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Let $A$ be an finite dimensional algebra over a field K. Given a short exact sequence $0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0$ of $A$-modules, we know the functor $Hom_A(X,-)$ is a left exact functor, where $X$ is a $A$-module. So the sequence $0 \rightarrow Hom_A(X,L) \rightarrow Hom_A(X,M) \rightarrow Hom_A(X,N)$ is still exact. Now Suppose there is a infinite exact sequence of $A$-modules $0 \rightarrow M_0 \rightarrow M_1 \rightarrow M_2 \rightarrow \dots$, I think the sequence $0 \rightarrow Hom_A(X,M_0) \rightarrow Hom_A(X,M_1) \rightarrow Hom_A(X,M_2) \rightarrow \dots$ is also exact. Is that right?

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  • $\begingroup$ I don't think that the infinite exact sequence stays exact, otherwise would be $\mathrm{Hom}_A(X,-)$ not only left exact, but exact on the nose, which is not true. You might find counterexamples for this doing research in MSE. $\endgroup$ – user213008 Nov 13 '16 at 13:20
  • $\begingroup$ @user213008 Thank you. I have tried it. And I think you are right. $\endgroup$ – Xiaosong Peng Nov 14 '16 at 1:33

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