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I have a question about derived equivalences:

Let $k$ be a field and $A$ and $B$ two finite dimensional algebras over $k$. Let $F : D^-(A) \to D^-(B)$ be an equivalence of triangulated categories. Let $K^b(proj(A))$ (resp. $K^b(proj(B))$ ) be the subcategory consisting of the bounded complexes of finitely generated projective $A$-modules (resp. $B$-modules).

It is known that the equivalence $F$ restricts to an equivalence between these two categories (because they correspond to the categories of compact objects).

It is probably obvious, but I was wondering if the functor $F$ has to send a bounded complex of finitely generated injective modules to a bounded complex of finitely generated injective modules ?

In particular, is there a way to characterize the category of bounded complexes of finitely generated injective modules in the derived category $D^-(A)$ ?

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The objects of $D^b(\text{mod}(A))$ (i.e., the subcategory of $D^-(A)$ consisting of bounded complexes of finitely generated modules) are characterized (up to isomorphism) as the objects $Y$ such that, for every $X$ in $K^b(\text{proj}(A))$, $\operatorname{Hom}(X,Y[t])$ is finite dimensional for all $t$ and zero for all but finitely many $t$.

Then the bounded complexes of finitely generated injectives are characterized (up to isomorphism) as the objects $Z$ such that, for every $Y$ in $D^b(\text{mod}(A))$, $\operatorname{Hom}(Y,Z[t])$ is finite dimensional for all $t$ and zero for all but finitely many $t$.

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  • $\begingroup$ Thank you very much ! The only thing that I don't get is why the condition $Hom(X,Y[t])$ is finite dimensional for every $t$ implies that the terms of the complex $X$ are all finite dimensional. $\endgroup$ – commut Nov 27 '15 at 16:41
  • $\begingroup$ @commut Do you mean the terms of $Y$. If $\operatorname{Hom}(X,Y[t])$ is finite dimensional for all $X$ in $K^b(\text{proj}(A))$, then in particular this is true for $X=A$, and $\operatorname{Hom}(A,Y[t])$ is just the degree $t$ (or $-t$ depending on your conventions) homology of $Y$. So this implies that $Y$ has finite-dimensional homology in all degrees. And $\operatorname{Hom}(A,Y[t])$ zero for all but finitely many $t$ implies that the homology of $Y$ vanishes in all but finitely many degrees. $\endgroup$ – Jeremy Rickard Nov 27 '15 at 19:56
  • $\begingroup$ Yes I want to say Y. Thank you very much ! $\endgroup$ – commut Nov 27 '15 at 20:37

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