
A reference for this section is [Section 1, Bondal-Kapranov].

Lemma 13.36.1. Let $\mathcal{D}$ be a triangulated category. Let $S \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ be a subset invariant under all shifts. Consider a distinguished triangle

$X \to Y \to Z \to X[1]$

of $\mathcal{D}$. The following are equivalent

1. $\mathop{\mathrm{Hom}}\nolimits (A, Z) = 0$ for all $A \in S$, and

2. $\mathop{\mathrm{Hom}}\nolimits (A, X) = \mathop{\mathrm{Hom}}\nolimits (A, Y)$ for all $A \in S$.

Proof. By Lemma 13.4.2 the functor $\mathop{\mathrm{Hom}}\nolimits (A, -)$ is homological and hence we get a long exact sequence as in (13.3.5.1). Assume (1) and let $A \in S$. Then we consider the exact sequence

$\mathop{\mathrm{Hom}}\nolimits (A[1], Z) \to \mathop{\mathrm{Hom}}\nolimits (A, X) \to \mathop{\mathrm{Hom}}\nolimits (A, Y) \to \mathop{\mathrm{Hom}}\nolimits (A, Z)$

Since $A[1] \in S$ we see that the first and last groups are zero. Thus we get (2). Assume (2) and let $A \in S$. Then we consider the exact sequence

$\mathop{\mathrm{Hom}}\nolimits (A, X) \to \mathop{\mathrm{Hom}}\nolimits (A, Y) \to \mathop{\mathrm{Hom}}\nolimits (A, Z) \to \mathop{\mathrm{Hom}}\nolimits (A[-1], X) \to \mathop{\mathrm{Hom}}\nolimits (A[-1], Y)$

and we conclude that $\mathop{\mathrm{Hom}}\nolimits (A, Z) = 0$ as desired. $\square$

Lemma 13.36.2. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{D}'$ be a full triangulated subcategory of $\mathcal{D}$. For an object $X$ of $\mathcal{D}$ consider the property $P(X)$: there exists a distinguished triangle $X' \to X \to Y \to X'[1]$ in $\mathcal{D}$ with $X'$ in $\mathcal{D}'$ and $\mathop{\mathrm{Hom}}\nolimits (A, Y) = 0$ for all $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}')$.

1. If $X_1 \to X_2 \to X_3 \to X_1[1]$ is a distinguished triangle and $P$ holds for two out of three, then it holds for the third.

2. If $P$ holds for $X_1$ and $X_2$, then it holds for $X_1 \oplus X_2$.

Proof. Let $X_1 \to X_2 \to X_3 \to X_1[1]$ be a distinguished triangle and assume $P$ holds for $X_1$ and $X_2$. Choose distinguished triangles

$X'_1 \to X_1 \to Y_1 \to X'_1[1] \quad \text{and}\quad X'_2 \to X_2 \to Y_2 \to X'_2[1]$

as in condition $P$. Since $\mathop{\mathrm{Hom}}\nolimits (X'_1, X'_2) = \mathop{\mathrm{Hom}}\nolimits (X'_1, X_2)$ by Lemma 13.36.1 there is a unique morphism $X'_1 \to X'_2$ such that the diagram

$\xymatrix{ X'_1 \ar[d] \ar[r] & X_1 \ar[d] \\ X'_2 \ar[r] & X_2 }$

commutes. Choose an extension of this to a diagram

$\xymatrix{ X'_1 \ar[r] \ar[d] & X_1 \ar[r] \ar[d] & Q_1 \ar[r] \ar[d] & X'_1[1] \ar[d] \\ X'_2 \ar[r] \ar[d] & X_2 \ar[r] \ar[d] & Q_2 \ar[r] \ar[d] & X'_2[1] \ar[d] \\ X'_3 \ar[r] \ar[d] & X_3 \ar[r] \ar[d] & Q_3 \ar[r] \ar[d] & X'_3[1] \ar[d] \\ X'_1[1] \ar[r] & X_1[1] \ar[r] & Q_1[1] \ar[r] & X'_1[2] }$

as in Proposition 13.4.22. As $Q_1 \to Q_2 \to Q_3 \to Q_1[1]$ is a distinguished triangle we see that $\mathop{\mathrm{Hom}}\nolimits (A, Q_3) = 0$ for all $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}')$ because this is true for $Q_1$ and $Q_2$ and because $\mathop{\mathrm{Hom}}\nolimits (A, -)$ is a homological functor (Lemma 13.4.2). Since $\mathcal{D}'$ is a full triangulated subcategory, we see that $X'_3$ is isomorphic to an object of $\mathcal{D}'$. Thus $X_3$ satisfies $P$. The other cases of (1) follow from this case by translation. Part (2) is a special case of (1) via Lemma 13.4.10. $\square$

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

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

2. 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. $\square$

Lemma 13.36.4. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{D}' \subset \mathcal{D}$ be a full triangulated subcategory. The following are equivalent

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

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

$K \to X \to X' \to K[1]$

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

Proof. Omitted. Dual to Lemma 13.36.3. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).