Equality of morphisms $f,g:K\rightarrow X$ of schemes, where $K$ is a reduced scheme. Suppose that $f,g:K\rightarrow X$ are morphisms of schemes, where $K$ is a reduced scheme. I want to show that $f=g$ if and only if for all $x\in K$, $f(x)\equiv g(x)$. 
Here $f(x)\equiv g(x)$ means that $f(x)=g(x)$ and the induced maps on residue fields $\kappa(f(x))\rightarrow\kappa(x)$ induced by $f_{x}^{\#}$ and $g_{x}^{\#}$ are equal. 
Notice that the direction $f=g$ implies $f(x)\equiv g(x)$ for all $x\in K$ is trivial. Thus we are left to show the other implication. 
So suppose that $f(x)\equiv g(x)$ for all $x\in K$, then $f=g$ as on the underlying topological spaces. We are left to show that the homomorphisms $f_{V}^{\#},g_{V}^{\#}:\mathcal{O}_{X}(V)\rightarrow \mathcal{O}_{K}(f^{-1}(V))$ are equal for each $V$ open in $X$. Note that the induced maps $\overline{f_{x}^{\#}},\overline{g_{x}^{\#}}:\mathcal{O}_{X,f(x)}/\mathfrak{m}_{f(x)}\rightarrow\mathcal{O}_{K,x}/\mathfrak{m}_{x}$ are equal. 
Question: How can I use the reducedness assumption to finish the prove here? 
Recall: Reducedness implies that for all opens $U\subset K$ the rings $\mathcal{O}_{K}(U)$ has no nilpotent, or equivalently for all $x\in K$ the stalks $\mathcal{O}_{K,x}$ have no nilpotent elements. 
Second Question (curiousity): Are there any trivial examples of non-reduced schemes $K$ for which this statement is false?
 A: Let $ U = f^{-1}(V) = g^{-1}(V) $. We wish to show that the two pullback maps $ f^*, g^* : \mathcal{O}_X(V) \rightarrow \mathcal{O}_K(U) $ are equal. Any local section $ s \in \mathcal{O}_X(V) $ can be viewed as a morphism $ s : V \rightarrow \mathbb{A}^{1}_{\mathbb{Z}} $, so we wish to show that the two morphisms $ s \circ f, s \circ g : U \rightarrow \mathbb{A}^{1}_{\mathbb{Z}} $ are equal. Noting that $ (s \circ f)(u) \equiv (s \circ g)(u) $ for all $ u \in U $, we have reduced to the case when $ X = \mathbb{A}^{1}_{\mathbb{Z}} $. Further, since $ U $ is covered by open affine subsets, we reduce to the case when $ K = \operatorname{spec}(B) $, $ B $ a reduced ring.
So let $ f,g : \operatorname{spec}(B) \rightarrow \mathbb{A}^{1}_{\mathbb{Z}} $ with $ f \equiv g $. Let $ \phi, \psi : \mathbb{Z}[T] \rightarrow B $ be the corresponding ring maps and we need to show $ \phi = \psi $. Suppose there is an element $ p \in \mathbb{Z}[T] $ such that $ \phi(p) - \psi(p) = b \neq 0 $. Since $ B $ is reduced, there is a prime ideal $ \mathfrak{q } $ of $ B $ such that $ b \notin \mathfrak{q} $. Let $ \mathfrak{p} = \phi^{-1}(\mathfrak{q}) = \psi^{-1}(\mathfrak{q}) $. The induced maps on the residue fields $ \phi_{\mathfrak{p}}, \psi_{\mathfrak{p}} : \mathbb{Z}[T]_{\mathfrak{p}} / \mathfrak{p} \mathbb{Z}[T]_{\mathfrak{p}} \rightarrow B_{\mathfrak{q}} / \mathfrak{q} B_{\mathfrak{q}} $ are equal by hypothesis. So $ \phi_{\mathfrak{p}}(p) = \psi_{\mathfrak{p}}(p) $ i.e., $ b \in \mathfrak{q}B_{\mathfrak{q}} \cap B = \mathfrak{q} $. This is a contradiction, finishing the proof.
KReiser's example for the second question is to take $ K = X = \operatorname{spec} (k[T]/T^2) $. The two maps on the rings are the identity map and $ T \rightarrow 0 $. Then of course the maps agree on the topological spaces and on the residue field $ k $, both maps induce the identity map.
