Here's a proof (with a slight modification of **Lemma (A)**):

If \$$s : X \rightarrow A\$$ has a left inverse \$$r\$$, then \$$s(x) = s(y) \implies rs(x) = rs(y) \implies x = y\$$, so \$$s\$$ is injective.

If \$$r : A \rightarrow X\$$ has a right inverse \$$s\$$, then for any \$$x \in X\$$ we have \$$r(s(x)) = rs(x) = x\$$, so \$$r\$$ is surjective.

The converses are slightly tricksier.

If \$$s : X \rightarrow A\$$ is injective, and \$$a \in A\$$ is in the image of \$$s\$$, we can define \$$r(a)\$$ to be the unique \$$X\$$-element sent to \$$a\$$. If \$$a\$$ is _not_ in the image of \$$s\$$, we can define it to be whatever we like – but we must define it to be something, so \$$X\$$ cannot be empty. If \$$X\$$ is empty then we need \$$A\$$ empty too in order to get a left inverse.

in summary: an injection \$$s : X \rightarrow A\$$ always has a left inverse _unless_ \$$X\$$ is empty and \$$A\$$ is non-empty.

If \$$r : A \rightarrow X\$$ is surjective, then for each \$$x \in X\$$ we can associate a non-empty set \$$A_x\$$ of \$$A\$$-elements sent to \$$x\$$. If we can pick one element from each \$$A_x\$$, we can build a right inverse – but this requires (and for that matter implies) the Axiom of Choice.

in summary: a surjection \$$r : A \rightarrow X\$$ always has a right inverse _providing_ we accept the Axiom of Choice.