Reduction to $k=\bar{k}$ in rigidity lemma I try to understand a reduction step the proof of rigidity lemma as proved in Moonen's and van der Geer's Abelian varieties (Lemma 1.11 on page 12):
Lemma.
Let $X$, $Y$ and $Z$ be algebraic varieties over a field $k$. Suppose that $X$
is complete. If $f : X × Y \rightarrow Z$ is a morphism with the property that, for some $y \in Y (k)$, the fibre $X \times_k \{y\}$ is mapped to a point $z \in Z(k)$ then $f$ factors through the projection $pr_Y : X \times_k Y → Y$.
The proof starts with reduction to $k=\bar{k}$. Why the reduction to the case $k=\bar{k}$ is allowed? 
I supposed at first glance a pure category theoretical argument in mind, but it fails: Assume we can show that after base change $ - \times \bar{k}$ we prove that $(X \times Y) \times \overline{k} \to Z \times \overline{k}$  factors through the projection to $Y \times \bar{k}$, does the original morphism $f : X × Y \rightarrow Z$ factor through projection to $Y$? I don't see why this can be deduced from universal property of fiber product. 
p.s.: I noticed that here was asked amongst other things the same question, but the accepted answer don't satisfy me: a "diagram chase" as suggested in an answer gives a factorisation of $f$ through $Y$ on underlying sets. There is no hint why the obtained map between sets is a morphism (in category of algebraic varieties).
 A: In term of $k$-algebras: Suppose $\lambda \colon A  \to B \otimes C$ is a $k$-algebra homomorphism ($\lambda$ corresponds [locally] to $f$). Choose $k$-bases $b_i$ and $c_j$ for $B$ and $C$, with $1$ belonging to (both) bases, and identify $B\otimes_k C$ with $B\otimes_k C \otimes_k k$. The terms $b_i\otimes c_j$ form a $k$-basis for $B\otimes C$.
If $a \in A$, then the choice of bases determines $x_{i,j}\in k$ uniquely (almost all zero) to satisfy $$ \lambda (a)  = \sum b_i\otimes c_j \otimes x_{i,j}.$$
We want, in the above expression, that  the only basis elements $b_i\otimes c_j$ with non-zero coefficients $x_{i,j}$ be of the form $1\otimes c_j$.
Now, identify $A$ with its image $A\otimes 1$ in  $A\otimes \overline k$, and $B\otimes_k C$ with its image in $ B\otimes_k C\otimes_k \overline k$. The elements $b_i\otimes c_j$ still form a $\overline k$ basis for $B\otimes C \otimes \overline k$. By hypothesis, we have the desired expression in $B\otimes C \otimes \overline k$:
$$ \lambda (a) = (\lambda \otimes 1) (a \otimes 1 ) = \sum 1 \otimes c_j \otimes x_j,$$
with $x_j \in \overline k$. But, since the two expressions for $\lambda (a)$ are given in terms of the same basis, the expressions must be the same. Hence, the only non-zero coefficients $x_{i,j}$ correspond to $b_i = 1$.
