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.

  • $\begingroup$ Just out of curiosity: Why are you not happy with this solution? $\endgroup$ Sep 30 '20 at 8:03
  • $\begingroup$ Because for showing the first result one uses 2 times the Rank-Nullity Theorem and after that again one time. So it seems to be just far from the shortest/direct way to prove it. $\endgroup$
    – RobbiTobbi
    Sep 30 '20 at 8:07

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 $V=\text{im}(f)+\ker(g)$. $\Box$

  • $\begingroup$ Thank you, V finite dimensional was fine for my application, but this is much more straight forward. $\endgroup$
    – RobbiTobbi
    Sep 30 '20 at 8:31
  • 1
    $\begingroup$ Just a side note: Instead of introducing another variable, we could just write $v=f(u)+(f(u)-v)$ and note that $f(u)\in\operatorname{im}(f)$ and $f(u)-v\in\operatorname{ker}(g)$. This gets rid of the unspecified "some $v_g$", since we pinned down what this $v_g$ is anyway. $\endgroup$
    – Christoph
    Sep 30 '20 at 8:35

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