Cohomology classes and representations

Question

Let $X$ be some topological space. Then by the universal coefficient theorem, for any abelian group $G$, we have a surjection $$H^n(X,G)\rightarrow \text{Hom}(H_n(X,\mathbb{Z}),G).$$ In a good amount of cases, namely when $\text{Ext}^1(H_n(X,\mathbb{Z}),G)\neq 0$, this surjection is not injective. However, this makes me curious if there exists some group $H$, such that we have an injection $$0\rightarrow H^n(X,G)\rightarrow \text{Hom}(H,G)$$ preferably where $H$ has some "geometric" meaning. In other words, can cohomology be seen as a subset of representations of some groups? If it helps I'm quite happy to assume that $X$ is a smooth compact manifold.

Answer 1

In general, there is no such $H$.

Specifically, if we set $X = \mathbb{R}P^2$, then $H^2(X;\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}$, where the $\mathbb{Z}/2\mathbb{Z}$ comes from the $\operatorname{Ext}$ term.

There is no group $H$ for which $\mathbb{Z}/2\mathbb{Z}$ injects into $\operatorname{Hom}(H,\mathbb{Z})$. This is because if $f:H\rightarrow \mathbb{Z}$ is non-trivial, then $d_n\circ f:H\rightarrow \mathbb{Z}$ is also non-trivial, where $d_n:\mathbb{Z}\rightarrow \mathbb{Z}$ is multiplication by $n$. That is, every element of $\operatorname{Hom}(H,\mathbb{Z})$ has infinite order.