Lemma 13.39.5. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A} \subset \mathcal{D}$ be a full triangulated subcategory. The following are equivalent

the inclusion functor $\mathcal{A} \to \mathcal{D}$ has a right adjoint, and

for every $X$ in $\mathcal{D}$ there exists a distinguished triangle

\[ A \to X \to B \to X'[1] \]

in $\mathcal{D}$ with $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ and $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}^\perp )$.

If this holds, then $\mathcal{A}$ is saturated (Definition 13.6.1) and if $\mathcal{A}$ is strictly full in $\mathcal{D}$, then $\mathcal{A} = {}^\perp (\mathcal{A}^\perp )$.

**Proof.**
Assume (1) and denote $v : \mathcal{D} \to \mathcal{A}$ the right adjoint. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$. Set $A = v(X)$. We may extend the adjunction mapping $A \to X$ to a distinguished triangle $A \to X \to B \to A[1]$. Since

\[ \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(A', A) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(A', v(X)) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {D}(A', X) \]

for $A' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$, we conclude that $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}^\perp )$ by Lemma 13.39.2.

Assume (2). We will contruct the adjoint $v$ explictly. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$. Choose $A \to X \to B \to A[1]$ as in (2). Set $v(X) = A$. Let $f : X \to Y$ be a morphism in $\mathcal{D}$. Choose $A' \to Y \to B' \to A'[1]$ as in (2). Since $\mathop{\mathrm{Hom}}\nolimits (A, A') = \mathop{\mathrm{Hom}}\nolimits (A, Y)$ by Lemma 13.39.2 there is a unique morphism $f' : A \to A'$ such that the diagram

\[ \xymatrix{ A \ar[d]_{f'} \ar[r] & X \ar[d]^ f \\ A' \ar[r] & Y } \]

commutes. Hence we can set $v(f) = f'$ to get a functor. To see that $v$ is adjoint to the inclusion morphism use Lemma 13.39.2 again.

Proof of the final statement. In other to prove that $\mathcal{A}$ is saturated we may replace $\mathcal{A}$ by the strictly full subcategory having the same isomorphism classes as $\mathcal{A}$; details omitted. Assume $\mathcal{A}$ is strictly full. If we show that $\mathcal{A} = {}^\perp (\mathcal{A}^\perp )$, then $\mathcal{A}$ will be saturated by Lemma 13.39.3. Since the incusion $\mathcal{A} \subset {}^\perp (\mathcal{A}^\perp )$ is clear it suffices to prove the other inclusion. Let $X$ be an object of ${}^\perp (\mathcal{A}^\perp )$. Choose a distinguished triangle $A \to X \to B \to A[1]$ as in (2). As $\mathop{\mathrm{Hom}}\nolimits (X, B) = 0$ by assumption we see that $A \cong X \oplus B[-1]$ by Lemma 13.4.11. Since $\mathop{\mathrm{Hom}}\nolimits (A, B[-1]) = 0$ as $B \in \mathcal{A}^\perp $ this implies $B[-1] = 0$ and $A \cong X$ as desired.
$\square$

## Comments (0)