The Stacks project

Composition of site functors respects continuity of site functors.

Lemma 7.14.4. Let $\mathcal{C}_ i$, $i = 1, 2, 3$ be sites. Let $u : \mathcal{C}_2 \to \mathcal{C}_1$ and $v : \mathcal{C}_3 \to \mathcal{C}_2$ be continuous functors which induce morphisms of sites. Then the functor $u \circ v : \mathcal{C}_3 \to \mathcal{C}_1$ is continuous and defines a morphism of sites $\mathcal{C}_1 \to \mathcal{C}_3$.

Proof. It is immediate from the definitions that $u \circ v$ is a continuous functor. In addition, we clearly have $(u \circ v)^ p = v^ p \circ u^ p$, and hence $(u \circ v)^ s = v^ s \circ u^ s$. Hence functors $(u \circ v)_ s$ and $u_ s \circ v_ s$ are both left adjoints of $(u \circ v)^ s$. Therefore $(u \circ v)_ s \cong u_ s \circ v_ s$ and we conclude that $(u \circ v)_ s$ is exact as a composition of exact functors. $\square$

Comments (1)

Comment #7345 by Alejandro González Nevado on

SS: Composition of site functors respects continuity of site functors.

There are also:

  • 1 comment(s) on Section 7.14: Morphisms of sites

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Your email address will not be published.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 03CB. Beware of the difference between the letter 'O' and the digit '0'.