Let $R$ be a commutative ring, $I$ be an ideal of $R$ with $I^2 = 0$ and let $M, N$ be projective $R$-modules. Then $M/IM$ and $N/IN$ are projective $R/IR$-modules and we get a map $\mathrm{Hom}_R(M, N) \to \mathrm{Hom}_{R/I}(M/IM, N/IN)$. I want to prove that the restriction of this map to the set of isomorphisms is surjective, i.e. that every isomorphism $f \colon M/IM \to N/IN$ comes from an isomorphism $M \to N$. This is a step for proving that $K_0(R) = K_0(R/I)$.
So far, I found a map $h \colon M/IM \to N$ with $\pi_M \circ h = f$ which arises from the fact that $M$ is projective. But $h$ is not an isomorphism. Does anybody know how to proceed?
Let $N$ be an arbitrary module. The set of modules $M$ for which the map $\hom(M,N) \to \hom(M/IM,N/IM)$ is surjective is closed under direct sums, direct summands, and contains $R$. It follows that it contains all projective modules.