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I know that one can define a chain complex for a CW complex X by taking the chain groups $C_n(X)$ as the free group generated by the $n$-cells, $C_n(X;\mathbb{Z}) = \mathbb{Z}\langle \sigma_1,\sigma_2,...,\sigma_l\rangle$, along with the usual boundary homomorphisms. I also know that we get a cochain complex, with coefficients in some abelian group $G$, by setting $C^n(X;G) = \hom(C_n(X), G)$, along with the usual boundary homomorphisms. My question is do we always have to assume the chain groups are of the form above; i.e., what if I wanted to compute the homology of a space $X$ with coefficients in $\mathbb{F}_2$ and then turn around and compute the cohomology of this space with coefficients in $\mathbb{F}_2$, can I simply set $C^n(X;\mathbb{F}_2) = \hom(C_n(X;\mathbb{F}_2), \mathbb{F}_2)$ and compute from there or does it only work as $C^n(X;\mathbb{F}_2) = \hom(C_n(X;\mathbb{Z}),\mathbb{F}_2)$? If so, why?

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$\hom(C_n(X;\mathbb{Z}),\mathbb{F}_2)\cong \hom(C_n(X;\mathbb{F}_2),\mathbb{F}_2)$ –  Grumpy Parsnip Apr 6 '13 at 1:18
    
Ok, but what if we consider say $\mathbb{F}_3$ for the chain groups and $\mathbb{F}_2$ for the cochain coefficients? Here we want have the isomorphism. So, can we either use $\mathbb{Z}$ for the chain and anything we want for the cochains or the same coefficients in both places, but nothing else in between? –  user71210 Apr 6 '13 at 1:34
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Of course you can define the group $\hom(C_n(X;G),H)$ for arbitrary abelian groups $G$ and $H$, but to what end? (Note that $\hom(C_n(X;F_2),F_3)=0$, for example.) –  Brad Apr 6 '13 at 1:41
    
So it's possible but you run the risk of losing a lot of information? –  user71210 Apr 6 '13 at 2:11
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