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.

Condition (2) is always satisfied since 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.