= Solution
A <left inverse> of $f$ is a function $\ell:B\to A$ satisfying $\ell\circ f=\operatorname{id}_A$. If such $\ell$ exists and $f(a)=f(a')$, applying $\ell$ gives $a=a'$, so $f$ is an <injection>. Conversely, suppose $f$ is an <injection> and fix $a_0\in A$, possible because $A$ is nonempty. Define
$$
\ell(b)=\begin{cases}a,&b=f(a)\text{ for some }a\in A,\\a_0,&b\notin f(A).\end{cases}
$$
The first value is unique by injectivity, and $\ell(f(a))=a$ for every $a$. Thus
$$
\boxed{f\text{ is injective}\iff f\text{ has a left inverse}.}
$$
No choice from a family of nonempty sets is needed here: each preimage in the first case is unique, and the second case uses one fixed element.
A <right inverse> is a function $r:B\to A$ with $f\circ r=\operatorname{id}_B$. Its existence makes every $b$ the image of $r(b)$, proving that $f$ is a <surjection>. Conversely, if $f$ is a <surjection>, every <fiber of a function> $f^{-1}(\{b\})$ is nonempty. Using the usual <axiom of choice> for arbitrary sets, choose $r(b)\in f^{-1}(\{b\})$ for each $b$. Then $f(r(b))=b$, giving
$$
\boxed{f\text{ is surjective}\iff f\text{ has a right inverse}\quad\text{assuming choice}.}
$$
The converse in this second equivalence is the <right-inverse characterization of the axiom of choice>; it is not an unrestricted choice-free theorem about arbitrary sets.
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