Exact short sequences of vector spaces If possible, how could one prove that every short exact sequence $0 \to A \xrightarrow f B \xrightarrow g C \to 0$ of vector spaces (here $A$, $B$ and $C$) splits without using any basis of $A$, $B$ or $C$. I was not able to exhibit a morphism $h \colon B \to A$ such that $h \circ f=Id_A$ without considering a basis.
 A: You can prove this using Zorn's lemma on the pairs (D,h) where D is a subspace of B and h is a partial section.
And yes, you do need Zorn's lemma: without it, there may exist vector spaces none of whose nontrivial subspaces has a complement (Herrlich, Axiom of Choice, LNM 1876, Disaster 4.43).
A: One can show the short exact sequence by proving that a field is a semisimple ring—alhough one could argue that somewhere hidden in that proof a basis is considered...
I cannot imagine a negative answer to your question in any other form than a model of ZF without the axiom of choice, so that there are vector spaces without bases, and in which maybe there are short exact sequences of vector spaces which do not split.
A: So, for example, $0 \to \mathbb{Q} \to \mathbb{R} \to \mathbb{R}/\mathbb{Q} \to 0$ ... the inclusion of the rationals into the reals, these are all vector spaces over the rationals.  Can you write down your split?  Not constructively!
A: Without using any basis of A, B or C specifically, but rather using the existence of bases for all vector spaces, we may observe that all vector spaces are free as modules; and all free modules are projective.
Finally, the projectivity property gives us a splitting by lifting the identity map on C along the surjection from B by projectivity of C. This gives us a splitting, thus splitting the entire sequence.
ETA With the arguments in the comments to Ben Websters answer, it is pointed out that without AC, and without at least the existence of bases, things can fail badly. Obviously, if we disallow AC, this answer fails as badly.
A: Apply $\mathrm{Hom}(C,-)$ to your short exact sequence.  It remains exact, so the identity map from $C$ to $C$ has at least one preimage.  In fact, the splittings are exactly its preimages.
I would guess the problem with the axiom of choice is that you need to actually choose one of them.  Or it hidden somewhere else?
A: It should feel natural that splittings and bases go together; in the opposite direction, suppose you only consider short exact sequences $0\rightarrow 0 \rightarrow V\rightarrow k^\beta\rightarrow 0$ where by $k^\beta$ I mean of course the free vector space on the set $\beta$ taken with its standard basis $\{x\in\beta\}$.  In this case a splitting is (essentially the same thing) as a basis for $V$.  Why would you want to avoid that sort of thing?
