A classical morphism pulls Zariski closed sets back to closed sets

If $r$ is regular on an open $U$ and $r(x)\mathrm{\ne }0$, write $r=a\mathrm{/}b$ near $x$ with $b$ nowhere zero. Intersect that neighbourhood with $D(a)$; it is a neighbourhood of $x$ where $r$ is nowhere zero. Thus the nonvanishing set of $r$ is open, and its zero set is relatively closed in $U$.

Write $C=Y\cap V(S)$. Each coordinate polynomial restricted to $Y$ is globally regular (its denominator is 1), so F3 makes its pullback regular. Therefore ${\varphi }^{-1}(C)$ is the intersection of their closed zero sets by step 1.1. This proves continuity, including empty and full closed sets.

For $x\in {\varphi }^{-1}(W)$, take a neighbourhood $W_0\subseteq W$ of $\varphi (x)$ where $s=g\mathrm{/}h$ with $h$ nowhere zero. On the open ${\varphi }^{-1}(W_0)$, $g\circ \varphi ,h\circ \varphi$ are regular and the latter is nowhere zero. Near $x$, write them as $a\mathrm{/}b,c\mathrm{/}d$ with $b,d$ nonzero. Shrink further to where $c\mathrm{\ne }0$ using step 1.1. Then $s\circ \varphi =ad\mathrm{/}(bc)$ is a valid local quotient, proving the local pullback property.