Let $(V, \partial)$ be a chain complex of $F$-vector spaces. I'm having trouble proving $\forall n \in \mathbb{Z}$: $$H^n(V^*, \partial^*) \cong H_n(V, \partial)^*$$ where the * denotes the dual vector space (i.e. $V^* = Hom(V, F)$).
Essentially this question is very similar to this 5 year old SE post. I understand the argument using the universal coefficient theorem (although one answer points out that this only works in the finitely generated case). Another answer suggests however, that the UCT is not needed here since the $Hom(\cdot, F)$ functor is exact in the case when $F$ is a field. This is what I want to understand. How does the argument with the exactness of the $Hom$-functor work? And under what circumstances do I still have an isomorphism in the non-finitely generated case?
I also found this question rather insightful, however I'm not sure how to translate it to the case of a contravariant functor (like $Hom(\cdot, F)$).
Thanks to @Najib Idrissi for linking to my answer! I hope it might be useful. It also explains why contravariance does not matter, but it does not matter for purely formal reasons: if you have a contravariant functor $F$, it is actually covariant functor $\mathbf{A}^\circ \to \mathbf{B}$ or $\mathbf{A} \to \mathbf{B}^\circ$, so it still preserves kernels, cokernels, and images, with the difference that kernels in the opposite category are cokernels, and vice versa. Note that the opposite of an abelian category is still abelian, but with all things dualized.
Let me just add a couple of points that you are asking about.
The contravariant functor $\operatorname{Hom}_k (-,k)\colon k\text{-Vect}^\circ \to k\text{-Vect}$ is exact by some linear algebra, and you can check this yourself. Alternatively ("Mathematics Made Difficult" :-), it is the same as saying that $k$ is an injective object in the category of $k$-modules. But over a field, every module is injective (hint: every short exact sequence of vector spaces splits), i.e. any functor $\operatorname{Hom}_k (-, W)$ is exact for any $k$-vector space $W$.
The assumption that the complex consists of finite dimensional vector spaces is not needed at any point to show that $$H^n (V_\bullet^\vee) \cong H_n (V_\bullet)^\vee.$$ It is needed to further say that $$H_n (V_\bullet)^\vee \cong H_n (V_\bullet),$$ since a finite dimensional vector space is isomorphic to its dual. However, this isomorphism is not canonical, so in my opinion it is morally wrong to use it (unless you are doing calculations and what you are interested in is that $\dim_k H^n (V_\bullet^\vee) = \dim_k H_n (V_\bullet)$).