Lemma 26.6.1. Let $X$ be a locally ringed space. Let $Y$ be an affine scheme. Let $f \in \mathop{\mathrm{Mor}}\nolimits (X, Y)$ be a morphism of locally ringed spaces. Given a point $x \in X$ consider the ring maps

\[ \Gamma (Y, \mathcal{O}_ Y) \xrightarrow {f^\sharp } \Gamma (X, \mathcal{O}_ X) \to \mathcal{O}_{X, x} \]

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

\[ \xymatrix{ \Gamma (X, \mathcal{O}_ X) \ar[r] & \mathcal{O}_{X, x} \\ \Gamma (Y, \mathcal{O}_ Y) \ar[r] \ar[u] & \mathcal{O}_{Y, f(x)} \ar[u] } \]

(see the discussion of $f$-maps below Sheaves, Definition 6.21.7). Since the right vertical arrow is local we see that $\mathfrak m_{f(x)}$ is the inverse image of $\mathfrak m_ x$. The result follows.
$\square$

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