Lemma 59.73.13. Let $X$ be an integral normal scheme with function field $K$. Let $E$ be a set.

1. Let $g : \mathop{\mathrm{Spec}}(K) \to X$ be the inclusion of the generic point. Then $g_*\underline{E} = \underline{E}$.

2. Let $j : U \to X$ be the inclusion of a nonempty open. Then $j_*\underline{E} = \underline{E}$.

Proof. Proof of (1). Let $x \in X$ be a point. Let $\mathcal{O}^{sh}_{X, \overline{x}}$ be a strict henselization of $\mathcal{O}_{X, x}$. By More on Algebra, Lemma 15.45.6 we see that $\mathcal{O}^{sh}_{X, \overline{x}}$ is a normal domain. Hence $\mathop{\mathrm{Spec}}(K) \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{X, \overline{x}})$ is irreducible. It follows that the stalk $(g_*\underline{E})_{\overline{x}}$ is equal to $E$, see Theorem 59.53.1.

Proof of (2). Since $g$ factors through $j$ there is a map $j_*\underline{E} \to g_*\underline{E}$. This map is injective because for every scheme $V$ étale over $X$ the set $\mathop{\mathrm{Spec}}(K) \times _ X V$ is dense in $U \times _ X V$. On the other hand, we have a map $\underline{E} \to j_*\underline{E}$ and we conclude. $\square$

Comment #9028 by DU Changjiang on

there is a typo in the proof: in the third line, a right parenthesis is missing. The right one should be $\left(g_{*}\underline{E}\right)_{\underline{x}}$

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