Lemma 40.11.2. Notation and assumptions as in Situation 40.11.1. If $a(R_1)$ is open in $R_2$, then $a(R_1)$ is closed in $R_2$.

**Proof.**
Let $r_2 \in R_2$ be a point in the closure of $a(R_1)$. We want to show $r_2 \in a(R_1)$. Pick $k \subset k'$ and $r_2' \in R'_2$ adapted to $(U, R_2, s_2, t_2, c_2)$ and $r_2$ as in Lemma 40.10.5. Let $R_ i'$ be the restriction of $R_ i$ via the morphism $U' = \mathop{\mathrm{Spec}}(k') \to U = \mathop{\mathrm{Spec}}(k)$. Let $a' : R'_1 \to R_2'$ be the base change of $a$. The diagram

is a fibre square. Hence the image of $a'$ is the inverse image of the image of $a$ via the morphism $p_2 : R'_2 \to R_2$. By Lemma 40.10.4 the map $p_2$ is surjective and open. Hence by Topology, Lemma 5.6.4 we see that $r_2'$ is in the closure of $a'(R'_1)$. This means that we may assume that $r_2 \in R_2$ has the property that the maps $k \to \kappa (r_2)$ induced by $s_2$ and $t_2$ are isomorphisms.

In this case we can use Lemma 40.10.6. This lemma implies $c(r_2, a(R_1))$ is an open neighbourhood of $r_2$. Hence $a(R_1) \cap c(r_2, a(R_1)) \not= \emptyset $ as we assumed that $r_2$ was a point of the closure of $a(R_1)$. Using the inverse of $R_2$ and $R_1$ we see this means $c_2(a(R_1), a(R_1))$ contains $r_2$. As $c_2(a(R_1), a(R_1)) \subset a(c_1(R_1, R_1)) = a(R_1)$ we conclude $r_2 \in a(R_1)$ as desired. $\square$

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