Let \(A\) be a ring and let \((X, \OO_X)\) be a scheme. Given a morphism \(f: X \to \Spec A\), we have an associated map on sheaves \(f^{\sharp}: \OO_{\Spec A} \to f_* \OO_X\). Taking global sections we obtain a homomorphism \(A \to \Gamma(X, \OO_X)\). Thus there is a natural map \[ \alpha: \Hom_{\Sch}(X, \Spec A) \to \Hom_{\Ring}\qty{A, \Gamma(X, \OO_X)}. \] Show that \(\alpha\) is bijective.
Solution
We use the affine case: for rings \(A,B\) there is a natural bijection \[ \operatorname{Hom}_{\mathrm{Sch}}(\operatorname{Spec}B,\operatorname{Spec}A) \cong \operatorname{Hom}_{\mathrm{Ring}}(A,B). \]
1. Let\[ \varphi:A\longrightarrow\Gamma(X,\mathcal O_X) \]be a ring homomorphism. Choose an affine open cover\[ X=\bigcup_iU_i, \qquad U_i=\operatorname{Spec}B_i. \]Restriction gives ring homomorphisms\[ \varphi_i: A\xrightarrow{\varphi} \Gamma(X,\mathcal O_X) \longrightarrow \Gamma(U_i,\mathcal O_X) =B_i. \]
The equality \[ \Gamma(U_i,\mathcal O_X)=B_i \] is the defining global-section property of the affine chart \(U_i\cong\operatorname{Spec}B_i\). Composition with restriction therefore gives the displayed maps.
2. Each \(\varphi_i\) determines a unique morphism of affine schemes\[ f_i:U_i\longrightarrow\operatorname{Spec}A. \]
Apply the affine anti-equivalence to the ring map \[ \varphi_i:A\to B_i. \]
3. For every pair \(i,j\), the morphisms \(f_i\) and \(f_j\) agree on the overlap\[ U_i\cap U_j. \]
Cover the open subscheme \[ U_i\cap U_j \] by affine opens \[ W=\operatorname{Spec}C. \]
The restriction \[ f_i|_W:W\to\operatorname{Spec}A \] corresponds, in the affine anti-equivalence, to the composite \[ A \xrightarrow{\varphi} \Gamma(X,\mathcal O_X) \longrightarrow \Gamma(U_i,\mathcal O_X) \longrightarrow \Gamma(W,\mathcal O_X)=C. \] The restriction \(f_j|_W\) corresponds to the same composite, because both restriction chains are the restriction of the same global section to \(W\).
Hence \[ f_i|_W=f_j|_W \] for every affine \(W\) in a cover of \(U_i\cap U_j\). Morphisms of schemes are local on the source, so \[ f_i|_{U_i\cap U_j}=f_j|_{U_i\cap U_j}. \]
4. The local morphisms \(f_i\) glue uniquely to a morphism\[ \boxed{ f_\varphi:X\longrightarrow\operatorname{Spec}A. } \]
By 3 the morphisms \(f_i\) agree on every pairwise overlap. Morphisms of locally ringed spaces glue over an open cover of the source: the underlying continuous maps glue because they agree on overlaps, and for every open subset \(W\) of the target the maps \[ \mathcal O_{\operatorname{Spec}A}(W) \longrightarrow \mathcal O_X(U_i\cap f_i^{-1}(W)) \] glue, by the sheaf axiom for \(\mathcal O_X\), to the required map on \[ f_\varphi^{-1}(W). \] The induced stalk maps are local because this can be checked on the open cover \(\{U_i\}\), where they are the stalk maps of the scheme morphisms \(f_i\).
Hence the \(f_i\) glue to a scheme morphism \(f_\varphi\). Uniqueness follows because two morphisms agreeing on every \(U_i\) agree globally.
5. The homomorphism on global sections induced by \(f_\varphi\) is exactly the original map \(\varphi\).
Let \[ \alpha(f_\varphi):A\to\Gamma(X,\mathcal O_X) \] be the pullback on global sections.
For every \(i\), restricting \(\alpha(f_\varphi)(a)\) to \(U_i\) gives the global-section pullback of \[ f_\varphi|_{U_i}=f_i. \] By construction of \(f_i\), this is \[ \varphi_i(a) =\varphi(a)|_{U_i}. \] Thus the two global sections \[ \alpha(f_\varphi)(a), \qquad \varphi(a) \] have equal restrictions to every member of the open cover \(\{U_i\}\). Since \(\mathcal O_X\) is a sheaf, they are equal. This holds for every \(a\in A\), so \[ \alpha(f_\varphi)=\varphi. \] Hence \(\alpha\) is surjective.
6. The map \(\alpha\) is injective.
Suppose \[ f,g:X\longrightarrow\operatorname{Spec}A \] induce the same ring homomorphism \[ \varphi:A\longrightarrow\Gamma(X,\mathcal O_X). \]
Choose an affine open cover \[ X=\bigcup_iU_i, \qquad U_i=\operatorname{Spec}B_i. \] The restrictions \[ f|_{U_i},g|_{U_i}:U_i\to\operatorname{Spec}A \] induce the same ring homomorphism \[ A \xrightarrow{\varphi} \Gamma(X,\mathcal O_X) \longrightarrow B_i, \] because pullback on global sections commutes with restriction to an open subset of the source.
By the affine anti-equivalence, \[ f|_{U_i}=g|_{U_i} \] for every \(i\). Since the \(U_i\) cover \(X\), the two morphisms are equal globally: \[ f=g. \] Thus \(\alpha\) is injective.
7. Therefore\[ \boxed{ \operatorname{Hom}_{\mathrm{Sch}}(X,\operatorname{Spec}A) \cong \operatorname{Hom}_{\mathrm{Ring}} \bigl(A,\Gamma(X,\mathcal O_X)\bigr). } \]
Step 5 proves surjectivity and 6 proves injectivity. The constructions use only restriction and the affine anti-equivalence, so the bijection is natural in both \(X\) and \(A\).
8. Q.E.D.
Step 7 is exactly the bijectivity of \(\alpha\) required by the exercise.