Lemma 39.7.7. Let $k$ be a field. Let $\psi : G' \to G$ be a morphism of group schemes over $k$. If $\psi (G')$ is open in $G$, then $\psi (G')$ is closed in $G$.

Proof. Let $U = \psi (G') \subset G$. Let $Z = G \setminus \psi (G') = G \setminus U$ with the reduced induced closed subscheme structure. By Lemma 39.7.2 the image of

$Z \times _ k G' \longrightarrow Z \times _ k U \longrightarrow G$

is open (the first arrow is surjective). On the other hand, since $\psi$ is a homomorphism of group schemes, the image of $Z \times _ k G' \to G$ is contained in $Z$ (because translation by $\psi (g')$ preserves $U$ for all points $g'$ of $G'$; small detail omitted). Hence $Z \subset G$ is an open subset (although not necessarily an open subscheme). Thus $U = \psi (G')$ is closed. $\square$

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