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Let $F$ be a left-exact functor from an abelian category $A$ to an abelian category $B$. Let $A'$ be a subcategory of $A$ (not necessarily abelian) and $B'$ a subcategory of $B$ which is abelian such that $R^iF(X)$ is an object of $B'$, for any object of $A'$. I am looking for a way to associate to objects of $A'$ some object in the derived category of $B'$ in a natural way. As a possible candidate, I suggest $RF(X)$, where $X$ will be an object of $A'$. Under what hypothesis on $F$ (and/or $A$,$A'$,$B$,$B'$) such assertion will hold? If you can find any other natural way to construct elements in $D(B')$ (starting from the data $(F,A,A’,B,B’)$, I am also interested.

Let me give one counter-example:

Let $A=\mathrm{Ab}$, the category of abelian group, $A'=A$ subcategory of $\mathrm{Ab}$ containing the object $\mathbb{Z}/p^2$. $B = B' = $ the category of $\mathbb{F}_p$-vector spaces, and $F = \ker$ of the multiplication by $p$. Then, $R^0F = F$ and $R^1F = \operatorname{coker}$ of the multiplication by $p$ have their image in $B$. However, $RF(\mathbb{Z}/p^2)=[\mathbb{Z}/p^2 \dashrightarrow \mathbb{Z}/p^2]$ is not in $D(B)$.

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  • $\begingroup$ The question is hard to read. Could you use LaTeX? $\endgroup$ Oct 13, 2021 at 9:12
  • $\begingroup$ Dear @user12770, I converted your formulas to LaTeX. Please check if everything is as you intended. $\endgroup$
    – M.G.
    Oct 13, 2021 at 9:45
  • $\begingroup$ I don't think your counterexample is a counterexample: taking the reduction modulo p is a chain map from your complex to $\mathbb Z/p\xrightarrow{0}\mathbb Z/p$ which induces an isomorphism on homology groups and therefore is an equivalence in the derived category. A similar example with $B$ having cohomological dimension $\ge 2$ and $B'$ not closed under extensions should work, though (e.g. $A = B = \operatorname{Mod}(k[x,y]),F = \operatorname{Hom}(-,k), B'$ the subcategory where $x$ and $y$ act as the zero map, $A'$ containing the object $k$). $\endgroup$ Oct 13, 2021 at 10:49
  • $\begingroup$ Thank you very much to M.G. And to Bertram Arnold for their input! $\endgroup$
    – user12770
    Oct 13, 2021 at 13:56
  • $\begingroup$ This is a naive remark but taking $RF$ works if you take it to have values in $D_B'(B)$, the subcategory with cohomology sheaves in B'. I think that some counterexamples should lie in the coherent world : have you tried looking at the difference between the bounded derived category of coherent sheaves v.s. the subcategory of bounded sheaves of O_X-modules with coherent cohomology ? $\endgroup$ Oct 19, 2021 at 12:31

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