Proof

It is clear that the third condition implies the first two: take the section/retract to be given by the canonical injection/projection maps that come with a direct sum.

Conversely, suppose we have a retract $r \colon B \to A$ of $i \colon A \to B$. Write $P \colon B \stackrel{r}{\to} A \stackrel{i}{\to} B$ for the coresponding idempotent.

Then every element $b \in B$ can be decomposed as $b = (b - P(b)) + P(b)$ hence with $b - P(b) \in ker(r)$ and $P(b) \in im(i)$. Moreover this decomposition is unique since if $b = i(a)$ while at the same time $r(b) = 0$ then $0 = r(i(a)) = a$. This shows that $B \simeq im(i) \oplus ker(r)$ is a direct sum and that $i \colon A \to B$ is the canonical inclusion of $im(i)$. By exactness it then follows that $ker(r) \simeq ker(p)$ and hence that $B \simeq A \oplus C$ with the canonical inclusion and projection.

The implication that the second condition also implies the third is formally dual to this argument.

Proof

Consider the first case. The other is formally dual.

By the properties of a short exact sequence the morphism $A \to B$ here is a monomorphism. By definition of injective object, if $A$ is injective then it has the right lifting property against monomorphisms and so there is a morphism $q : B \to A$ that makes the following diagram commute:

Hence $q$ is a retract as in def. 1.