I am looking for a proof of the following fact: let $C$ be a chain complex of real vector spaces, $C^*$ the dual cochain complex. Then $H^n(C^*) \cong H_n(C)^*$. That is, taking homology commutes with taking duals.
This fact is mentioned and used in Roberto Frigerio's book "Bounded cohomology of discrete groups", and referred to as the Universal Coefficient theorem. However, all the references I could find for the Universal Coefficient Theorem in cohomology are either in the algebraic topology setting, or in a general homological algebra setting, but with free abelian groups instead of vector spaces.
I should mention that although the book is about group cohomology, I really am looking for the general homological algebra statement: the book uses this fact for arbitrary chain complexes.
The statement which, from what I could find, I imagine is the general statement I should be looking for, is the following: let $R$ be a principal ideal domain, $M$ an $R$-module, $C$ a chain complex of free $R$-modules and $C^*$ its dual cochain complex. Then there is a short exact sequence:
$$0 \to \operatorname{Ext}^1_R(H_{n-1}(C), M) \to H^n(C^*) \to \operatorname{Hom}_R(H_n(C), R) \to 0.$$
This implies the statement I am looking for since the first term vanishes when $R = M = \mathbb{R}$.
Best Answer
If $k$ is a field and you're looking at $k-\mathbf{Vect}$, the category of $k$-vector spaces and linear maps, then $\hom(-,k) : k-\mathbf{Vect}\to k-\mathbf{Vect}$ is (contravariant) exact, and therefore commutes with homology, no need for the full universal coefficient theorem in this setting.
Note that the proof you found about free abelian groups can most likely be adapted verbatim to free $R$-modules for a general ring $R$, so even if you want the general statement of the universal coefficient theorem, the version for abelian groups is most likely enough for all rings (you just need to change "free abelian group" to "free $R$-module" in a few sentences)