# How does Matlis duality behave w.r.t. Hopfian and Co-hopfian modules?

Let $$(R,\mathfrak m, k)$$ be a Noetherian, complete, local ring. Let $$E$$ be an injective hull of $$k$$. We know that the Matlis duality functor $$D(-):= Hom_R(-, E)$$ gives an anti-equivalence between the category of Noetherian and Artinian $$R$$-modules.

My question is : How does Matlis duality behave w.r.t. Hopfian and Co-hopfian modules ? Under what suitable assumptions on $$M$$, can we conclude that $$M$$ is Hopfian (resp. Co-hopfian) if (or only if) $$D(M)$$ is Co-hopfian (resp. Hopfian) ?

NOTE: Hopfian means every surjective endomorphism is injective. Co-hopfian means every injective endomorphism is surjective

## 1 Answer

If $$D(M)$$ is Hopfian (Co-Hopfian), then $$M$$ is Co-Hopfian (Hopfian resp.). Let me prove one of these, the other being similar. So, assume that $$D(M)$$ is Hopfian and let $$f:M\to M$$ be injective. Then, we have an exact sequence, $$0\to D(M/f(M))\to D(M)\to D(M)\to 0$$, and thus by assumption, $$D(M/f(M))=0$$. Since $$E$$ is the injective hull of $$k$$, easy to see that $$M/f(M)=0$$.