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 $\mathcal{C}_1 \to \mathcal{C}_3$.
Composition of site functors respects continuity of site functors.
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.
$\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (1)
Comment #7345 by Alejandro González Nevado on
There are also: