Lemma 85.2.3. Let $f : Y \to X$ be a morphism of simplicial spaces. Then the functor $u : X_{Zar} \to Y_{Zar}$ which associates to the open $U \subset X_ n$ the open $f_ n^{-1}(U) \subset Y_ n$ defines a morphism of sites $f_{Zar} : Y_{Zar} \to X_{Zar}$.

**Proof.**
It is clear that $u$ is a continuous functor. Hence we obtain functors $f_{Zar, *} = u^ s$ and $f_{Zar}^{-1} = u_ s$, see Sites, Section 7.14. To see that we obtain a morphism of sites we have to show that $u_ s$ is exact. We will use Sites, Lemma 7.14.6 to see this. Let $V \subset Y_ n$ be an open subset. The category $\mathcal{I}_ V^ u$ (see Sites, Section 7.5) consists of pairs $(U, \varphi )$ where $\varphi : [m] \to [n]$ and $U \subset X_ m$ open such that $Y(\varphi )(V) \subset f_ m^{-1}(U)$. Moreover, a morphism $(U, \varphi ) \to (U', \varphi ')$ is given by a $\psi : [m'] \to [m]$ such that $X(\psi )(U) \subset U'$ and $\varphi \circ \psi = \varphi '$. It is our task to show that $\mathcal{I}_ V^ u$ is cofiltered.

We verify the conditions of Categories, Definition 4.20.1. Condition (1) holds because $(X_ n, \text{id}_{[n]})$ is an object. Let $(U, \varphi )$ be an object. The condition $Y(\varphi )(V) \subset f_ m^{-1}(U)$ is equivalent to $V \subset f_ n^{-1}(X(\varphi )^{-1}(U))$. Hence we obtain a morphism $(X(\varphi )^{-1}(U), \text{id}_{[n]}) \to (U, \varphi )$ given by setting $\psi = \varphi $. Moreover, given a pair of objects of the form $(U, \text{id}_{[n]})$ and $(U', \text{id}_{[n]})$ we see there exists an object, namely $(U \cap U', \text{id}_{[n]})$, which maps to both of them. Thus condition (2) holds. To verify condition (3) suppose given two morphisms $a, a': (U, \varphi ) \to (U', \varphi ')$ given by $\psi , \psi ' : [m'] \to [m]$. Then precomposing with the morphism $(X(\varphi )^{-1}(U), \text{id}_{[n]}) \to (U, \varphi )$ given by $\varphi $ equalizes $a, a'$ because $\varphi \circ \psi = \varphi ' = \varphi \circ \psi '$. This finishes the proof. $\square$

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