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?
Best Answer
As I said in my comment above:
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.