Conditions for Uniqueness of Left Inverse Mapping

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Theorem

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

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:

$T \setminus \Img f = \O$
and we have that $g = f^{-1}$.

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

$\Box$


If $S$ is a singleton then there can only be one mapping $g: T \to S$:

$\forall t \in T: \map g t = s$

$\Box$


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

Then:

$T \setminus \Img f \ne \O$

Let $t \in T \setminus \Img f$.

We can now choose any $x_0 \in S$ such that $\map g t = x_0$.

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

Hence the result.

$\blacksquare$


Warning

In some expositions of set theory, the case where $S$ is a singleton is not mentioned.

This should be considered a mistake.

Such sources include:

1967: George McCarty: Topology: An Introduction with Application to Topological Groups
1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability


Examples

Arbitrary Example

Let $S = \set {0, 1}$.

Let $T = \set {a, b, c}$.

Let $f: S \to T$ be defined as:

$\forall x \in S: \map f x = \begin {cases} a & : x = 0 \\ b & : x = 1 \end {cases}$

Then $f$ has $2$ distinct left inverses.


Also see


Sources

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