Lemma 114.18.1. Let $X \to Y$ be a morphism of schemes.

1. The category $\mathcal{C}_{X/Y}$ is fibred over $X_{Zar}$.

2. The category $\mathcal{C}_{X/Y}$ is fibred over $Y_{Zar}$.

3. The category $\mathcal{C}_{X/Y}$ is fibred over the category of pairs $(U, V)$ where $U \subset X$, $V \subset Y$ are open and $f(U) \subset V$.

Proof. Ad (1). Given an object $U \to A$ of $\mathcal{C}_{X/Y}$ and a morphism $U' \to U$ of $X_{Zar}$ consider the object $i' : U' \to A$ of $\mathcal{C}_{X/Y}$ where $i'$ is the composition of $i$ and $U' \to U$. The morphism $(U' \to A) \to (U \to A)$ of $\mathcal{C}_{X/Y}$ is strongly cartesian over $X_{Zar}$.

Ad (2). Given an object $U \to A/V$ and $V' \to V$ we can set $U' = U \cap f^{-1}(V')$ and $A' = V' \times _ V A$ to obtain a strongly cartesian morphism $(U' \to A') \to (U \to A)$ over $V' \to V$.

Ad (3). Denote $(X/Y)_{Zar}$ the category in (3). Given $U \to A/V$ and a morphism $(U', V') \to (U, V)$ in $(X/Y)_{Zar}$ we can consider $A' = V' \times _ V A$. Then the morphism $(U' \to A'/V') \to (U \to A/V)$ is strongly cartesian in $\mathcal{C}_{X/Y}$ over $(X/Y)_{Zar}$. $\square$

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