Lemma 4.26.8. Let $\mathcal{C}$ be a category and let $S$ be a left multiplicative system of morphisms of $\mathcal{C}$.

1. The rules $X \mapsto X$ and $(f : X \to Y) \mapsto (f : X \to Y, \text{id}_ Y : Y \to Y)$ define a functor $Q : \mathcal{C} \to S^{-1}\mathcal{C}$.

2. For any $s \in S$ the morphism $Q(s)$ is an isomorphism in $S^{-1}\mathcal{C}$.

3. If $G : \mathcal{C} \to \mathcal{D}$ is any functor such that $G(s)$ is invertible for every $s \in S$, then there exists a unique functor $H : S^{-1}\mathcal{C} \to \mathcal{D}$ such that $H \circ Q = G$.

Proof. Parts (1) and (2) are clear. (In (2), the inverse of $Q(s)$ is the equivalence class of the pair $(\text{id}_ Y, s)$.) To see (3) just set $H(X) = G(X)$ and set $H((f : X \to Y', s : Y \to Y')) = G(s)^{-1} \circ G(f)$. Details omitted. $\square$

## Comments (1)

Comment #325 by arp on

Typo: In the definition of $H$ in the proof, it should say $G(s)^{-1} \circ G(f)$ instead of $H(s)^{-1} \circ H(f)$.

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• 8 comment(s) on Section 4.26: Localization in categories

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