The Stacks project

Lemma 4.27.8. 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 in the proof, it should say instead of .

There are also:

  • 19 comment(s) on Section 4.27: Localization in 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Your email address will not be published.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 04VG. Beware of the difference between the letter 'O' and the digit '0'.