Does anybody know where to find an English proof of the following proposition

Let $(X, \mathcal{O}_X)$ be a locally ringed space, $Y = \operatorname{Spec} A$ an affine scheme. Then the natural map $$\begin{align}\operatorname{Hom}(X, Y ) &\to \operatorname{Hom}(A, Γ(X, \mathcal{O}_X)),\\ (f, f^\flat) &\mapsto f^\flat_Y ,\end{align}$$ is a bijection.

This is Prop. 3.4 in Görtz's and Wedhorn's Algebraic Geometry I (there the proof is only given for the case where $X$ is a scheme) and Proposition 1.6.3 in the 1971 edition of Grothendieck's Eléments de Géométrie Algébrique (full proof given there, but it's in French and AFAIK there is no translation).


Here's my effort at a translation. The original text uses $S = Y$, as I will.

Let $\DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\O}{\mathcal{O}} A = \Gamma(S, \O_S)$, and consider a ring homomorphism $\varphi: A \to \Gamma(X, \O_X)$. For each $x \in X$, the set of $f \in A$ such that $\varphi(f)(x) = 0$ (0, 4.1.9) is a prime ideal of $A$, because $\DeclareMathOperator{\m}{\mathfrak{m}} \O_x/\m_x = \kappa(x)$ is a field; it is thus an element of $\DeclareMathOperator{\Spec}{Spec} S = \Spec(A)$, that we will again denote ${^a \varphi}(x)$. Moreover, for each $f \in A$, we have by definition (0, 4.1.13) $\newcommand{\varphia}{{^a \varphi}} \varphia^{-1}(D(f)) = X_{\varphi(f)}$, which shows that $\varphia$ is a continuous map from $X$ to $S$. Define next a homomorphism $\newcommand{\varphitilde}{\tilde{\varphi}} \varphitilde: \O_S \to \varphia_*(\O_X)$ of $\O_S$-modules: for each $f \in A$, we have $\Gamma(D(f), \O_S) = A_f$ (1.3.6); for each $s \in A$, we will make correspond to $s/f \in A_f$ the element $$ (\varphi(s)|X_{\varphi(f)})(\varphi(f)|X_{\varphi(f)})^{-1} $$ of $\Gamma(X_{\varphi(f)}, \O_X) = \Gamma(D(f), \varphia_*(\O_X))$, and we prove immediately, by passing from $D(f)$ to $D(fg)$ for each $g \in A$, that this defines a homomorphism of $\O_S$-modules; thus we have obtained a morphism $(\varphia, \varphitilde)$ of ringed spaces. Furthermore, with the same notation, and setting $y = \varphia(x)$, we see immediately (0, 3.7.1) that we have $\varphitilde_x^\#(s_y/f_y) = (\varphi(s)_x)(\varphi(f)_x)^{-1}$; as the relation $s_y \in \m_y$ is by definition equivalent to $\varphi(s)_x \in \m_x$, we see that $\varphitilde_x^\#$ is a local homomorphism $\O_y \to \O_x$, in other words $(\varphia, \varphitilde)$ is a morphism of locally ringed spaces. Thus we have defined a canonical map \begin{equation} \sigma : \Hom(\Gamma(S, \O_S), \Gamma(X, \O_X)) \to \Hom_{\text{loc}}(X,S) \, . \tag{} \end{equation}

It remains to prove that $\rho_{\text{loc}}$ and $\sigma$ are mutually inverse. However, the definition given above for $\varphitilde$ shows immediately that $\Gamma(\varphitilde) = \varphi$, and hence $\rho_{\text{loc}} \circ \sigma$ is the identity. To see that $\sigma \circ \rho_\text{loc}$ is the identity, we begin with a morphism $(\psi, \theta): X \to S$ of locally ringed spaces, and set $\varphi: \Gamma(\theta)$; the hypothesis that $\theta_x^\#$ is local allows us to deduce that this homomorphism, by passing to quotients, is a monomorphism of fields $\theta^x : \kappa(\theta(\psi(x)) \to \kappa(x)$ such that, for each section $f \in A = \Gamma(S, \O_S)$, we have $\theta^x(f(\psi(x)) = \varphi(f)(x)$; the relation $f(\psi(x)) = 0$ is thus equivalent to $\varphi(f)(x) = 0$, which shows that $\varphia = \psi$ by virtue of the definition of $\varphia$. On the other hand, the definitions imply that the diagram [see diagram in text] is commutative, and the same is true of the analogous diagram where $\theta_x^\#$ is replaced by $\varphitilde_x^\#$, hence $\varphitilde_x^\# = \theta_x^\#$ (Bourbaki, Alg. comm., chap. II, $\S 2$, no. 1, prop. 1), which implies that $\varphitilde = \theta$.


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