Affine morphisms are contravariantly equivalent to coordinate-ring homomorphisms

Let $\phi :X\to Y$ be a morphism. By [L1], pullback sends global regular functions on $Y$ to global regular functions on $X$, and [L2] identifies those two rings with $k[Y]$ and $k[X]$. Therefore $\phi$ determines a $k$-algebra homomorphism ${\phi }^{\ast }:k[Y]\to k[X]$, $h\mapsto h\circ \phi$. This is the desired pullback homomorphism attached to $\phi$.

Conversely, let $\alpha :k[Y]\to k[X]$ be a $k$-algebra homomorphism. Write ${\bar{y}}_1,\dots ,{\bar{y}}_m$ for the coordinate classes of $Y$ in [L3]. By [L2], each $\alpha ({\bar{y}}_i)$ is a global regular function on $X$, hence an honest function $X\to k$. Define ${\phi }_{\alpha }(x):=(\alpha ({\bar{y}}_1)(x),\dots ,\alpha ({\bar{y}}_m)(x))\in {\mathbf{A}}_k^m$. This produces a candidate point of affine $m$-space for each $x\in X$.

For every polynomial $g\in I(Y)$, its class is $0$ in $k[Y]$ by [L3]. Hence $g(\alpha ({\bar{y}}_1),\dots ,\alpha ({\bar{y}}_m))=\alpha (g)=0$ in $k[X]$. Evaluating at $x\in X$ gives $g({\phi }_{\alpha }(x))=0$. Since [L4] says that the common zero set of $I(Y)$ is exactly $Y$, the point ${\phi }_{\alpha }(x)$ lies in $Y$. Thus ${\phi }_{\alpha }:X\to Y$ is well defined.

Let $h\in k[Y]$. By [L3], $h$ is a polynomial expression in the coordinate classes ${\bar{y}}_i$. Pulling that expression back along ${\phi }_{\alpha }$ replaces each ${\bar{y}}_i$ by $\alpha ({\bar{y}}_i)$, so ${\phi }_{\alpha }^{\ast }(h)=\alpha (h)$. In particular ${\phi }_{\alpha }^{\ast }(h)$ is regular on $X$ by [L2], and [L1] shows that ${\phi }_{\alpha }$ is a morphism.

Starting from a morphism $\phi$, the map constructed from ${\phi }^{\ast }$ has the same pullback on the coordinate classes ${\bar{y}}_i$, so it has the same value as $\phi$ at every point of $X$. Starting from a homomorphism $\alpha$, step 2.2 shows that the pullback of the constructed morphism is exactly $\alpha$. Therefore the two constructions are inverse to each other.

For the identity morphism on $X$, the pullback is the identity on $k[X]$ by the formula in step 1.1. If $\psi :Y\to Z$ and $\phi :X\to Y$ are morphisms, then for every $h\in k[Z]$ one has $(\psi \circ \phi {)}^{\ast }(h)=h\circ \psi \circ \phi ={\phi }^{\ast }({\psi }^{\ast }(h))$, so pullback reverses composition. Together with step 3.1 this proves the stated contravariant correspondence.