Let $X,Y,Z$ be schemes. I'm trying to prove that:
If $X\to Y\to Z$ is a closed immersion and $Y\to Z$ is separated, then $X\to Y$ is a closed immersion
Here's where I'm at: Let $f:X\to Y$, $g:Y\to Z$. Since $g\circ f$ is a closed immersion, in particular $f$ is injective. Furthermore, $(g\circ f)^\#_x=f_x^\#\circ g_x^\#$ is surjective, hence $f_x^\#$ is surjective for all $x$.
The only thing left is to show that $f$ is closed. By assumption, if $F\subset X$ is closed, then $(g\circ f)(F)$ is closed.
I don't know how to conclude $f(F)$ is closed using the fact that $g$ is separated.
Is this even possible? Any suggestions are welcome, thank you!
First, the graph $\Gamma_f = (1_X,f) : X \to X \times_Z Y$ is the pullback of the diagonal $\Delta = (1_Y, 1_Y) : Y \to Y \times_Z Y$ along the map $f \times_Z 1_Y : X \times_Z Y \to Y \times_Z Y$ (check this directly). Since closed immersions are stable under base change and $\Delta$ is a closed immersion by assumption, $\Gamma_f$ is a closed immersion.
Now, $f$ factors as $\pi_2 \circ \Gamma_f$, where $\pi_2 : X \times_Z Y \to Y$ is the projection on the second factor, so it suffices to show that $\pi_2$ is a closed immersion. To do this, note that $\pi_2$ can be identified with map $(g \circ f) \times_Z 1_Y : X \times_Z Y \to Z \times_Z Y$ via the canonical identification $Y \cong Z \times_Z Y$. But $g \circ f$ is a closed immersion by assumption, and closed immersions are stable under base change, so $\pi_2$ is a closed immersion as well.
We conclude that $f = \pi_2 \circ \Gamma_f$ is the composition of two closed immersions, hence also a closed immersion.