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

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.

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

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.

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

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.