In Hartshorne's "Algebraic Geometry", on p.99, at the end of the proof of Lemma 4.5, the author mentions that in general, if $\xi \in X \times_Y X$ and $p_1(\xi) = p_2(\xi)$, where $p_1$ and $p_2$ are the projection morphisms onto the first and second factors respectively, then it is not true in general that $\xi$ belongs necessarily to $\Delta(X)$, where
$\Delta: X \to X \times_Y X$
is the diagonal morphism.
I haven't thought much about it, but is it easy to construct a counterexample? I ask for examples of schemes $X$ and $Y$, with a morphism $f: X \to Y$, such that there is a point $\xi \in X \times_Y X$ with $p_1(\xi) = p_2(\xi)$, but $\xi \notin \Delta(X)$.
Edit: If we take $X = \mathbb{A}^1_\mathbb{C}$ to be the affine line over $\mathbb{C}$. Then $X \times_\mathbb{C} X$ is the affine plane $\mathbb{A}^2_\mathbb{C}$. Let $\xi$ be the generic point in $\mathbb{A}^2_\mathbb{C}$ of the line $y=2x$. Then is it true that $p_1(\xi) = p_2(\xi)$, and that $\xi \notin \Delta(X)$? Let $f_1: \mathbb{C}[x] \to \mathbb{C}[x,y]$ and $f_2: \mathbb{C}[y] \to \mathbb{C}[x,y]$ be the natural ring homomorphisms, corresponding to the 2 projections. Then $f_i^{-1}(y-2x) = (0)$, for $i=1,2$, which confirms my suspicions, that $p_1(\xi) = p_2(\xi)$. Moreover, $\xi$ does not belong to $\Delta(X)$. Thus I indeed found a counterexample.
Let $Y=\operatorname{Spec}(\mathbb R), X=\operatorname{Spec}(\mathbb C)$. Since $X$ consists of one point, $\Delta(X)$ is just one point, and therefore also $p_1(q)=p_2(q)$ for each $q\in X\times_Y X$. Note that $$X\times_Y X=\operatorname{Spec}(\mathbb C\otimes_{\mathbb R}\mathbb C)=\operatorname{Spec}(\mathbb C\times\mathbb C)=\operatorname{Spec}(\mathbb C)\amalg\operatorname{Spec}(\mathbb C)$$ consists of two points, so $\Delta(X)\neq X\times_Y X$.