# Short exact sequence of vector spaces splits always

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?

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

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.

• sext step. I guess you mean ‘next step’. – Bernard Sep 20 '19 at 10:02
• @Berard Well, that's an unfortunate misspelling. However, if I fix it, the quote will be inaccurate... The dilemmas one faces. – Arthur Sep 20 '19 at 10:18
• Thanks, I feel so stupid for overseeing this! :) – rae306 Sep 20 '19 at 16:34

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 $$y=p(x)=p(x'+x'')=p(x')+p(x'')=p(x')$$.