Rank inequality for a composition

Kernel of the composition. $\ker(g\circ f)=\{x: g(f(x))=0\}=f^{-1}(\ker g)$.

Dimension bound. Let $N=f^{-1}(\ker g)$ and let $h:N\to\ker g$ be the restriction of $f$ to $N$ (it takes values in $\ker g$ by definition of $N$). Then $\ker h=\ker f$, because $\ker f\subseteq N$. By rank-nullity applied to $h$,
$$\dim N=\dim\ker f+\dim\operatorname{Im} h\le\dim\ker f+\dim\ker g.$$
Thus $\dim\ker(g\circ f)\le\dim\ker f+\dim\ker g$.

Inequality. By rank-nullity for $g\circ f$, $f$ and $g$:
$$\operatorname{rank}(g\circ f)=\dim E-\dim\ker(g\circ f)\ge\dim E-\dim\ker f-\dim\ker g,$$
and $\dim E-\dim\ker f=\operatorname{rank}(f)$, $-\dim\ker g=\operatorname{rank}(g)-\dim F$. Hence
$$\operatorname{rank}(g\circ f)\ge\operatorname{rank}(f)+\operatorname{rank}(g)-\dim F.$$

Conclusion. This is Sylvester's rank inequality. For matrices $A\in\mathcal{M}_{p,n}$, $B\in\mathcal{M}_{n,q}$ it reads $\operatorname{rank}(AB)\ge\operatorname{rank}A+\operatorname{rank}B-n$.