A classical fibre is base change to a point

Every point of a classical algebraic set over the page's algebraically closed field $k$ is the image of a unique morphism from the one-point algebraic set $\{\ast \}$; such maps are regular. Test the given pullback universal property on $\{\ast \}$. It identifies morphisms $\{\ast \}\to X{\times }_S\{s\}$ with pairs of point maps whose composites to $S$ agree. Projection therefore gives a canonical bijection of its underlying set with $f^{-1}(s)$, rather than assuming that identification as part of the construction.

Choose affine opens $W\subseteq X$ and $V\subseteq S$ with $s\in V$ and $f(W)\subseteq V$. Let ${\mathfrak{m}}_s\subseteq k[V]$ be the point ideal and let $J=k[W]f^{\ast }({\mathfrak{m}}_s)$ be its generated ideal in $k[W]$. The equations $f^{\ast }(h)=0$ for $h\in {\mathfrak{m}}_s$ cut out exactly $W\cap f^{-1}(s)$. With its reduced classical algebraic-set structure this locus has coordinate ring $k[W]\mathrm{/}\sqrt{J}$. These local structures agree on restrictions and give the reduced closed fibre $F\subseteq X$.

If a pair of regular maps from a classical test object commutes over $s$, its map to $X$ has image in $F$. On the affine charts of step 2.1, pullback kills $J$ and therefore $\sqrt{J}$, since regular functions on a classical algebraic set form a reduced ring. The map consequently factors regularly through $F$, uniquely because $F\to X$ is inclusion. These local factorizations agree on overlaps. Thus $F$ has the same pullback universal property, and the unique projection-compatible isomorphism with the given constructed pullback proves the coordinate-ring assertion. The empty fibre corresponds to the unit ideal and zero coordinate ring.