If A and B are finite dimensional k-algebras, k is a field. $_{A}G\in A-mod$ is a Gorenstein projective module, then we have $RHom_{A}(G,A)\simeq Hom(G,A)$ since $Ext_{A}^{i}(G,A)=0$ for any $i\in \mathbb{Z}^{+}$. It is easy to prove that for $P\in K^{b}(A-proj)$ we have $RHom(G,P)\simeq Hom(G,P)$ in $D(k)$. If $_{A}X_{B}$ is a bi-module complex and $_{A}X\in K^{b}(A-proj)$, we know $RHom_{A}(G,X)\simeq Hom(G,X)$ in $D(k)$. My question is if $RHom_{A}(G,X)\simeq Hom(G,X)$ in $D(B^{op})$?
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Yes. You can compute $RHom_A(G,X)$ as Hom-complex from a projective resolution $P^\bullet$ of $G$ to $X$. You have a natural map $Hom_A(G,X) \to RHom_A(G,X)$ induced by thinking of a map $G\to X_k$ as a map of the projective resolution $P^\bullet\to X$. You know this map is an quasi-isomorphism (apply the spectral sequence to the double complex of Homs, taking homology with respect to the complex $P^\cdot$ first; the next page of the sequence is $Hom_A(G,X)$ by your observation on higher Exts), so you just need to show that it's a B-module map. But that's clear from the construction.
answered Feb 1, 2021 at 1:34
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$\begingroup$ I'm sorry but here X is a complex，how can I prove the complex you get is isomorphic to Hom（G，X）in D（B）？ $\endgroup$ Commented Feb 1, 2021 at 2:16
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$\begingroup$ Complexes still have injective resolutions; I think what I wrote before was making this too complicated, though. The point is just that you have to write down what the map from Hom -> RHom is carefully enough, and it will be clear it's a B-module homomorphism. $\endgroup$– Ben Webster ♦Commented Feb 1, 2021 at 2:45
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$\begingroup$ You could also apply the argument in the current answer to a complex of injectives quasi-isomorphic to X built from an injective resolution of the individual modules X_k. $\endgroup$– Ben Webster ♦Commented Feb 1, 2021 at 2:47
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