Surjection iff Right Inverse/Non-Uniqueness/Examples/Arbitrary Example
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Example of Surjection iff Right Inverse: Non-Uniqueness
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.
Proof
Let $g_0: T \to S$ and $g_1: T \to S$ be the mappings defined as:
| \(\ds \map {g_0} a\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \map {g_1} a\) | \(=\) | \(\ds 1\) |
We have that:
- $\Dom {g_0} = \Cdm f = \Dom {g_1}$
and that $f$ is a surjection.
Hence we can construct:
| \(\ds \map {f \circ g_0} a\) | \(=\) | \(\ds \map f 0\) | Definition of $g_0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a\) | Definition of $f$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds f \circ g_0\) | \(=\) | \(\ds I_T\) | the identity mapping on $T$ |
and:
| \(\ds \map {f \circ g_1} a\) | \(=\) | \(\ds \map f 1\) | Definition of $g_1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a\) | Definition of $f$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds f \circ g_1\) | \(=\) | \(\ds I_T\) | the identity mapping on $T$ |
Hence both $g_0$ and $g_1$ are distinct right inverses of $f$.
$\blacksquare$
Sources
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Exercise $\text{G}$