Conditions for Uniqueness of Left Inverse Mapping/Examples/Arbitrary Example
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Example of Conditions for Uniqueness of Left Inverse Mapping
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.
Proof
Let $g_0: T \to S$ be the mapping defined as:
| \(\ds \map {g_0} a\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \map {g_0} b\) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\ds \map {g_0} c\) | \(=\) | \(\ds 0\) |
Let $g_1: T \to S$ be the mapping defined as:
| \(\ds \map {g_0} a\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \map {g_0} b\) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\ds \map {g_0} c\) | \(=\) | \(\ds 1\) |
We have that:
- $\Cdm {g_0} = \Dom f = \Cdm {g_1}$
and that $f$ is an injection.
Hence we can construct:
| \(\ds \map {g_0 \circ f} 0\) | \(=\) | \(\ds \map {g_0} a\) | Definition of $g_0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) | Definition of $f$ | |||||||||||
| \(\ds \map {g_0 \circ f} 1\) | \(=\) | \(\ds \map {g_0} b\) | Definition of $g_0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) | Definition of $f$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds g_0 \circ f\) | \(=\) | \(\ds I_S\) | the identity mapping on $S$ |
and:
| \(\ds \map {g_1 \circ f} 0\) | \(=\) | \(\ds \map {g_1} a\) | Definition of $g_1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) | Definition of $f$ | |||||||||||
| \(\ds \map {g_1 \circ f} 1\) | \(=\) | \(\ds \map {g_1} b\) | Definition of $g_1$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) | Definition of $f$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds g_1 \circ f\) | \(=\) | \(\ds I_S\) | the identity mapping on $S$ |
Hence both $g_0$ and $g_1$ are distinct left 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 $\text{G}$