Surjection iff Right Inverse/Non-Uniqueness

Theorem

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

Let $f: S \to T$ be a surjection.

A right inverse of $f$ is in general not unique.

Uniqueness occurs if and only if $f$ is a bijection.

Proof

If $f$ is not an injection then:

$\exists y \in T: \exists x_1, x_2 \in S: \map f {x_1} = y = \map f {x_2}$

Hence we have more than one choice in $\map {f^{-1} } {\set y}$ for how to map $\map g y$.

That is, $\map g y$ is not unique.

This does not happen if and only if $f$ is an injection.

Hence the result.

$\blacksquare$

Examples

Arbitrary Example

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

Let $T = \set a$.

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

$\forall x \in S: \map f x = a$

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

Sources

• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $6$