Lemma 66.33.1. Let $X$ be an algebraic space over $\mathbf{Z}$. Let $T$ be an affine scheme. The map

which maps $f$ to $f^\sharp $ (on global sections) is bijective.

Lemma 66.33.1. Let $X$ be an algebraic space over $\mathbf{Z}$. Let $T$ be an affine scheme. The map

\[ \mathop{\mathrm{Mor}}\nolimits (X, T) \longrightarrow \mathop{\mathrm{Hom}}\nolimits (\Gamma (T, \mathcal{O}_ T), \Gamma (X, \mathcal{O}_ X)) \]

which maps $f$ to $f^\sharp $ (on global sections) is bijective.

**Proof.**
We construct the inverse of the map. Let $\varphi : \Gamma (T, \mathcal{O}_ T) \to \Gamma (X, \mathcal{O}_ X)$ be a ring map. Choose a presentation $X = U/R$, see Spaces, Definition 65.9.3. By Schemes, Lemma 26.6.4 the composition

\[ \Gamma (T, \mathcal{O}_ T) \to \Gamma (X, \mathcal{O}_ X) \to \Gamma (U, \mathcal{O}_ U) \]

corresponds to a unique morphism of schemes $g : U \to T$. By the same lemma the two compositions $R \to U \to T$ are equal. Hence we obtain a morphism $f : X = U/R \to T$ such that $U \to X \to T$ equals $g$. By construction the diagram

\[ \xymatrix{ \Gamma (U, \mathcal{O}_ U) & \Gamma (X, \mathcal{O}_ X) \ar[l] \\ & \Gamma (T, \mathcal{O}_ T) \ar[lu]^{g^\sharp } \ar[u]^{\varphi }_{f^\sharp } } \]

commutes. Hence $f^\sharp $ equals $\varphi $ because $U \to X$ is an étale covering and $\mathcal{O}_ X$ is a sheaf on $X_{\acute{e}tale}$. The uniqueness of $f$ follows from the uniqueness of $g$. $\square$

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