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!
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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.
For the infinite rank case see this SE post.
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$\begingroup$ 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 ? $\endgroup$– palioJul 25, 2015 at 12:26
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$\begingroup$ 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 ? $\endgroup$– palioJul 25, 2015 at 12:28
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$\begingroup$ Also i don't understand your sentence "and dual of a (finite dimensional ) vector space is isomorphic to itself" could you please explain what do you mean ? $\endgroup$– palioJul 25, 2015 at 12:36
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$\begingroup$ I meant to say rank of the module, which in this case is same as the dimension of the vector space. $\endgroup$– ChesterXJul 25, 2015 at 14:19
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$\begingroup$ 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. $\endgroup$– ChesterXJul 25, 2015 at 14:24