X)$ is always free abelian group, and therefore $\text{Ext}^1(H_0(X),G)=0$ regardless of the choice of $G$.

$(b)$ $H^n(X,G)$ is isomorphic to $\operatorname{Hom}(H_n(X), G),$ for any space $X.$

That is false. For a counterexample pick $X$ such that $H_1(X)=\mathbb{Z}_2$ and $H_2(X)=0$, for example the real projective plane $\mathbb{R}\mathbb{P}^2$. Then $\text{Ext}^1(H_1(X),G)=G/2G$ from general properties of $\text{Ext}$. In particular if we choose $G=\mathbb{Z}$, then $$\text{Ext}^1(H_1(X),\mathbb{Z})=\mathbb{Z}_2$$ $$\text{Hom}(H_2(X),\mathbb{Z})=0$$

and therefore $H^2(X)=\mathbb{Z}_2$ which is not isomorphic to $\text{Hom}(H_2(X),\mathbb{Z})=0$.

$(c)$ $H^1(\mathcal{C},G)$ is isomorphic to $\operatorname{Hom}(H_1(\mathcal{C}), G),$ for any chain complex $\mathcal{C}$ of free abelian groups.

Take chain complex $\mathcal{C}$ associated with $\mathbb{R}\mathbb{P}^2$ from the previous example (which of course is of free abelian groups). Define a new chain complex $\mathcal{D}$ by $\mathcal{D}_n=\mathcal{C}_{n+1}$, and analogously we shift boundary operators. So it is just $\mathcal{C}$ with all indexes moved to the left by $1$.

Then $H_n(\mathcal{D})=H_{n+1}(\mathcal{C})$ and $H^n(\mathcal{D},G)=H^{n+1}(\mathcal{C}, G)$. And therefore $H^1(\mathcal{D},G)$ is not isomorphic to $\text{Hom}(H_1(\mathcal{D}), G)$ because $H^2(\mathcal{C},G)$ is not isomorphic to $\text{Hom}(H_2(\mathcal{C}), G)$ by previous counterexample.

Note that in general this construction cannot arise from a topological space (and that's why it doesn't contradict point (a)), for which $\mathcal{C}_n=0$ whenever $n<0$.