I have problems to understand a proof in [this paper](https://www.pierrickdartois.fr/homepage/wp-content/uploads/2022/04/seminar_report.pdf) 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](http://van-der-geer.nl/%7Egerard/AV.pdf) 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?



Proof of rigidity lemma
If $\varphi\_{\bar{k}}(X\_{\bar{k}} \times\_{\bar{k}} Y\_{\bar{k}})=\{z\_0\}$ then since $\varphi(X\times Y) \subset \varphi\_{\bar{k}}(X\_{\bar{k}} \times\_{\bar{k}} Y\_{\bar{k}})$ we also have $\varphi(X\times Y)\subset\{z\_0\}$.


