Prove: $ T ◦ S $ is surjective iff $ $ T is surjective and $ ImS + KerT = V $ Let $  S: U \rightarrow V $ and $ T: V \rightarrow W $ be linear transformations over field $ F $.
Prove that $ T ◦ S $ is surjective iff $  $ T is surjective and $ ImS + KerT = V $
My attempt:
$ \rightarrow $ : Suppose that $ T ◦ S $ is surjective. Let $ w \in W $ be arbitrary. Then there exists $ u \in U $ such that $ w=T(S(u)) $. Since $ S(u) \in V $, denote $ v = S(u) $ and so $ w = T(v) $ and so we immediately see that $ T $ is surjective.
Next, by dimension theorem, we know that,
$ dim U=dimKer(T ◦ S) + dimIm(T ◦ S) $ , $ dim U = dimKer(S) + dimIm(S)  $ ,  $ dim V = dimKer(T) + dimIm(T)  $,
Since $ T ◦ S $ is surjective and we showed  $ T $ is surjective, then: $ dimIm(T ◦ S) =  dimW $  , $ dimImT = dimW $
[ Basically after this I'm pretty much stuck. I've reached the equation $ dimV = dimKer(T) + dimImS +  dimKerS  -  dimKer(T ◦ S)  $ , so In order to proceed I wanted to show that $ dimKerS  =  dimKer(T ◦ S) $ I tried to do this since It is immediate that $ ker(S) \subseteq Ker(T ◦ S) $ but it is left to show that $ Ker(T ◦ S)  \subseteq ker(S) $ and this part stumbled me   ]
$ \leftarrow $ : Suppose T is surjective and $ ImS + KerT = V $. So  $ dimImT = dimW $ [ I got a few more equations on dimensionality but they didn't help, so I got stuck here. ]
 A: For the "$\Leftarrow$" part: Let $T$ be surjective and let's assume $V = \operatorname{Ker}(T) + \operatorname{Im}(S)$. Let $w \in W$. We need to find $u \in U$ such that $T(S(u))=w$.
Since T is surjective, $\exists v \in V$ such that $w=T(v)$. Since $V = \operatorname{Ker}(T) + \operatorname{Im}(S)$, there are some $v_1 \in \operatorname{Ker}(T), v_2 \in \operatorname{Im}(S)$ such that $v=v_1+v_2$. Can you continue from here?
A: I think it might be the easiest to verify the equality by direct computation via elements instead of dimension (the latter approach is actually a bit problematic if the vector spaces are no longer finite-dimensional)
For $\Rightarrow$: Let $v\in V$ be arbitrary. We wish to express it as a sum of elements in $\ker T$ and $\operatorname{im} S$. For this notice that since $TS$ is surjective there is some $u\in U$ with $T(v)=T(S(u))$, so we have $v-S(u)\in\ker T$. Now $$v=(v-S(u))+S(u)$$ is the desired decomposition.
The reverse implication is similar. Take $w\in W$. By assumption there is some $v\in V$ with $T(v)=w$. Furthermore we have $v=x+y$ with $x\in\ker T,y\in \operatorname{im} S$, i.e. $x=S(u)$ for some $u\in U$. Check that $T(S(u))=w$.
(I left out some details in the answer, which you might need to fill in)
