Lemma 13.36.3. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{D}' \subset \mathcal{D}$ be a full triangulated subcategory. The following are equivalent

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

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

\[ X' \to X \to Q \to X'[1] \]

in $\mathcal{D}$ with $X' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}')$ and $\mathop{\mathrm{Hom}}\nolimits (Y', Q) = 0$ for all $Y' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}')$.

If this holds, then $\mathcal{D}'$ is saturated (Definition 13.6.1).

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

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

for $Y' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}')$, we conclude that $\mathop{\mathrm{Hom}}\nolimits (Y', Q) = 0$ by Lemma 13.36.1 applied with $S = \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}')$.

Assume (2). We will contruct the adjoint $v$ explictly. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$. Choose $X' \to X \to Q_ X \to X'[1]$ as in (2). Set $v(X) = X'$. Let $f : X \to Y$ be a morphism in $\mathcal{D}$. Choose $Y' \to Y \to Q_ Y \to Y'[1]$ as in (2). Since $\mathop{\mathrm{Hom}}\nolimits (X', Y') = \mathop{\mathrm{Hom}}\nolimits (X', Y)$ by Lemma 13.36.1 there is a unique morphism $f' : X' \to Y'$ such that the diagram

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

commutes. Hence we can set $v(f) = f'$ to get a functor. To see that $v$ is adjoint to the inclusion morphism use Lemma 13.36.1 again.

To prove the final statement, suppose that $X_1 \oplus X_2$ is an object of $\mathcal{D}$. Choose a distinguished triangle

\[ X'_1 \to X_1 \to Q_1 \to X'_1[1] \]

as in (2). Then the map $X_1 \oplus X_2 \to Q_1$ is zero because $X_1 \oplus X_2$ is an object of $\mathcal{D}'$. This implies that $X_1 \to Q_1$ is zero too. Hence $X_1 \cong X'_1 \oplus Q_1[-1]$ (Lemma 13.4.10 and rotation). This means that $Q_1[-1]$ is a direct summand of $X_1 \oplus X_2$. Hence the identity of $Q_1$ factors through an object of $\mathcal{D}'$, which by the same arguments implies that $Q_1$ is zero as desired.
$\square$

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