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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. ]

hazelnut_116
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2 Answers2

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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?

J. De Ro
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David
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  • I improved the formatting a bit. I hope you don't mind. – J. De Ro Jan 13 '21 at 15:04
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    Here's the continuation: since $ v_2 \in Im(S) $ then there exists $ q \in U $ such that $ v_2 = S(q) $ therefore $ T(v) = T(v_1 + v_2 ) = T(v_2) $
    ( $ T(v_1) = 0 $ since $ v_1 \in kerT $ ) ,so $ T(v) = T(S(q)) $ ,hence for arbitrary $ w \in W $ there exists $ q \in U $ such that $ w = T(s(q)) $ , hence $ T◦S $ is surjective.     – hazelnut_116 Jan 13 '21 at 15:47
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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)

leoli1
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  • Nice, can you explain how you thought about doing the decomposition? does it relates to some "rule-of-thumb" in linear algebra? – hazelnut_116 Jan 13 '21 at 15:51
  • I would say the "rule-of-thumb" in that case is to try to use everything that is given at that point, i.e. we only know that $TS$ is surjective. So it is probably good to pick a preimage $u$ of $T(v)$ under $TS$ since we cannot really do much otherwise. Now we have an element in $U$, but we want a decomposition in $V$, so it probably makes sense to look at $S(u)$. Our goal is to find something for $v$, so we look at the relation between $v$ and $S(u)$, and in this case it turns out that this works. – leoli1 Jan 13 '21 at 15:59