A locally ringed space is a topological space $X$ equipped with a sheaf of commutative rings ${\mathcal{O}}_X$ whose stalk ${\mathcal{O}}_{X,x}$ at every point is a local ring. A morphism of locally ringed spaces $(f,f^{\#}):(Y,{\mathcal{O}}_Y)\to (X,{\mathcal{O}}_X)$ consists of a continuous map $f:Y\to X$ and a morphism of sheaves of rings $f^{\#}:{\mathcal{O}}_X\to f_{\ast }{\mathcal{O}}_Y$, with every induced stalk map ${\mathcal{O}}_{X,f(y)}\to {\mathcal{O}}_{Y,y}$ a local homomorphism: the inverse image of the target maximal ideal is the source maximal ideal. Here $f_{\ast }{\mathcal{O}}_Y$ is the direct image sheaf, with sections ${\mathcal{O}}_Y(f^{-1}U)$ on $U$.

Let $\varphi :A\to B$ be a ring homomorphism. Define the map between the spectra of rings byThe inverse image is a prime ideal, and $f^{-1}(D(a))=D(\varphi (a))$. Thus $f$ is a continuous map, because the principal open subsets form a basis. On these principal open subsets, define the structure sheaf map using the localizationsThese maps commute with restrictions and consequently give a morphism of sheaves. At $\mathfrak{q}$, putting $\mathfrak{p}={\varphi }^{-1}\mathfrak{q}$, the stalk map is $A_{\mathfrak{p}}\to B_{\mathfrak{q}}$. An element $b/s$ belongs to its inverse image of $\mathfrak{q}{B}_{\mathfrak{q}}$ exactly when $b\in \mathfrak{p}$, so this is a local homomorphism. We have constructed a morphism of locally ringed spaces.

Conversely, take a morphism of locally ringed spaces $f:\mathrm{Spec}B\to \mathrm{Spec}A$. Its map on global sections gives a ring homomorphismLocality of the stalk map implies, for every $a\in A$,Hence the point map must be $\mathfrak{q}\mapsto {\varphi }^{-1}\mathfrak{q}$. On $D(a)$, its structure sheaf map must send $b/1$ to $\varphi (b)/1$ and the inverse of $a$ to the inverse of $\varphi (a)$. Thus it is precisely the localization map constructed above. Agreement on this basis proves agreement of the morphisms of sheaves. In particular, the inducing ring homomorphism exists and is unique.

If $\varphi$ is surjective, write $B=A/I$. The correspondence between prime ideals of $A/I$ and prime ideals of $A$ containing $I$ identifies $\mathrm{Spec}B$ homeomorphically with the closed subset $V(I)$: the principal open subsets correspond on both sides. Moreover, the stalk maps are the surjections $A_{\mathfrak{p}}\to (A/I{)}_{\mathfrak{p}/I}$, so ${\mathcal{O}}_X\to f_{\ast }{\mathcal{O}}_Y$ is surjective.

For the converse, it is important not to assume that global sections preserve an arbitrary surjection of sheaves. Here the affine module-sheaf equivalence supplies the extra information: on every $D(a)$,Thus $f_{\ast }{\mathcal{O}}_Y$ is the quasi-coherent sheaf associated with the $A$-module $B$, and the cokernel sheaf of ${\mathcal{O}}_X\to f_{\ast }{\mathcal{O}}_Y$ is associated with the module $C=\mathrm{coker}(A\to B)$, since localization of a module is exact. Surjectivity of the morphism of sheaves gives $C_{\mathfrak{p}}=0$ for every prime ideal $\mathfrak{p}$. This implies $C=0$: if $0\ne c\in C$, its proper annihilator lies in a maximal ideal $\mathfrak{m}$, and then $c/1\ne 0$ in $C_{\mathfrak{m}}$. Therefore $A\to B$ is surjective. We have proved the closed immersion criterion for affine schemes: