How do I understand this sentence in Hartshorne Chapter 2? It is in Hartshorne Chapter 2 theorem 4.3.(at the top of page 99)
The theorem (Valuative Criterion of Separatedness) claims that a morphism $f: X \rightarrow Y$ is separated if and only if the following holds. For any field $K$ and any valuation ring $R$ with quotient field $K$, let $T=\operatorname{Spec}R$ and $U=\operatorname{Spec}K$, let $i: U\rightarrow T$ be the morphism induced by inclusion $R\subset K$, given morphisms $g_1: U\rightarrow X$ and $g_2: T\rightarrow Y$ such that $f\circ g_1=g_2\circ i$ (i.e. the diagram commutes). Then there is at most one morphism from $T$ to $X$ making the diagram still commuted.
In the proof of the theorem, Hartshorne first notes that if $h$ and $h'$ are two different morphism from $T$ to $X$, then by universal property of fibre product, we have an induced morphism $h'':T\rightarrow X\otimes X$, then he claimed that 'since the restrictions of $h$ and $h'$ to $U$ are the same, the generic point $t_1$ of $T$ has image in the diagonal $\Delta (X)$.' Where this diagonal map，$\Delta: X\rightarrow X\otimes X$, is induced by two identities maps from $X$ to $X$.
I am confused that why the image is in the diagonal $\Delta (X)$? I think Hartshorne is saying $h(t_1)=h'(t_1)$, where $t_1$ is the generic point of $T$, because they restricted to the same map from $U$ to $X$, however I am not sure does this mean $h''(t_1)=\Delta(h(t_1))$?
 A: The restrictions of $h$ and $h'$ to $U$ are the same, let's call them $a: U\to X$. Then the induced map $U\to X\times X$ given by restricting $h''$ is the same as the map to the product induced by the restrictions of $h$ and $h'$; i.e., it is the product of $a$ with itself. But $a\times a: U\to X\times X$ factors as $$U\xrightarrow{a}X\xrightarrow{\Delta}X\times X,$$ so the image of $U$ lies in the diagonal. Explicitly, this does mean that $h''(t_1)=\Delta(h(t_1))=\Delta(h'(t_1))$.
A: Let $g:U\rightarrow X\times X$ to be the morphism induced by $g_1:U\rightarrow X$ (i.e. $g=g_1\times g_1$). Let $\pi_1 : X\times X \rightarrow X$ and $\pi_2 : X\times X \rightarrow X$ be the projection to each of the $X$ respectively. Then by definition, $\pi_1 \circ g =g_1$.
Now notice $\pi_1 \circ \Delta \circ h \circ i =id \circ g_1 = g_1$, then by universal property of $g$ (i.e. $g$ is the UNIQUE map such that $\pi_1 \circ g =g_1$), we have $\Delta \circ h \circ i = g$.
Similarly, notice $\pi_1 \circ h'' \circ i = h \circ i = g_1$, therefore $h'' \circ i = g$.
This means $h ''\circ i = \Delta \circ h \circ i$, which is the same as saying $h''(t_1)=\Delta(h(t_1))$.
