The Stacks project

The universal property of the Verdier quotient.

Lemma 13.6.8. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{B}$ be a full triangulated subcategory of $\mathcal{D}$. Let $Q : \mathcal{D} \to \mathcal{D}/\mathcal{B}$ be the quotient functor.

  1. If $H : \mathcal{D} \to \mathcal{A}$ is a homological functor into an abelian category $\mathcal{A}$ such that $\mathcal{B} \subset \mathop{\mathrm{Ker}}(H)$ then there exists a unique factorization $H' : \mathcal{D}/\mathcal{B} \to \mathcal{A}$ such that $H = H' \circ Q$ and $H'$ is a homological functor too.

  2. If $F : \mathcal{D} \to \mathcal{D}'$ is an exact functor into a pre-triangulated category $\mathcal{D}'$ such that $\mathcal{B} \subset \mathop{\mathrm{Ker}}(F)$ then there exists a unique factorization $F' : \mathcal{D}/\mathcal{B} \to \mathcal{D}'$ such that $F = F' \circ Q$ and $F'$ is an exact functor too.

Proof. This lemma follows from Lemma 13.5.7. Namely, if $f : X \to Y$ is a morphism of $\mathcal{D}$ such that for some distinguished triangle $(X, Y, Z, f, g, h)$ the object $Z$ is isomorphic to an object of $\mathcal{B}$, then $H(f)$, resp. $F(f)$ is an isomorphism under the assumptions of (1), resp. (2). Details omitted. $\square$


Comments (1)

Comment #987 by on

Suggested slogan: Quotients of triangulated categories have a universal property for homological functors and exact functors.

There are also:

  • 2 comment(s) on Section 13.6: Quotients of triangulated categories

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Your email address will not be published.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 05RJ. Beware of the difference between the letter 'O' and the digit '0'.