Let $f: U\longrightarrow V$ and $g: V \longrightarrow W$ be linear maps.
If $g\circ f$ is surjective, then ker(g)+im(f)=V. Is there an easy way to see/proof this? I managed to show it by first showing
dim(ker g $\cap$ im f) = dim ker(g $\circ$ f) − dim ker f
and then using the Rank-Nullity Theorem and dimension arguments, but that seems to be to much work.
Best Answer
The dimension argument is ok as long as you assume that $V$ is finite dimensional. Otherwise it is wrong (you cannot subtract cardinal numbers, and there's no way around it).
But the statement is true for any $V$.
Proof. Let $v\in V$. Since $g\circ f$ is surjective, then
$$(g\circ f)(u)=g(v)\text{ for some }u\in U$$
therefore $g(f(u)-v)=0$. In particular $f(u)-v\in\ker(g)$. Meaning
$$v=f(u)+v_g\text{ for some }v_g\in\ker(g)$$
By the arbitrary choice of $v$ we conclude that $V=\text{im}(f)+\ker(g)$. $\Box$