Lemma 40.11.5. Notation and assumptions as in Situation 40.11.1. If the image $a(R_1)$ is a locally closed subset of $R_2$ then it is a closed subset.

**Proof.**
Let $k \subset k'$ be a perfect closure of the field $k$. Let $R_ i'$ be the restriction of $R_ i$ via the morphism $U' = \mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$. Note that the morphisms $R_ i' \to R_ i$ are universal homeomorphisms as compositions of base changes of the universal homeomorphism $U' \to U$ (see diagram in statement of Lemma 40.10.4). Hence it suffices to prove that $a'(R_1')$ is closed in $R_2'$. In other words, we may assume that $k$ is perfect.

If $k$ is perfect, then the closure of the image is a groupoid scheme $Z \subset R_2$, by Lemma 40.11.4. By the same lemma applied to $\text{id}_{R_1} : R_1 \to R_1$ we see that $(R_2)_{red}$ is a groupoid scheme. Thus we may apply Lemma 40.11.2 to the morphism $a|_{(R_2)_{red}} : (R_2)_{red} \to Z$ to conclude that $Z$ equals the image of $a$. $\square$

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