Lemma 13.5.7 (05R7)—The Stacks project

Lemma 13.5.7. Let $\mathcal{D}$ be a pre-triangulated category. Let $S$ be a multiplicative system compatible with the triangulated structure. Let $Q : \mathcal{D} \to S^{-1}\mathcal{D}$ be the localization functor, see Proposition 13.5.6.

If $H : \mathcal{D} \to \mathcal{A}$ is a homological functor into an abelian category $\mathcal{A}$ such that $H(s)$ is an isomorphism for all $s \in S$, then the unique factorization $H' : S^{-1}\mathcal{D} \to \mathcal{A}$ such that $H = H' \circ Q$ (see Categories, Lemma 4.27.8) is a homological functor too.
If $F : \mathcal{D} \to \mathcal{D}'$ is an exact functor into a pre-triangulated category $\mathcal{D}'$ such that $F(s)$ is an isomorphism for all $s \in S$, then the unique factorization $F' : S^{-1}\mathcal{D} \to \mathcal{D}'$ such that $F = F' \circ Q$ (see Categories, Lemma 4.27.8) is an exact functor too.

Proof. This lemma proves itself. Details omitted. $\square$

Comments (1)

Comment #9852 by Elías Guisado on November 06, 2024 at 10:51

I don't know if this is relevant, but maybe one could stress out the fact that in (2) the choice for the isomorphism $F'\circ[1]\cong[1]\circ F'$ is uniquely determined by the isomorphism $F\circ[1]\cong [1]\circ F$ . Specifically, if $(F,\xi):\mathcal{D}\to\mathcal{D}'$ is a triangulated functor sending all morphism in $S$ to isos in $\mathcal{D}'$ then there is a unique triangulated functor $(F',\xi'):S^{-1}\mathcal{D}\to\mathcal{D}'$ factoring $(F,\xi)$ through $(Q,\zeta)$ (where $\zeta:Q\circ[1]=[1]\circ Q$ is the identity natural transformation).

There are also:

4 comment(s) on Section 13.5: Localization of triangulated categories

Comments (1)

Post a comment