Inducing a module homomorphism

Question

Suppose that I have a homomorphism of $A$-modules $M \xrightarrow{f} N$ ($A$ commutative with 1), a ring homomorphism $A \xrightarrow{g} A'$, and two $A'$-modules $M'$ and $N'$. Suppose further that I have two abelian group homomorphisms $M \xrightarrow{\phi} M'$ and $N \xrightarrow{\psi} N'$ which are compatible with the action of $A$. By this I mean that $\phi(ax)=g(a)\phi(x)$ for all $a \in A$ and $x \in M$, and similarly for $\psi$.

Question: Do these data imply the existence of an $A'$-module homomorphism $M' \to N'$?

Answer 1

I guess, you would expect the morphism $M'\to N'$ to be nonzero.

Well, this is not the case in general. Consider e.g. $A=A'=M=N=N'=\Bbb Z$ and $M':=\Bbb Z/2\Bbb Z$, and take the identity for every morphism except for $\phi$.

However, if we had $\phi$ in the other direction: $M'\overset\phi\to M$, and $g$ is surjective, then we get the composite map $\psi\circ f\circ\phi:M'\to N'$ which will be compatible with the action of $A'$.

$\psi(f(\phi(ga\cdot x))) = \psi(f(a\cdot \phi(x)))=\psi(a\cdot f(\phi(x)))=g(a)\cdot \psi(f(\phi(x))).$