# With field coefficients homology and cohomology coincide

With field coefficients the universal coefficients theorem takes the form: $$H^n(X;F)=Hom_{F-modules}(H_n(X;F),F)$$. Now in all computations I have seen with field coefficients we have $$H^n(X;F)=H_n(X;F)$$. Is this true for any field or just for specific fields that we are usually working with, so my question can be reformulated as follows: for which fields $F$ do we have that $Hom_{F-modules}(H_n(X;F),F)=H_n(X;F)$? Thank you for your help!

Firstly, why are you writing $\hom_{F-modules}(H_n(X;F),F)$? UCT implies $$H^n(X;F)=\hom_{F}(H_n(X;F),F)$$.
Now, every $F$-module is a $F$-vector space and dual of a (finite dimensional) vector space is isomorphic to itself. So, $H^n(X;F)=\hom_F(H_n(X;F),F)=H_n(X;F)$, whenever $H_n(X;F)$ has finite rank.
• I'm writing so to say that it is the set of $F-$module homomorphisms from the $F-$module $H_n(X;F)$ to the $F-$module $F$. This is to differentiate it from group homomorphism or any other less structure so we need homomorphisms that respect $F-$module structure. If this is what you mean buy your simpler notation by putting $F$ instead of $F-$modules ? Jul 25, 2015 at 12:26
• Also $H_n(X;F)$ is an $F-$ module so when you talk about finite rank, you mean finite dimension as an $F-$vector space ? Jul 25, 2015 at 12:28
• As for your second comment, that statement is a fact from linear algebra. Also for an $F$-vector space $V$, $\hom_F(V,F)$ is same as $V^*$, the dual space. Jul 25, 2015 at 14:24