Using the axiom of choice, or more directly, the statement that every linearly independent set of vectors in a vector space may be extended to a basis, it is easy to prove that every short exact sequence of vector spaces splits.
Is the converse also true? Over ZF, does the splitting of each short exact sequence of vector space suffice to prove that every vector space has a basis?
One thing I tried was the following: suppose that $V$ is a vector space over $k$, and let $k[V]$ be the vector space freely generated by the elements of $V$, and let $k[V] \to V$ be the "projection". Then you get a section $V \to k[V]$... but it doesn't seem there's any way to guarantee the image of the section contains any standard basis vectors.
