What assumptions on $f: X \to Y$ such that $\Delta: X \to X \times_Y X$ is a closed immersion?

Question

Let $X, Y$ be separated schemes and $f: X \to Y$ a morphism of schemes. Is it then always the case that $f$ is separated (i.e. $\Delta: X \to X \times_Y X$ is a closed immersion)?

Answer 1

Lemma: Let $f:X\to Y$ and $g:Y\to Z$ be morphisms of schemes. If $g\circ f$ is separated, then $f$ is separated.

Proof: Consider the factorization of the diagonal morphism $X\to X\times_Y X\to X\times_Z X$. The second map is an immersion: $X\times_Y X\to X\times_Z X$ is the base change of $\Delta:Y\to Y\times_Z Y$ along $X\times_Z X\to Y\times_Z Y$, and the base change of an immersion is an immersion. As the composite $X\to X\times_Y X\to X\times_Z X$ is a closed immersion by assumption, we see that $X\to X\times_Y X$ must also be a closed immersion. $\blacksquare$

To apply this to your situation, all you need is that $X$ is separated (or more generally, $X$ and $Y$ are $Z$-schemes with $f$ a $Z$-morphism and $X$ separated over $Z$).