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I asked a question here:https://mathoverflow.net/questions/252724/how-to-prove-hom-ap-0-x-hom-ap-1-x-are-injective-right-r-modules. But there are no responses and I really want to know how to solve it. So I repost it here. Hope for any help.

Let $A$ be a finite dimensional k-algebra. Suppose $X$ is a left $A$-module such that every indecomposable projective or injective $A$-left module is isomorphic to a direct summand of $X$. Let $\dots \rightarrow P_1 \rightarrow P_0 \rightarrow X \rightarrow 0$ be a projective resolution of $X$. $R:=End_A(X)$. Then we get the exact sequence $$0 \rightarrow Hom_A(X,X)=R\rightarrow Hom_A(P_0,X)\rightarrow Hom_A(P_1,X)$$ of right $R$-modules and $R$-homomorphisms.

Suppose $X$ is projective and injective as right $R$-module. Since $P_0,P_1\in add(X)$, we know $Hom_A(P_0,X),Hom_A(P_1,X)$ are projective right $R$-modules. My question is how to prove $Hom_A(P_0,X),Hom_A(P_1,X)$ are also injective right $R$-modules?

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Let $P$ be a projective $A$-module. Then we have $P\oplus Q=A^{(I)}$, for some set $I$ (free module), so $\def\H{\operatorname{Hom}_A}\H(P,X)$ is a direct summand of $\H(A^{(I)},X)\cong(\H(A,X))^I\cong X^I$ (direct product).

A direct product of injective modules is injective, as well as direct summands thereof.

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  • $\begingroup$ The set of indices is I because it deals with an injective module? ;o) $\endgroup$ – Bernard Oct 24 '16 at 10:05

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