Conditions for Uniqueness of Left Inverse Mapping

Theorem
Let $S$ and $T$ be sets such that $S \ne \varnothing$.

Let $f: S \to T$ be an injection.

Then a left inverse mapping of $f$ is in general not unique.

Uniqueness occurs under either of two circumstances:


 * $(1): \quad: S$ is a singleton
 * $(2): \quad: f$ is a bijection.

Proof
If $f$ is a bijection, then by definition $f$ is also a surjection.

Then:
 * $T \setminus \operatorname{Im} \left({f}\right) = \varnothing$
 * and we have that $g = f^{-1}$.

As $f^{-1}$ is uniquely defined $g$ is itself unique.

If $S$ is a singleton then there can only be one mapping $g: T \to S$:
 * $\forall t \in T: g \left({t}\right) = s$

If $f$ is not a bijection, then as it is an injection it can not be a surjection.

Then:
 * $T \setminus \operatorname{Im} \left({f}\right) \ne \varnothing$

Let $t \in T \setminus \operatorname{Im} \left({f}\right)$.

We can now choose any $x_0 \in S$ such that $g \left({t}\right) = x_0$.

If $S$ is not a singleton, such an $x_0$ is not unique.

Hence the result.