Dominant maps pull back function fields functorially

Take a representative $\varphi :U\to Y$. Pullback maps $B=k[Y]$ into ${\mathcal{O}}_X(U)$, and F2 embeds the latter into $K=k(X)$. If b has zero image, F2 says $b\circ \varphi$ is the zero function on U. Its closed zero locus in Y contains the dense image of $\varphi$, so b vanishes everywhere on Y and is zero as a coordinate function. Thus the composite $B\to K$ is injective and fixes k.

F4 extends that injection uniquely to ${\mathrm{\Phi }}^{\ast }:k(Y)\to K$ by $a\mathrm{/}b\mapsto {\varphi }^{\ast }(a)\mathrm{/}{\varphi }^{\ast }(b)$. Here $b\mathrm{\ne }0$ has nonzero pullback by step 1.1, so the quotient is defined. If two representatives agree on a nonempty open, each pulled-back coordinate function agrees there; restriction compatibility and injectivity in F2 make their images in K equal. Uniqueness in F4 then identifies their field maps.

Let s be regular on a nonempty target open W, with a quotient expression $s=a\mathrm{/}b$ on a nonempty subopen. By F3 the inverse image of that subopen is nonempty open. F6 pulls s back regularly there, and its value is $(a\circ \varphi )\mathrm{/}(b\circ \varphi )$. F2 identifies this fraction with the field element in step 2.1. By restriction compatibility, this also proves agreement on the whole nonempty inverse-image domain.

For a composable dominant rational map $\mathrm{\Psi }:Y⇢Z$, F5 supplies a nonempty composition domain. For a coordinate function c on Z, substitution there gives $c\circ (\psi \circ \varphi )=(c\circ \psi )\circ \varphi$. Step 3.1 identifies the right side with ${\mathrm{\Phi }}^{\ast }({\mathrm{\Psi }}^{\ast }(c))$. F2 gives equality in k(X), and F4 extends it from coordinate-ring elements to all fractions. Thus $(\mathrm{\Psi }\circ \mathrm{\Phi }{)}^{\ast }={\mathrm{\Phi }}^{\ast }\circ {\mathrm{\Psi }}^{\ast }$. Identity pullback fixes every fraction.