Hopfian modules and equivalence of categories of modules

For a ring with unity (not necessarily commutative) $$R$$, let $$R$$-$$Mod$$ denote the category of left $$R$$-modules.

Let $$R,S$$ be two rings with unity and $$T: R$$-Mod $$\to S$$-Mod be an equivalence of categories ($$T$$ be co-variant). Is it true that $$M$$ is a Hopfian $$R$$-module if and only if $$T(M)$$ is an Hopfian $$S$$-module ?

In general, if $$R$$-Mod and $$S$$-Mod are equivalent as categories, then do we have a one-to-one correspondence between Hopfian $$R$$-modules and Hopfian $$S$$-modules ?

Yes, this is trivial: a module $$M$$ is Hopfian iff every epimorphism $$M\to M$$ is an isomorphism, which is preserved by any equivalence of categories.
• So here's my confusion: Let $M \in R$-Mod be Hopfian. Want to show $T(M)$ is Hopfian . Now $T$ being an equivalence, I know $T$ is additive and exact. So let $T(M) \to T(M)\to 0$, where the map is say $g$. We want to show $g$ is injective. How do I know $g=T(f)$ for some $f\in Hom(M,M)$ ? And even when I know this, how do I get back to $M \to M \to 0$ in $R$-Mod ? – uno May 16 at 18:37