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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 ?

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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.

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  • $\begingroup$ 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 ? $\endgroup$ – uno May 16 at 18:37
  • $\begingroup$ "Equivalence" means much more than "additive and exact"! In particular, an equivalence of categories is fully faithful. $\endgroup$ – Eric Wofsey May 16 at 19:16

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