Proposition 13.40.10. Let \mathcal{D} be a triangulated category. Let \mathcal{A} \subset \mathcal{D} and \mathcal{B} \subset \mathcal{D} be subcategories. The following are equivalent
\mathcal{A} is right admissible and \mathcal{B} = \mathcal{A}^\perp ,
\mathcal{B} is left admissible and \mathcal{A} = {}^\perp \mathcal{B},
\mathop{\mathrm{Hom}}\nolimits (A, B) = 0 for all A \in \mathcal{A} and B \in \mathcal{B} and for every X in \mathcal{D} there exists a distinguished triangle A \to X \to B \to A[1] in \mathcal{D} with A \in \mathcal{A} and B \in \mathcal{B}.
If this is true, then \mathcal{A} \to \mathcal{D}/\mathcal{B} and \mathcal{B} \to \mathcal{D}/\mathcal{A} are equivalences of triangulated categories, the right adjoint to the inclusion functor \mathcal{A} \to \mathcal{D} is \mathcal{D} \to \mathcal{D}/\mathcal{B} \to \mathcal{A}, and the left adjoint to the inclusion functor \mathcal{B} \to \mathcal{D} is \mathcal{D} \to \mathcal{D}/\mathcal{A} \to \mathcal{B}.
Proof.
The equivalence between (1), (2), and (3) follows in a straightforward manner from Lemmas 13.40.7 and 13.40.8 (small detail omitted). Denote v : \mathcal{D} \to \mathcal{A} the right adjoint of the inclusion functor i : \mathcal{A} \to \mathcal{D}. It is immediate that \mathop{\mathrm{Ker}}(v) = \mathcal{A}^\perp = \mathcal{B}. Thus v factors over a functor \overline{v} : \mathcal{D}/\mathcal{B} \to \mathcal{A} by the universal property of the quotient. Since v \circ i = \text{id}_\mathcal {A} by Categories, Lemma 4.24.4 we see that \overline{v} is a left quasi-inverse to \overline{i} : \mathcal{A} \to \mathcal{D}/\mathcal{B}. We claim also the composition \overline{i} \circ \overline{v} is isomorphic to \text{id}_{\mathcal{D}/\mathcal{B}}. Namely, suppose we have X fitting into a distinguished triangle A \to X \to B \to A[1] as in (3). Then v(X) = A as was seen in the proof of Lemma 13.40.7. Viewing X as an object of \mathcal{D}/\mathcal{B} we have \overline{i}(\overline{v}(X)) = A and there is a functorial isomorphism \overline{i}(\overline{v}(X)) = A \to X in \mathcal{D}/\mathcal{B}. Thus we find that indeed \overline{v} : \mathcal{D}/\mathcal{B} \to \mathcal{A} is an equivalence. To show that \mathcal{B} \to \mathcal{D}/\mathcal{A} is an equivalence and the left adjoint to the inclusion functor \mathcal{B} \to \mathcal{D} is \mathcal{D} \to \mathcal{D}/\mathcal{A} \to \mathcal{B} is dual to what we just said.
\square
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