The 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?