If $f:X\rightarrow Y$ is a surjective morphism of schemes and g:X'\rightarrow Y is another morphism of schemes, one can show that p_{2}:X\times_{Y}X'\rightarrow X' is also surjective. The proof that I have seen (in Mumford's Red Book, for instance) goes as follows: Given P\in X', we can find points $R\in Y$ and $P\in X$ and inclusions $k\left(R\right)\hookrightarrow k\left(P\right)$ and $k\left(R\right)\hookrightarrow k\left(Q\right)$. The desired result can then be established if we can say that there exists a field $F$ containing each of $k\left(P\right)$ and $k\left(Q\right)$. However, I don't see how we can conclude that such a field exists.
It is easy to just give a different proof without looking at a field extensions of $k\left(P\right)$ and $k\left(Q\right)$, but I wanted to know how the sketch of a proof above can be used. Thanks!