Lemma 59.44.4. Let $f : X \to Y$ be an integral morphism of schemes which defines a homeomorphism of $X$ with a closed subset of $Y$. Then property (C) holds.

Proof. Let $g : U \to X$ be an étale morphism. We need to find an object $V \to Y$ of $Y_{\acute{e}tale}$ and a surjective morphism $X \times _ Y V \to U$ over $X$. Suppose that for every $u \in U$ we can find an object $V_ u \to Y$ of $Y_{\acute{e}tale}$ and a morphism $h_ u : X \times _ Y V_ u \to U$ over $X$ with $u \in \mathop{\mathrm{Im}}(h_ u)$. Then we can take $V = \coprod V_ u$ and $h = \coprod h_ u$ and we win. Hence given a point $u \in U$ we find a pair $(V_ u, h_ u)$ as above. To do this we may shrink $U$ and assume that $U$ is affine. In this case $g : U \to X$ is locally quasi-finite. Let $g^{-1}(g(\{ u\} )) = \{ u, u_2, \ldots , u_ n\}$. Since there are no specializations $u_ i \leadsto u$ we may replace $U$ by an affine neighbourhood so that $g^{-1}(g(\{ u\} )) = \{ u\}$.

The image $g(U) \subset X$ is open, hence $f(g(U))$ is locally closed in $Y$. Choose an open $V \subset Y$ such that $f(g(U)) = f(X) \cap V$. It follows that $g$ factors through $X \times _ Y V$ and that the resulting $\{ U \to X \times _ Y V\}$ is an étale covering. Since $f$ has property (B) , see Lemma 59.43.4, we see that there exists an étale covering $\{ V_ j \to V\}$ such that $X \times _ Y V_ j \to X \times _ Y V$ factor through $U$. This implies that $V' = \coprod V_ j$ is étale over $Y$ and that there is a morphism $h : X \times _ Y V' \to U$ whose image surjects onto $g(U)$. Since $u$ is the only point in its fibre it must be in the image of $h$ and we win. $\square$

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