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

Definition 13.40.1. Let $\mathcal{D}$ be an additive category. Let $\mathcal{A} \subset \mathcal{D}$ be a full subcategory. The right orthogonal $\mathcal{A}^\perp$ of $\mathcal{A}$ is the full subcategory consisting of the objects $X$ of $\mathcal{D}$ such that $\mathop{\mathrm{Hom}}\nolimits (A, X) = 0$ for all $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. The left orthogonal ${}^\perp \mathcal{A}$ of $\mathcal{A}$ is the full subcategory consisting of the objects $X$ of $\mathcal{D}$ such that $\mathop{\mathrm{Hom}}\nolimits (X, A) = 0$ for all $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$.

Lemma 13.40.2. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A} \subset \mathcal{D}$ be a full subcategory invariant under all shifts. Consider a distinguished triangle

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

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

1. $Z$ is in $\mathcal{A}^\perp$, and

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

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 $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. 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 $A[1] \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ we see that the first and last groups are zero. Thus we get (2). Assume (2) and let $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. 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 13.40.3. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B} \subset \mathcal{D}$ be a full subcategory invariant under all shifts. Consider a distinguished triangle

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

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

1. $X$ is in ${}^\perp \mathcal{B}$, and

2. $\mathop{\mathrm{Hom}}\nolimits (Y, B) = \mathop{\mathrm{Hom}}\nolimits (Z, B)$ for all $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$.

Proof. Dual to Lemma 13.40.2. $\square$

Lemma 13.40.4. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A} \subset \mathcal{D}$ be a full subcategory invariant under all shifts. Then both the right orthogonal $\mathcal{A}^\perp$ and the left orthogonal ${}^\perp \mathcal{A}$ of $\mathcal{A}$ are strictly full, saturated1, triangulated subcagories of $\mathcal{D}$.

Proof. It is immediate from the definitions that the orthogonals are preserved under taking shifts, direct sums, and direct summands. Consider a distinguished triangle

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

of $\mathcal{D}$. By Lemma 13.4.16 it suffices to show that if $X$ and $Y$ are in $\mathcal{A}^\perp$, then $Z$ is in $\mathcal{A}^\perp$. This is immediate from Lemma 13.40.2. $\square$

Lemma 13.40.5. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A}$ 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 $A \to X \to B \to A[1]$ in $\mathcal{D}$ with $A$ in $\mathcal{A}$ and $B$ in $\mathcal{A}^\perp$.

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

$A_1 \to X_1 \to B_1 \to A_1[1] \quad \text{and}\quad A_2 \to X_2 \to B_2 \to A_2[1]$

as in condition $P$. Since $\mathop{\mathrm{Hom}}\nolimits (A_1, A_2) = \mathop{\mathrm{Hom}}\nolimits (A_1, X_2)$ by Lemma 13.40.2 there is a unique morphism $A_1 \to A_2$ such that the diagram

$\xymatrix{ A_1 \ar[d] \ar[r] & X_1 \ar[d] \\ A_2 \ar[r] & X_2 }$

commutes. Choose an extension of this to a diagram

$\xymatrix{ A_1 \ar[r] \ar[d] & X_1 \ar[r] \ar[d] & Q_1 \ar[r] \ar[d] & A_1[1] \ar[d] \\ A_2 \ar[r] \ar[d] & X_2 \ar[r] \ar[d] & Q_2 \ar[r] \ar[d] & A_2[1] \ar[d] \\ A_3 \ar[r] \ar[d] & X_3 \ar[r] \ar[d] & Q_3 \ar[r] \ar[d] & A_3[1] \ar[d] \\ A_1[1] \ar[r] & X_1[1] \ar[r] & Q_1[1] \ar[r] & A_1[2] }$

as in Proposition 13.4.23. By TR3 we see that $Q_1 \cong B_1$ and $Q_2 \cong B_2$ and hence $Q_1, Q_2 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}^\perp )$. As $Q_1 \to Q_2 \to Q_3 \to Q_1[1]$ is a distinguished triangle we see that $Q_3 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A}^\perp )$ by Lemma 13.40.4. Since $\mathcal{A}$ is a full triangulated subcategory, we see that $A_3$ is isomorphic to an object of $\mathcal{A}$. 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 13.4.11. $\square$

Lemma 13.40.6. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B}$ 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 $A \to X \to B \to A[1]$ in $\mathcal{D}$ with $B$ in $\mathcal{B}$ and $A$ in ${}^\perp \mathcal{B}$.

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. Dual to Lemma 13.40.5. $\square$

Lemma 13.40.7. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{A} \subset \mathcal{D}$ be a full triangulated subcategory. The following are equivalent

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

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

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

in $\mathcal{D}$ with $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ 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 13.40.2.

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 13.40.2 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 13.40.2 again.

Proof of the final statement. In order 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 13.40.4. 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. $\square$

Lemma 13.40.8. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B} \subset \mathcal{D}$ be a full triangulated subcategory. The following are equivalent

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

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

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

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

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

Proof. Dual to Lemma 13.40.7. $\square$

Definition 13.40.9. Let $\mathcal{D}$ be a triangulated category. A right admissible subcategory of $\mathcal{D}$ is a strictly full triangulated subcategory satisfying the equivalent conditions of Lemma 13.40.7. A left admissible subcategory of $\mathcal{D}$ is a strictly full triangulated subcategory satisfying the equivalent conditions of Lemma 13.40.8. A two-sided admissible subcategory is one which is both right and left admissible.

Let $\mathcal{A}$ be a right admissible subcategory of the triangulated category $\mathcal{D}$. Then we observe that for $X \in \mathcal{D}$ the distinguished triangle

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

with $A \in \mathcal{A}$ and $B \in \mathcal{A}^\perp$ is canonical in the following sense: for any other distinguished triangle $A' \to X \to B' \to A'[1]$ with $A' \in \mathcal{A}$ and $B' \in \mathcal{A}^\perp$ there is an isomorphism $(\alpha , \text{id}_ X, \beta ) : (A, X, B) \to (A', X, B')$ of triangles. The following proposition summarizes what was said above.

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

1. $\mathcal{A}$ is right admissible and $\mathcal{B} = \mathcal{A}^\perp$,

2. $\mathcal{B}$ is left admissible and $\mathcal{A} = {}^\perp \mathcal{B}$,

3. $\mathop{\mathrm{Hom}}\nolimits (A, B) = 0$ 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 straighforward 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. $\square$

[1] Definition 13.6.1.

Comment #6021 by Sandeep on

Two typos: In the statements of Lemmas 13.39.5 and 13.39.6, the distinguished triangles should be $A \to X \to B \to A[1]$ and $B\to X\to A \to A[1]$ respectively.

Comment #6022 by Sandeep on

Sorry, I mean $B\to X\to A \to B[1]$ .

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