Lemma 115.19.1. Let $X \to Y$ be a morphism of schemes.
The category $\mathcal{C}_{X/Y}$ is fibred over $X_{Zar}$.
The category $\mathcal{C}_{X/Y}$ is fibred over $Y_{Zar}$.
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 $(X/Y)_{Zar}$. $\square$
Your email address will not be published.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 (0)