Lemma 13.5.8 (0GSL)—The Stacks project

Lemma 13.5.8. Let $\mathcal{D}$ be a pre-triangulated category and let $\mathcal{D}' \subset \mathcal{D}$ be a full, pre-triangulated subcategory. Let $S$ be a saturated multiplicative system of $\mathcal{D}$ compatible with the triangulated structure. Assume that for each $X$ in $\mathcal{D}$ there exists an $s : X' \to X$ in $S$ such that $X'$ is an object of $\mathcal{D}'$ . Then $S' = S \cap \text{Arrows}(\mathcal{D}')$ is a saturated multiplicative system compatible with the triangulated structure and the functor

$(S')^{-1}\mathcal{D}' \longrightarrow S^{-1}\mathcal{D}$

is an equivalence of pre-triangulated categories.

Proof. Consider the quotient functor $Q : \mathcal{D} \to S^{-1}\mathcal{D}$ of Proposition 13.5.6. Since $S$ is saturated we have that a morphism $f : X \to Y$ is in $S$ if and only if $Q(f)$ is invertible, see Categories, Lemma 4.27.21. Thus $S'$ is the collection of arrows which are turned into isomorphisms by the composition $\mathcal{D}' \to \mathcal{D} \to S^{-1}\mathcal{D}$ . Hence $S'$ is is a saturated multiplicative system compatible with the triangulated structure by Lemma 13.5.4. By Lemma 13.5.7 we obtain the exact functor $(S')^{-1}\mathcal{D}' \to S^{-1}\mathcal{D}$ of pre-triangulated categories. By assumption this functor is essentially surjective. Let $X', Y'$ be objects of $\mathcal{D}'$ . By Categories, Remark 4.27.15 we have

$\mathop{\mathrm{Mor}}\nolimits _{S^{-1}\mathcal{D}}(X', Y') = \mathop{\mathrm{colim}}\nolimits _{s : X \to X'\text{ in }S} \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(X, Y')$

Our assumption implies that for any $s : X \to X'$ in $S$ we can find a morphism $s' : X'' \to X$ in $S$ with $X''$ in $\mathcal{D}'$ . Then $s \circ s' : X'' \to X'$ is in $S'$ . Hence the colimit above is equal to

$\mathop{\mathrm{colim}}\nolimits _{s'' : X'' \to X'\text{ in }S'} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(X'', Y') = \mathop{\mathrm{Mor}}\nolimits _{(S')^{-1}\mathcal{D}'}(X', Y')$

This proves our functor is also fully faithful and the proof is complete. $\square$

Comments (5)

Comment #8347 by Elías Guisado on April 24, 2023 at 11:50

Typos: in the subscripts of the colimits, instead of $S$ and $S'$ , it should be $(S/X')^\text{opp}$ and $(S'/X')^\text{opp}$ , respectively.

Also, to justify equality of the two colimits maybe it would be nice to explicitly invoke 4.17.2 plus the following observation: if $F:\mathcal{I}\to\mathcal{J}$ is a fully faithful functor such that $\mathcal{J}$ is filtered and condition (1) from Definition 4.17.1 is satisfied, then $F$ is cofinal and $\mathcal{I}$ is filtered. Here we apply the observation to the inclusion $S'/X'\to S/X'$ .

Comment #8383 by Elías Guisado on May 08, 2023 at 08:52

I just found out that the observation I mentioned was already included in the SP with tag 4.19.3.

Comment #8954 by Stacks project on April 11, 2024 at 17:13

Going to leave as is.

Comment #9818 by Elías Guisado on October 24, 2024 at 06:29

Alternative proof: Refer to #9817. It is only left to check MS5 and MS6. This is justified in the four first sentences of the current proof.

Comment #9822 by Elías Guisado on October 24, 2024 at 08:37

We can actually generalize this result to:

Lemma. Let $\mathcal{D}$ be a pre-triangulated category. Let $S$ be a multiplicative system in $\mathcal{D}$ compatible with the triangulated structure. Let $\mathcal{D}'\subset\mathcal{D}$ be a full pre-triangulated subcategory. Suppose at least one of the following assertions hold:

For every morphism $X'\to X$ in $S$ with $X'\in\mathcal{D}'$ there is a morphism $X\to X''$ in $S$ with $X''\in\mathcal{D}'$ .

For every $X\in\mathcal{D}$ there is $X\to X'$ in $S$ with $X'\in\mathcal{D}'$ .

Then $S'=S\cap\operatorname{Mor}(\mathcal{D}')$ is a multiplicative system in $\mathcal{D}'$ compatible with the triangulated structure, and if $S$ is saturated then so is $S'$ . Moreover, the triangulated functor $(S')^{-1}\mathcal{D}'\to S^{-1}\mathcal{D}$ coming from Lemma 13.5.7 is fully faithful. Furthermore, if 2 holds, then $(S')^{-1}\mathcal{C}'\to S^{-1}\mathcal{C}$ is an equivalence of pre-triangulated categories.

Proof. By #9820, we must only show MS5 and MS6 for $S'$ . But this follows from the same axioms for $S$ and the fact that $\mathcal{D}'$ is a full and triangulated sub-category. $\square$

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