Injective module and the homology of a complex

Question

If $K$ is a complex of $R$-modules and $J$ is an injective $R$-module, prove that \begin{equation*} H^n(\operatorname{Hom}_R(K,J))\cong \operatorname{Hom}_R(H_n(K),J). \end{equation*}

Thank you in advance.

Answer 1

If $F:R-\textrm{Mod}\to R-\textrm{Mod}$ is an exact additive covariant functor, then $F$ commutes with (co)homology, that is, for every chain complex $K_\bullet$ we have $F(H_n(K_\bullet))\simeq H_n(F(K_\bullet))$.

Let $$K_\bullet: \cdots\to K_{n+1}\stackrel{f_{n+1}}\to K_n\stackrel{f_n}\to K_{n-1}\to\cdots0$$ be a chain complex. Then $H_n(K_\bullet)=\ker f_n/\operatorname{im}f_{n+1}$. Since $$0\to\operatorname{im}f_{n+1}\to\ker f_n\to H_n(K_\bullet)\to 0$$ is a short exact sequence and $F$ is exact we get that $$0\to F(\operatorname{im}f_{n+1})\to F(\ker f_n)\to F(H_n(K_\bullet))\to 0$$ is exact, and we are done.

Edit. If $A\stackrel{f}\to B$ then we have $0\to\ker f\to A\stackrel{f}\to B$ and $A\stackrel{f}\to B\to B/\operatorname{im}f\to 0$. We get $0\to F(\ker f)\to F(A)\stackrel{F(f)}\to F(B)$ and $F(A)\stackrel{F(f)}\to F(B)\to F(B/\operatorname{im}f)\to 0$. This gives $\ker F(f)=F(\ker f)$ and $\operatorname{im}F(f)=\ker F(\pi)=F(\ker\pi)=F(\operatorname{im}f)$.

We obtain similar properties for contravariant functors. If $A\stackrel{f}\to B$ then we have $0\to\ker f\to A\stackrel{f}\to B$ and $A\stackrel{f}\to B\to B/\operatorname{im}f\to 0$. We get $F(B)\stackrel{F(f)}\to F(A)\to F(\ker f)\to0$ and $0\to F(B/\operatorname{im}f)\to F(B)\stackrel{F(f)}\to F(A)$. This gives us $\operatorname{coker}F(f)=F(\ker f)$ and $\ker F(f)=F(\operatorname{coker}f)$.

Since $\operatorname{Hom}_R(-,J)$ is an exact contravariant functor one can rephrase your question as follows:

If $F:R-\textrm{Mod}\to R-\textrm{Mod}$ is an exact additive contravariant functor, then $F$ commutes with (co)homology, that is, for every chain complex $K_\bullet$ we have $F(H_n(K_\bullet))\simeq H^n(F(K_\bullet))$.

As before $H_n(K_\bullet)=\ker f_n/\operatorname{im}f_{n+1}$. Since $$0\to\operatorname{im}f_{n+1}\to\ker f_n\to H_n(K_\bullet)\to 0$$ is a short exact sequence and $F$ is exact we get that $$0\to F(H_n(K_\bullet))\to F(\ker f_n)\to F(\operatorname{im}f_{n+1}) \to 0$$ is exact. This shows that $$F(H_n(K_\bullet))=\ker(F(\ker f_n)\to F(\operatorname{im}f_{n+1}))=\ker(\operatorname{coker}F(f_n)\to F(\operatorname{im}f_{n+1}))=\ker(F(K_n)/\operatorname{im}F(f_n)\to F(\operatorname{im}f_{n+1}))=\ker F(f_{n+1})/\operatorname{im}F(f_n)=H^n(F(K_\bullet)).$$

(For the last but one equality we have used the sequence $0\to\operatorname{im}f_{n+1}\to K_n\to\operatorname{coker} f_{n+1}\to 0$.)