I want to prove that surjectivity is stable under base change: if $f:X\to S$ a surjective morphism of scheme and $\varphi:T\to S$ then $f_T:X\times_S T\to T$ is surjective.
Idea 1: I know that for all $t\in T$, $f_T^{-1}(t)\simeq (X\times_S T)\times_T \mathrm{Spec}(k(t))\simeq X\times_S \mathrm{Spec}(k(t))$ and (for the same reason) with $s=\varphi(t)$, $f^{-1}(s)\simeq X\times_S\mathrm{Spec}(k(s))\neq\emptyset$. But how deduce that $X\times_S \mathrm{Spec}(k(t))\neq\emptyset$? Maybe as $k(t)\simeq k(s)$ (topologically) so $\mathrm{Spec}(k(t))\twoheadrightarrow\mathrm{Spec}(k(s))$ and so $X\times_S \mathrm{Spec}(k(t))\twoheadrightarrow X\times_S \mathrm{Spec}(k(s))$?
Idea 2: let $s=\varphi(t)$ and $x\in X$ so that $f(x)=s$ then $f(s)=\varphi(t)$. In the purely set-theoretic construction of $X\times_S T$ this would implicate the existence of $z\in X\times_S T$ with $f_T(z)=t$. But the fibred-product of scheme is not so built…
Best Answer
Following your Idea 1 it only remains to show $X\times_S \mathrm{Spec} (k(t))$ is non-empty.
Let $s=\varphi(t)$, then the above fiber product is also $X_s\times_{\mathrm{Spec}(k(s))} \mathrm{Spec}(k(t))$. Take a non-empty affine open subset $U=\mathrm{Spec}(R)$ of $X_s$. It is enough to show $U\times_{\mathrm{Spec}(k(s))} \mathrm{Spec}(k(t))$ is non-empty.
The latter is an affine scheme given by $R\otimes_{k(s)} k(t)$. As $R\ne 0$, $R\otimes_{k(s)} k(t)\ne 0$ because we make a field extension (a vector basis of $R$ over $k(s)$ extends to a vector basis over $k(t)$). By Krull's theorem, $R\otimes_{k(s)} k(t)$ has a maximal (hence prime) ideal, so $U\times_{\mathrm{Spec}(k(s))} \mathrm{Spec}(k(t))\ne\emptyset$.