The Stacks project

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 construct the adjoint $v$ explicitly. 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. $\square$


Comments (0)

There are also:

  • 3 comment(s) on Section 13.40: Admissible subcategories

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0CQS. Beware of the difference between the letter 'O' and the digit '0'.