 Solution

A morphism of schemes $f:X\to Y$ defines a natural transformation $g_f:g_X\to g_Y$ by $u\mapsto f\circ u$. We construct the inverse correspondence.

Let $\alpha :g_X\to g_Y$ be a natural transformation, and choose an affine open cover $(j_i:U_i\hookrightarrow X)$. Regard $j_i$ as an element of $g_X(U_i)$ and setHere and below the equivalence of categories in the question lets us write the restricted functor of points on affine schemes, contravariantly, rather than on commutative rings, covariantly. If $W\subseteq U_i\cap U_j$ is an affine open subscheme, naturality for its two inclusions givesSuch $W$ cover the overlap even when that overlap is not an affine scheme. Consequently the $f_i$ agree and glue to a morphism of schemes $f:X\to Y$.

It remains to check all elements of the restricted functor of points, not only the chosen inclusions. Given an affine scheme $T$ and $u:T\to X$, cover each $u^{-1}(U_i)$ by affine open subschemes $V$. On such a $V$, factor $u{|}_V=j_i\circ v$. Applying naturality first to $v:V\to U_i$ and then to $V\hookrightarrow T$ givesThese local equalities prove ${\alpha }_T(u)=f\circ u$. The glued $f$ is unique, since its restrictions to all $U_i$ are prescribed. Henceso the assignment is fully faithful.

If $g_X\cong g_Y$, the natural isomorphism and its inverse correspond to morphisms of schemes $f:X\to Y$ and $h:Y\to X$. Their compositions induce the identity natural transformations, and faithfulness implies $hf={\mathrm{id}}_X$, $fh={\mathrm{id}}_Y$. Conversely, an isomorphism of schemes induces a natural isomorphism. Thus