Equality of two $K$-valued points for reduced $K$

Question

I'm reading "Red Book of varieties and schemes". There is definition 2, page 118. Let $f,g:K \to X $ be to $K-$valued points of scheme $X$, we say that they are equal at $x\in K$ $(f(x) \equiv g(x))$ if $f i_x=g i_x$, where $i_x: \operatorname{Spec}(k_x) \to K$ is canonical morphism. Then he wrote "It is easy to check that if K is reduced then $f=g$ iff $f(x)\equiv g(x)$ for all $x\in K$" How it could be proven?

Answer 1

Suppose $f(x) \equiv g(x)$ for all $x \in K$. Then in particular $f = g$ as continuous maps $K \to X$, so we only need to check that they're the same on sheaves as well. Pick some $x \in K$, and let $y := f(x)$. Then we want to check that $f^\#_y : \mathcal{O}_{X, y} \to \mathcal{O}_{K, x}$ agrees with $g^\#_y$. Since $f \circ i_x = g \circ i_x$ as morphisms of schemes, we have that the compositions $$\mathcal{O}_{X, y} \xrightarrow{f^\#_y} \mathcal{O}_{K, x} \xrightarrow{i^\#_x} \mathcal{O}_{\operatorname{Spec} k(x), x}$$ and $$\mathcal{O}_{X, y} \xrightarrow{g^\#_y} \mathcal{O}_{K, x} \xrightarrow{i^\#_x} \mathcal{O}_{\operatorname{Spec} k(x), x}$$ agree. Now $$\mathcal{O}_{\operatorname{Spec} k(x), x} = k(x) = \mathcal{O}_{K, x} / \mathfrak{m}_{K, x}$$ and $i^\#_x$ is the quotient map.

So, in other words, we're given two local homomorphisms $$\varphi, \psi : (R, \mathfrak{m}) \to (S, \mathfrak{n})$$ such that $$R \xrightarrow{\varphi} S \twoheadrightarrow S / \mathfrak{n}$$ agrees with $$R \xrightarrow{\psi} S \twoheadrightarrow S / \mathfrak{n}$$ and we claim that if $S$ is reduced then we must have $\varphi = \psi$.