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.
$\square$

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## Comments (1)

Comment #7345 by Alejandro González Nevado on

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