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.
$\square$

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