# Inverse Mapping is Unique

Jump to navigation
Jump to search

## Theorem

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

If $f$ has an inverse mapping, then that inverse mapping is unique.

That is, if:

- $f$ and $g$ are inverse mappings of each other

and

- $f$ and $h$ are inverse mappings of each other

then $g = h$.

## Proof 1

By the definition of inverse mapping:

\(\displaystyle g \circ f\) | \(=\) | \(\displaystyle I_S\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle h \circ f\) |

and:

\(\displaystyle f \circ g\) | \(=\) | \(\displaystyle I_T\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle f \circ h\) |

So:

\(\displaystyle h\) | \(=\) | \(\displaystyle h \circ I_T\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle h \circ \paren {f \circ g}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {h \circ f} \circ g\) | Composition of Mappings is Associative | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle I_S \circ g\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle g\) |

So $g = h$ and the inverse is unique.

$\blacksquare$

## Proof 2

We need to show that:

- $\forall t \in T: \map g t = \map h t$

So:

\(\displaystyle \map f {\map g t}\) | \(=\) | \(\displaystyle t\) | Definition of Inverse Mapping | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map h t\) | \(=\) | \(\displaystyle \map h {\map f {\map g t} }\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map h t\) | \(=\) | \(\displaystyle \map g t\) | as $\forall s \in S: \map h {\map f s} = s$ |

$\blacksquare$

Hence the result.

## Sources

- 1964: W.E. Deskins:
*Abstract Algebra*... (previous) ... (next): Exercise $1.3: \ 12 \ \text{(a)}$ - 1972: A.G. Howson:
*A Handbook of Terms used in Algebra and Analysis*... (previous) ... (next): $\S 2$: Sets and functions: Inverse images and inverse functions - 1993: Keith Devlin:
*The Joy of Sets: Fundamentals of Contemporary Set Theory*(2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.6$: Functions - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Proposition $3.2$: Remark