Equality of two $K$-valued points for reduced $K$ 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? 
 A: 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 $\varphi = \psi$.
A: Suppose $f(x) \equiv g(x)$ for all $x \in K$, then $f = g$ as continuous maps.
Let $U$ open in $X$ and let $s \in \Gamma(U, \mathcal{O}_X)$, we will check that $f_U^\sharp(s) -g_U^\sharp(s)=0$ in $\Gamma(f^{-1}U,\mathcal{O}_K)$.
Let $U_\alpha = \operatorname{Spec}(K_\alpha)$ be an affine cover of $K$. 
Then choose a $h \in K_\alpha$ such that $D(h) \subset f^{-1}U\cap U_\alpha$.
Then for all $y \in D(h)$ we can write $y = [P]$ for some prime ideal $P \subset K_\alpha$ and $[(r_{f^{-1}U, D(h)}(f_U^\sharp(s)-g_U^\sharp(s)),D(h))] \in \mathcal{O}_{K,y} = \mathcal{O}_{U_\alpha,y} = (K_\alpha)_{P}$. Here $r_{f^{-1}U,D(h)} : \Gamma(f^{-1}U,\mathcal{O}_K) \to \Gamma(D(h),\mathcal{O}_K)$ is the restriction map.
Now we use that $i_{y,[(0)]}^\sharp : \mathcal{O}_{K,y} \to \mathcal{O}_{\operatorname{Spec}(k(y)),[(0)]} = (K_\alpha)_P \to (K_\alpha)_P/(P (K_\alpha)_P)$ maps $[(r_{f^{-1}U, D(h)}(f_U^\sharp(s)-g_U^\sharp(s)),D(h))]$ to zero because $f \circ i_y = g \circ i_y$, thus $r_{f^{-1}U, D(h)}(f_U^\sharp(s)-g_U^\sharp(s)) \in P (K_\alpha)_P$.
Then because our choice of $y$ was arbitrary, we have $r_{f^{-1}U, D(h)}(f_U^\sharp(s)-g_U^\sharp(s)) \in \bigcap_{h \not \in P} P(K_\alpha)_P = \operatorname{Nil}((K_\alpha)_h)$.
Then because $K$ is reduced, we have $\operatorname{Nil}(K_\alpha)=0$ so also $\operatorname{Nil}((K_\alpha)_h) = 0$.
This means $r_{f^{-1}U, D(h)}(f_U^\sharp(s)-g_U^\sharp(s)) = 0$ in $(K_\alpha)_h= \Gamma(D(h),\mathcal{O}_K)$ and because our choice of $h$ was arbitrary, we also have $r_{f^{-1}U, f^{-1}U \cap U_\alpha}(f_U^\sharp(s)-g_U^\sharp(s)) = 0$ in $\Gamma(f^{-1}U \cap U_\alpha,\mathcal{O}_K)$. Then finally because our choice of $\alpha$ was arbitrary, we have $f_U^\sharp(s)-g_U^\sharp(s) = 0$ in $\Gamma(f^{-1}U,\mathcal{O}_K)$.
