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In the paper "Finite dimensional algebras and highest weight categories" of Cline, Parshall and Scott is stated as follows:

Let $\mathcal{B}$ be an abelian subcategory of a triangulated category $\mathcal{D}.$ Let $\mathcal{D'}=\mathcal{D}_{\mathcal{B}}$ be the full triangulated subcategory of $\mathcal{D}$ generated by $\mathcal{B}$ (in the sense that $\mathcal{D'}$ is the smallest strict triangulated subcategory of $\mathcal{D}$ containing $\mathcal{B}$). Assume that the inclusion $\mathcal{B}\subset \mathcal{D}$ is induced by an exact functor $i_*:D^b(\mathcal{B})\rightarrow \mathcal{D}$ which factors through $\mathcal{D'}$ and which is a full embedding. Then $i_*$ induces an equivalence $D^b(\mathcal{B})\cong \mathcal{D'}=\mathcal{D}_{\mathcal{B}}$ of triangulated categories.

From definition of $\mathcal{D'}$ we know that $\mathcal{D'}\rightarrow \mathcal{D}$ is a full embedding and $D^b(\mathcal{B})\rightarrow \mathcal{D}$ is a full embedding. I think it's also neccesary to use the information that $i_*$ factors through $\mathcal{D'}.$ I can to conclude that $D^b(\mathcal{B})\rightarrow \mathcal{D}$ is embedding, but it is, of course, not sufficient to conclude natural equivalence. I don't know how to obtain equivalence of categories given above.

I would be grateful for your advices and hints.

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  • $\begingroup$ I don't understand your problem : $D^b(B)\to D'$ is a full and faithful exact functor, its image is therefore a full triangulated subcategory containing $B$, so it's $D'$ : the functor is fully faithful and essentially surjective, hence an equivalence (of triangulated categories, since it is exact) EDIT : ah, maybe there are too many exact triangles in $D'$, is that it ? $\endgroup$ Oct 29, 2019 at 22:52
  • $\begingroup$ I can't see why it is full. It could be easy, but at the moment I don't see a reason for it. $\endgroup$
    – jpatrick
    Oct 29, 2019 at 22:54
  • $\begingroup$ You assumed it was ! $\endgroup$ Oct 29, 2019 at 23:00
  • $\begingroup$ I assumed it for $D^b(\mathcal{B})\rightarrow \mathcal{D}$, but $\mathcal{D'}$ is the full subcategory of $\mathcal{D}$, so it also follows for $D^b(\mathcal{B})\rightarrow \mathcal{D'}$ by simple manipulations on a definitions... It was very easy! And what is the problem about so many exact triangles? I didn't noticed this problem. $\endgroup$
    – jpatrick
    Oct 29, 2019 at 23:15
  • $\begingroup$ A priori there may be too many exact triangles in $D'$ : it's not clear a priori that if $F:T\to T'$ is an exact equivalence, then any quasi-inverse is too. The paper ( pcwww.liv.ac.uk/~arizzard/FourierMukai/…) says that a proof can be found in Huybrechts' "Fourier-Mukai transforms in algebraic geometry" so there is no actual problem, but it's not a priori immediate that it holds (I could get my hands on the book and indeed found the proof, it's not complicated, but nontrivial) $\endgroup$ Oct 29, 2019 at 23:31

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