Lemma 40.14.6. Let $(U, R, s, t, c)$ be a groupoid scheme over a scheme $S$ with $s, t$ integral. Let $g : U' \to U$ be an integral morphism such that every $R$-orbit in $U$ meets $g(U')$. Let $(U', R', s', t', c')$ be the restriction of $R$ to $U'$. If $u' \in U'$ is contained in an $R'$-invariant affine open, then the image $u \in U$ is contained in an $R$-invariant affine open of $U$.

**Proof.**
Let $W' \subset U'$ be an $R'$-invariant affine open. Set $\tilde R = U' \times _{g, U, t} R$ with maps $\text{pr}_0 : \tilde R \to U'$ and $h = s \circ \text{pr}_1 : \tilde R \to U$. Observe that $\text{pr}_0$ and $h$ are integral. It follows that $\tilde W = \text{pr}_0^{-1}(W')$ is affine. Since $W'$ is $R'$-invariant, the image $W = h(\tilde W)$ is set theoretically $R$-invariant and $\tilde W = h^{-1}(W)$ set theoretically (details omitted). Thus, if we can show that $W$ is open, then $W$ is a scheme and the morphism $\tilde W \to W$ is integral surjective which implies that $W$ is affine by Limits, Proposition 32.11.2. However, our assumption on orbits meeting $U'$ implies that $h : \tilde R \to U$ is surjective. Since an integral surjective morphism is submersive (Topology, Lemma 5.6.5 and Morphisms, Lemma 29.44.7) it follows that $W$ is open.
$\square$

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