# Let $f\circ g$ be a surjective composition of linear maps. Show that ker(g)+im(f)=V

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.

• Just out of curiosity: Why are you not happy with this solution? Sep 30 '20 at 8:03
• 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. 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 $$V=\text{im}(f)+\ker(g)$$. $$\Box$$

• Thank you, V finite dimensional was fine for my application, but this is much more straight forward. Sep 30 '20 at 8:31
• 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. Sep 30 '20 at 8:35