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Let $0\rightarrow E \stackrel{i}{\longrightarrow} F \stackrel{p} {\longrightarrow} G \rightarrow 0$ be an exact sequence of vector spaces. I want to prove that this exact sequence splits, i.e. that there exists $s:G\to S$ such that $p\circ s=\text{id}_G$.

We know that $i(E)\subset F$ is a linear subspace. Therefore there exists, according to some theorem in linear algebra, a supplement $S$ such that $F=S\oplus i(E)$. My professor claims that now $p|_S:S\to G$ is a linear isomorphism, and that we can set $s=(p|_S)^{-1}$. I am trying to understand why this is true.

The exactness of the sequence gives $i(E)=\ker p$, so by the direct sum composition $S\cap i(E)=\{0\}$, we have $\ker p|_S=\{0\}$, which in the finite case means that $p|_S$ is an isomorphism.

I can't see how to prove it in the general case for an infinite dimensional vector space. Can someone provide any help?

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    $\begingroup$ Do you know that $G$ has a basis? $\endgroup$ – Angina Seng Sep 20 '19 at 9:15
  • $\begingroup$ From the exactness at $F$, you have only used $\ker p\subseteq i(E)$. Presumably, the sext step would be to use $i(E)\subseteq \ker p$. $\endgroup$ – Arthur Sep 20 '19 at 9:21
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As I said in my comment above:

From the exactness at $F$, you have only used $\ker p\subseteq i(E)$. Presumably, the [next] step would be to use $i(E)\subseteq \ker p$.

So that's what I'll do. We want to show that $p|_S$ is surjective, so the most natural thing would be to take a $g\in G$ and then go from there.

Since $p$ is surjective, there is an $f\in F$ such that $p(f) = g$.

Let $f = f_i+f_S$ be the decomposition of $f$ in $i(E)\oplus S$, with $f_i \in i(E)$ and $f_S\in S$. Then by $i(E)\subseteq \ker p$, we get that $$g = p(f) = p(f_i+f_S) = p(f_i) + p(f_S) = p(f_S)$$ This shows that $p|_S$ is surjective.

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  • $\begingroup$ sext step. I guess you mean ‘next step’. $\endgroup$ – Bernard Sep 20 '19 at 10:02
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    $\begingroup$ @Berard Well, that's an unfortunate misspelling. However, if I fix it, the quote will be inaccurate... The dilemmas one faces. $\endgroup$ – Arthur Sep 20 '19 at 10:18
  • $\begingroup$ Thanks, I feel so stupid for overseeing this! :) $\endgroup$ – rae306 Sep 20 '19 at 16:34
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Since $p(F)=G$, $\left.p\right\rvert_S$ is surjective as well. In fact, let $y\in G$ and $x\in F$ be such that $p(x)=y$. Then, by $i(E)\oplus S=F$, there are some $x'\in S$ and $x''\in i(E)$ such that $x'+x''=x$. Therefore $y=p(x)=p(x'+x'')=p(x')+p(x'')=p(x')$.

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