The [splitting lemma](http://en.wikipedia.org/wiki/Splitting_lemma) says:



> 
> Given a short exact sequence with maps $q$ and $r$:
> 
> 
> $0 \rightarrow A \overset{q}{\rightarrow} B \overset{r}{\rightarrow} C \rightarrow 0$
> 
> 
> then the following are equivalent:
> 
> 
> 1. ...
> 2. there exists a map $u : C \rightarrow B$ such that $r \circ u = \mathrm{id}\_C$
> 3. $B \cong A \oplus C$
> 
> 
> 


Now I figured, since $r(B) = \ker 0$, $r$ is surjective. Hence, for every $c \in C$ we have some $b \in B$ such that $r(b) = c$. Simply set $u(c) = b$ for one of these $b$s and you have your map $u$.


However, it turns out that this construction assumes the axiom of choice; it chooses one element from each of an infinite number of sets. So my question is: assuming the axiom of choice, is condition (2) always satisfied? Because this would imply that (3) holds for any such short exact sequence. Or am I making some mistake here?



Splitting lemma under assumption of the  axiom of choice
I assume you are working in some fixed abelian category $\mathcal{A}$.


It is not true in general that every short exact sequence in $\mathcal{A}$ will split. The problem is that although you can pick a preimage for every 'element' $c\in C$ there is no guarantee that you can assemble this into a morphism in $\mathcal{A}$. It is true in the category of sets that every surjection splits if one assumes the axiom of choice but this is only a set map.


For instance in the category of abelian groups
$$0 \to \mathbb{Z}/2\mathbb{Z} \stackrel{2}{\to} \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$$
is exact but $\mathbb{Z}/4\mathbb{Z}$ is indecomposable.


Also if you are not in an abelian category it is not necessarily true that exact sequences display this symmetry. See for example [this](https://mathoverflow.net/questions/3757/when-does-splits-imply-cosplits) question where the notion is considered in the category of groups.


