Proof in English of Proposition on morphisms into affine schemes

Question

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).

Answer 1

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{1.6.3.2} \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 $\varphitilde = \theta$.