ct.category theory - Splitting lemma under assumption of the axiom of choice

This is right: a map $f:A rightarrow B$ is surjectiv $Longleftrightarrow$ $f$ has a right-inverse. The proof needs the axiom of choice, as you pointed out correctly. But this is just a map of sets.

EDIT: In the following I'm talking about groups (vector spaces, vector bundles, presheaves, sheaves,... should also do the job)

Every short exact seqeunce can be seen as a sequence
$$0 overset{inc_0}rightarrow A overset{inc}{rightarrow} B overset{pi}{rightarrow} B/A rightarrow 0$$
where $inc$ denotes the inclusion and $pi$ the projection.

But (as for example Charles Siegel pointed out) surjectivity gives you just a rightinverse $u:C rightarrow B$ as map of sets. So if you have further structures (let's say a group structure, vector space structure, etc.), this doesn't mean, that the map $u$ is an inverse with respect to these structures