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...