I need some hints to understand the conclusion of the proof of the following lemma from the Stacks Project:
Lemma $\mathbf{6.1.}\,$ Let $X$ be a locally ringed space. Let $Y$ be an affine scheme. Let $f\in\operatorname{Mor}(X,Y)$ be a morphism of locally ringed spaces. Given a point $x\in X$ consider the ring maps $$\Gamma(Y,\mathcal O_Y)\xrightarrow{f^{\#}}\Gamma(X,\mathcal O_X)\to\mathcal O_{X,x}$$ Let $\mathfrak p\subset\Gamma(Y,\mathcal O_Y)$ denote the inverse image of $\mathfrak m_x$. Let $y\in Y$ be the corresponding point. Then $f(x)=y$.
Proof. Consider the commutative diagram
$$\require{AMScd}\begin{CD}\Gamma(X,\mathcal O_X) @>>> \mathcal O_{X,x}\\ @AAA @AAA \\ \Gamma(Y,\mathcal O_Y) @>>> \mathcal O_{Y,f(x)}\end{CD}$$ (see the discussion of $f$-maps below Sheaves, Definition $21.7$). Since the right vertical arrow is local we see that $\mathbf m_{f(x)}$ is the inverse image of $\mathbf m_x$. The result follows.$\qquad\qquad\qquad\qquad\quad\square$
I don't understand the last sentence, i.e. "The result follows". If $m_{f(x)}$ (the maximal ideal of $\mathcal O_{Y,f(x)}$) is the inverse image of $m_x$, why should be $f(x)=y$?
Thanks in advice.
$Y$ is an affine scheme, so $f(x)$ is a prime ideal of $\mathscr{O}_Y(Y)$ and $\mathscr{O}_{Y,f(x)}$ is the localization of $\mathscr{O}_Y(Y)$ at $f(x)$. A basic fact about localization is that the inverse image in $\mathscr{O}_Y(Y)$ of the unique maximal ideal $\mathfrak{m}_{f(x)}$ of $\mathscr{O}_{Y,f(x)}$ is $f(x)$. Now it follows from the commutativity of the diagram, together with the fact you mentioned, which amounts to the map on stalks being local, that $f(x)$ is the inverse image in $\mathscr{O}_Y(Y)$ of $\mathfrak{m}_x$ under the map $\mathscr{O}_Y(Y)\rightarrow\mathscr{O}_X(X)\rightarrow\mathscr{O}_{X,x}$.