Lemma 40.10.7. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. If $U$ is the spectrum of a field, $W \subset R$ is open, and $Z \to R$ is a morphism of schemes, then the image of the composition $Z \times _{s, U, t} W \to R \times _{s, U, t} R \to R$ is open.

**Proof.**
Write $U = \mathop{\mathrm{Spec}}(k)$. Consider a field extension $k'/k$. Denote $U' = \mathop{\mathrm{Spec}}(k')$. Let $R'$ be the restriction of $R$ via $U' \to U$. Set $Z' = Z \times _ R R'$ and $W' = R' \times _ R W$. Consider a point $\xi = (z, w)$ of $Z \times _{s, U, t} W$. Let $r \in R$ be the image of $z$ under $Z \to R$. Pick $k' \supset k$ and $r' \in R'$ as in Lemma 40.10.5. We can choose $z' \in Z'$ mapping to $z$ and $r'$. Then we can find $\xi ' \in Z' \times _{s', U', t'} W'$ mapping to $z'$ and $\xi $. The open $c(r', W')$ (Lemma 40.10.6) is contained in the image of $Z' \times _{s', U', t'} W' \to R'$. Observe that $Z' \times _{s', U', t'} W' = (Z \times _{s, U, t} W) \times _{R \times _{s, U, t} R} (R' \times _{s', U', t'} R')$. Hence the image of $Z' \times _{s', U', t'} W' \to R' \to R$ is contained in the image of $Z \times _{s, U, t} W \to R$. As $R' \to R$ is open (Lemma 40.10.4) we conclude the image contains an open neighbourhood of the image of $\xi $ as desired.
$\square$

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