Characterization of $\mathrm{rank}(g)\leq \mathrm{rank}(f)$ Let $E$ and $F$ two finite-dimensional spaces and $f,g\in\mathcal{L}(E,F)$. The question is to prove that:

$\mathrm{rank}(g)\leq \mathrm{rank}(f)$ iff there's $h\in GL(F)$ and $k\in\mathcal{L}(E)$ such that $h\circ g=f\circ k$

It's pretty easy to prove the sufficient condition but I'm stuck on the necessary condition. Thanks for any help.

Suppose $\mathrm{rank}\, g \leq \mathrm{rank}\, f$. Because of this condition, one can easily construct $h \in GL(F)$ such that $h(g(E)) \subset f(E)$ by taking bases. In other words, $\tilde{g} = h \circ g$ gives a linear map $E \to f(E)$.
Let $S$ be a subspace of $E$ such that $E = S \oplus \ker f$, and consider the restriction $\overline{f} = f|_S$. Then the map $\overline{f}\colon S \to f(E)$ is an isomorphism (this is a special case of the first factorization theorem).
It is straightforward to check that $k = {\overline{f}}^{-1} \circ \tilde{g}$, trivially extended from a map $E \to S$ to an endomorphism $E \to E$, provides a good choice. Indeed, since $k(E) \subset S$ one has
$$f \circ k = \overline{f}\circ{\overline{f}}^{-1} \circ \tilde{g} = \tilde{g} = h \circ g.$$
