This comes from the situation given in Hartshorne's proof of Theorem II.4.3. Essentially we assume $f: X \to Y$ is a separated morphism of schemes where $X$ is assumed to be noetherian. We also have morphisms $h_1: T \to X$ and $h_2: T \to X$ where $T = \operatorname{Spec} R$ for a Valuation Ring $R$ with field $K$. Denote $U = \operatorname{Spec} K$. There is an morphism $i: U \to T$ induced by $R \subseteq K$.
We assume that there are also morphisms $U \to X$ and $T \to Y$ so that the corresponding diagram commutes, so that $f \circ h_1 = f \circ h_2$ inducing a morphism a morphism $h': T \to X \times_Y X$. In the proof, we know that if $t_1$ is the generic point of $T$ then it comes from $U$ so $h_1(t_1) = h_2(t_1)$. Then, Hartshorne concludes from this that $h'(t_1) \in \Delta(X)$ which is unclear to me.
There are some other questions which address this exact same situation with the agreement happening on open/closed sets but many of the answers reccomended against working with points but I'm not sure how to avoid that in this case, since we only know the agreement happens on a point, which is not closed and probably not open.
How can I show that $h'(t_1) \in \Delta(X)$ without these?
Thank you!
I think I figured it out. Excuse the large commutative diagrams since I don't know how to add them otherwise. Suppose $p_1, p_2: X \times_Y X \to X$ are the projection morphisms. We recall that if we have a map of $Y$-schemes $f: U \to X \times_Y X$ which satisfies $p_1 \circ f = p_2 \circ f$, then $f$ factors through the diagonal.
With this in mind, suppose that we have morphisms $i: U \to T$ where $h_1: T \to X$ and $h_2: T \to X$ are morphisms $Y$-schemes inducing $h: T \to X \times_Y X$. I want to show that $i(U) \subseteq h^{-1}(\Delta(X))$. We then observe the badly drawn commutative diagram
In particular, we note that $$p_1 \circ h \circ i = h_1 \circ i = h_2 \circ i = p_2 \circ h \circ i$$ so that $h \circ i$ factors through $\Delta$. This then yields the following commutative square.
From the commutativity of the square we can quickly conclude that $h(i(U)) \subseteq \Delta(X)$ so that $i(U) \subseteq h^{-1}(\Delta(X))$ as required.