Dominant maps pull back function fields functorially

Choose a representative morphism $\phi :U\to Y$ of $\eta$ on a nonempty affine open $U\subseteq X$, guaranteed by [L2]. By [L4], $\phi$ induces a $k$-algebra homomorphism ${\phi }^{\ast }:k[Y]\to k[U]$. Using [L6] and [L1], regard $k[U]$ as a subring of $k(X)$.

The map ${\phi }^{\ast }$ is injective. Let $0\mathrm{\ne }h\in k[Y]$. By [L5], the ring $k[Y]$ is a domain, so [L7] makes $D_Y(h)$ a nonempty open subset of $Y$. Let $\psi :\mathrm{Dom}(\eta )\to Y$ be the representative from [L3] whose image is dense in $Y$. By the definition of rational map, $\psi$ and $\phi$ agree on some nonempty open subset $W\subseteq U$. Since $W$ is a nonempty open subset of the classical affine variety $X$, [L8] makes $W$ dense in $X$, hence dense in the open subspace $\mathrm{Dom}(\eta )$. Therefore $\psi (W)$ is dense in $\psi (\mathrm{Dom}(\eta ))$, which is dense in $Y$. Because $\phi (W)=\psi (W)\subseteq \phi (U)$, the image $\phi (U)$ is dense in $Y$. It follows that $\phi (U)$ meets $D_Y(h)$, so for some $u\in U$ one has ${\phi }^{\ast }(h)(u)=h(\phi (u))\mathrm{\ne }0$. Hence ${\phi }^{\ast }(h)\mathrm{\ne }0$.

For $a\mathrm{/}b\in k(Y)=\mathrm{Frac}(k[Y])$, define ${\eta }^{\ast }(a\mathrm{/}b):={\phi }^{\ast }(a)\mathrm{/}{\phi }^{\ast }(b)\in \mathrm{Frac}(k[U])=k(X)$. This is well defined because step 2.1 makes ${\phi }^{\ast }(b)$ nonzero whenever $b\mathrm{\ne }0$.

If $\psi :V\to Y$ is another representative of $\eta$, then by the definition of rational map the two representatives agree on some nonempty affine open $W\subseteq U\cap V$. Their coordinate-ring pullbacks therefore agree on $k[Y]$, and after passing to fraction fields and using [L6] for $W$, the two constructions of step 3.1 agree in $k(X)$. Thus ${\eta }^{\ast }$ is independent of the chosen representative.

Let $\theta :Y⇢Z$ be another dominant rational map, represented on some nonempty affine open of $Y$. On a common domain where both pullbacks are defined, pullback of regular functions satisfies $(\theta \circ \eta {)}^{\ast }(h)={\eta }^{\ast }({\theta }^{\ast }(h))$ for every $h\in k[Z]$. By fraction-field arithmetic the same identity holds for every element of $k(Z)$. Therefore pullback on function fields is functorial.