Let $f$ be an arbitrary function. We define a right-inverse of $f$ as a function $g$ satisfying $f(g(y))=y$ for all $y$ in the range of $f$. Using the axiom of choice, it is can be proven that every function has a right-inverse:
Let $S=\{f^{-1}(\{y\}):y\in\operatorname{ran}(f)\}$. Let $\phi:S\to\bigcup S$ be a choice function such that $\phi(A)\in A$ for each $A\in S$. Let $g(y)=\phi\bigl(f^{-1}(\{y\})\bigr)$ for all $y$ in the range of $f$. This function is a right-inverse of $f$.
Of course, the axiom of choice is not always necessary to define such a function $g$. For example, if $f(x)=x^2$, then the function $g(x)=\sqrt{x}$ is an explicit example of a right-inverse of $f$. Is there an easy example of a function where we do need to invoke the axiom of choice to prove that it has a right-inverse?
Consider map from infinite sequences of real numbers to countable sets of reals given by mapping a sequence to the set of reals appearing in the sequence.
Clearly, this is a surjection, and an inverse function would allow you to uniformly choose an enumeration for each countable set of reals. However, without choice it is consistent that there is no such inverse.
Indeed, if there is such inverse, then there is a set which is not Lebesgue measurable, and there is a set without the Baire property.
Similarly, one can replace this by mapping the sequence to an ordinal $\alpha$ if the sequence's range is order isomorphic to $\alpha$ (under the standard order of the reals), or to $0$ if it isn't well ordered at all. An inverse function here would imply there is an injection from $\omega_1$ into the reals, which consistently does not exist without assuming choice.