The SES in question is $0\to \mathbb{Z}_2\to\mathbb{Z}_4\to\mathbb{Z}_2 \to 0$. For a given topological space $X$, I want to show that the following is short exact: $$0\to C^i(X;\mathbb{Z}_2)\to C^i(X;\mathbb{Z}_4)\to C^i(X;\mathbb{Z}_2)\to 0$$
I've already tried the method of showing that the original SES induces an SES on chain complexes $C_i(X;-)$, but I'm not sure how to get from there to cochain complexes.
Hint: this follows from the following fact : given any free abelian group $F$, $\hom(F,-)$ is exact, that is, for any short exact sequence $0\to A\to B\to C\to 0$, $0\to \hom(F,A)\to \hom(F,B)\to \hom(F,C)\to 0$ is exact.
This is very specific to free abelian groups (in fact with some work, one can prove that they're the only abelian groups for which it works - over a more general ring, it would be the projective modules that have this property, but for abelian groups, projective = free)