Surjection from Natural Numbers iff Right Inverse

Theorem
Let $S$ be a set.

Let $f: \N \to S$ be a mapping, where $\N$ denotes the set of natural numbers.

Then $f$ is a surjection $f$ admits a right inverse.

Necessary Condition
Suppose that $g: S \to \N$ is a right inverse of $f$.

That is, let $f \circ g = I_S$, the identity mapping on $S$.

We have that $I_S$ is a surjection.

By Surjection if Composite is Surjection, it follows that $f$ is a surjection.

Sufficient Condition
Now, suppose that $f: \N \to S$ is a surjection.

By the definition of a surjection, it follows that:
 * $\forall x \in S: f^{-1} \left({x}\right)$ is non-empty

where $f^{-1} \left({x}\right)$ denotes the preimage of $x$ under $f$.

From the Well-Ordering Principle, $\N$ is well-ordered.

Hence, we can define the mapping $g: S \to \N$ as:
 * $\forall x \in S: g \left({x}\right) = \min f^{-1} \left({x}\right)$

By the definition of a smallest element, it follows that:
 * $\forall x \in S: g \left({x}\right) \in f^{-1} \left({x}\right)$

That is:
 * $\forall x \in S: f \left({g \left({x}\right)}\right) = \left({f \circ g}\right) \left({x}\right) = x$

In other words, $g$ is a right inverse of $f$.

Also see

 * Surjection iff Right Inverse, which depends on the axiom of choice.