Surjectivity survives arbitrary base change

Let $s^{\mathrm{\prime }}\in S^{\mathrm{\prime }}$ and let $s$ be its image. Surjectivity gives $x\in X$ with $f(x)=s$. The residue fields $\kappa (x)$ and $\kappa (s^{\mathrm{\prime }})$ are nonzero vector spaces over $\kappa (s)$. Under Choice choose bases containing the element 1; F2 makes their tensor product free on the nonempty product of those bases, so it is a nonzero ring. The same basis argument shows that $A\to A{\otimes }_{k}K$, $a\mapsto a\otimes 1$, is injective for any $k$-algebra and field extension $K\mathrm{/}k$: choosing a basis of $K$ containing 1 identifies this map with inclusion of one summand (or use bases of both vector spaces).

By F3 the zero ideal of that nonzero ring lies in a maximal ideal, which is prime. F1 then supplies a point of $X{\times }_S{S}^{\mathrm{\prime }}$ projecting to $s^{\mathrm{\prime }}$. This proves surjectivity. If $S^{\mathrm{\prime }}$ is empty the assertion is vacuous; if $S$ is empty both sources are empty.

A nonempty $X$ maps surjectively to the one-point scheme $\mathrm{Spec}k$. Applying the result to $\mathrm{Spec}K\to \mathrm{Spec}k$ gives a point of $X_K$. Identity extensions also satisfy the argument; no algebraicity or reducedness hypothesis is used.