Scalar extension preserves isomorphisms

Question

Let $A$ and $B$ be rings and $\rho: A\to B$ be a ring homomorphism which makes $B$ into a left A-module. If $M$ and $N$ are right $A$-modules such that there is an isomorphism of right $A$-modules

$$f:M\cong N,$$

is the map $f\otimes_AB$ between the scalar extensions $M\otimes_AB$ and $N\otimes_AB$ an isomorphism of right $B$ modules? That is, is

$$f\otimes_AB: M\otimes_AB\to N\otimes_AB$$

an isomorphism of right $B$-modules?

I think that the answer is positive since the scalar extension $(-)\otimes_AB$ is a covariant functor from the category of right $A$-modules to the category of right $B$-modules. Since functors preserve isomorphisms, the statement follows naturally.

Is that a sufficient argumentation or is there something more to it?

Answer 1

That's a sufficient argument: once you know $- \otimes_A B$ is a functor you know it preserves isomorphisms.