Classical affine morphisms are contravariantly equivalent to coordinate-ring homomorphisms

If $\varphi :X\to Y$ is a morphism, F1 and F2 identify $s\mapsto s\circ \varphi$ as a map $k[Y]\to k[X]$. Pointwise addition, multiplication and constants show it is a unital $k$-algebra homomorphism.

Conversely let $\alpha :k[Y]\to k[X]$ be a unital $k$-algebra map, with $Y\subseteq k^m$ and coordinate classes $y_j$. Define $\varphi (x)=(\alpha (y_1)(x),\dots ,\alpha (y_m)(x))$. If $P\in I(Y)$ then polynomial evaluation and the homomorphism laws give $P(\varphi (x))=\alpha (P(y_1,\dots ,y_m))(x)=0$. Hence $\varphi (x)\in V(I(Y))=Y$.

Every global regular function of $Y$ is a polynomial in the $y_j$ by F1 and F3. Substitution in step 1.2 gives $s\circ \varphi =\alpha (s)$, a polynomial and hence regular function on $X$. Thus $\varphi$ is a morphism and its pullback is $\alpha$. If one starts with $\varphi$, its pulled-back coordinate values reconstruct exactly $\varphi (x)$, so the constructions are inverses.

For composable morphisms, $s\circ (\psi \circ \varphi )=(s\circ \psi )\circ \varphi$, so $(\psi \circ \varphi {)}^{\ast }={\varphi }^{\ast }\circ {\psi }^{\ast }$; F8 ensures the composites are morphisms, and the identity pulls each function to itself. This also gives naturality of the bijection. If $X$ is empty there is one map to any $Y$ and one unital homomorphism to $k[X]=0$. If $Y$ is empty and $X$ nonempty there is neither a set map nor a unital map $0\to k[X]$, since $0=1$ would force $k[X]=0$. Both empty gives one on each side.

Let $B$ be a nonzero finite-type domain. By F7 choose a surjection $k[T_1,\dots ,T_r]\to B$ with kernel $P$. It is proper since $B\mathrm{\ne }0$, and $uv\in P$ forces one image to be zero since $B$ is a domain; thus $P$ is prime. F6 gives a variety $V(P)$ with $I(V(P))=P$. Its coordinate ring is the presentation quotient $B$. Together with the bijection and composition law this proves the stated antiequivalence on domains.