I have a exact sequence $1 \longrightarrow A \overset{\phi}\longrightarrow B \overset{\psi}\longrightarrow C \longrightarrow 1$ of commutative rings $A,B,C$ and the the exact sequence is such that we only consider the multiplicative structure then if I take a free abelian group $X$, how would I form $0 \longrightarrow Hom(X,A) \overset{\phi'}\longrightarrow Hom(X,B) \overset{\psi'}\longrightarrow Hom(X,C) \longrightarrow 0$ since in the second sequence the groups are under addition. I have seen this done many times when both sequences are of additive groups and you just define $\phi'\circ \alpha=\phi \circ \alpha$ and this works as both $\alpha,\phi$ are additive. But I cant see how to do this when $\phi$ is multiplicative and $\phi'$ has to be additive.
Or more generaly if $A,B,C,X$ where all $G$-modules for some finite group $G$