I have problems to understand a proof in this paper by Pierrick Dartois on Abelian varieties:
Theorem 1.13 (rigidity lemma). Let $ \varphi: X \times_k Y \to Z$ be a morphism of $k$-schemes. Assume that $X$ is proper and geometrically integral, that $ X \times_k Y $ is geometrically irreducible, that $ Z$ is separated and that there exist $ k$-valued points $x_0 ∈ X(k), y_0 ∈ Y(k) $ and $z_0 ∈ Z(k)$ such that: $ \varphi(X \times \{y_0\}) = \{z_0\} = \varphi({x_0} \times Y)$ Then, $ \varphi(X \times Y) = \{ z_0 \}$.
The proof starts with:
Proof. One can easily see that the properties of the theorem are invariant under base change, so we can assume that $k$ is algebraically closed. Let $U_0$ be an open affine neighborhood of $z_0$ ...
Question: Why $ k$ can be assumed to be algebraically closed? Since all properties of the schemes in the assumptions are geometrically, the conditions not change if we replace the schemes by fibre products $ X \times \text{Spec} (\overline{k})$ with algebraic closure of $k$. But having proved the lemma for them, how can we derive the statement for schemes over not algebraically closed $k$?
Remark: I noticed that the proof strategy is rather similar with lemma 1.12 in these notes by by Moonen & Geer with one exception that there is not assumed the existence of a $k$-point $x_0 ∈ X(k)$. This can be only assured after base change to algebraic closure. But the problem remains the same: how to retain the claim after hsving proved it for algebraic closed base field?
