When is $F:L(U,V)\times L(V,W)\to L(U,W)$, $F(f,g)=f\circ g$ surjective? Let $U,V,W$ be finite dimensional vector spaces over a field $K$ and let $L(U,V)$ be the set of all linear applications from $U$ to $V$ (analogues for other pairs): Define an application $F:L(U,V)\times L(V,W)\to L(U,W)$ by $F(f,g)=f\circ g$. What is a necessary and sufficient condition for the dimension of $V$ to guarantee that $F$ is surjective?
The one thing I could figure out is that $\dim V≥\frac{\dim U\dim W}{\dim U+\dim W}$. Is this already enough?

First, note that composition is typically written in the opposite order of the arrows so $F(f,g) = g \circ f$. A sufficient and necessary condition for the map $F$ to be surjective is that $\dim V = \min(\dim U, \dim V)$.
If $\dim U \leq \dim W$ then you can construct a map $h \colon U \rightarrow W$ with $\dim \mathrm{im}(h) = \dim U$. If such a map $h$ can be factored as $h = g \circ f$ then $\dim \mathrm{im}(h) \leq \dim V$ showing that $\dim V \geq \dim U$. On the other hand, if $\dim V = \dim U$ then choose any isomorphism $f \colon U \rightarrow V$ and then $h = f \circ (f^{-1} \circ h)$.
If $\dim U > \dim W$ then you can construct a map $h \colon U \rightarrow W$ with $\dim \mathrm{im}(h) = \dim W$. If such a map can be factored as $h = g \circ f$ then $\dim \mathrm{im}(h) \leq \dim V$ showing that $\dim W \leq \dim V$. On the other hand, if $\dim W = \dim V$, choose any isomorphism $g \colon V \rightarrow W$ and then $h = (h \circ g^{-1}) \circ g$.
