Inverse Mapping is Unique

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:

$\ds g \circ f$

$=$

$\ds I_S$

$\ds $

$=$

$\ds h \circ f$

and:

$\ds f \circ g$

$=$

$\ds I_T$

$\ds $

$=$

$\ds f \circ h$

So:

$\ds h$

$=$

$\ds h \circ I_T$

$\ds $

$=$

$\ds h \circ \paren {f \circ g}$

$\ds $

$=$

$\ds \paren {h \circ f} \circ g$

Composition of Mappings is Associative

$\ds $

$=$

$\ds I_S \circ g$

$\ds $

$=$

$\ds g$

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:

$\ds \map f {\map g t}$

$=$

$\ds t$

Definition of Inverse Mapping

$\ds \leadsto \ \ $

$\ds \map h t$

$=$

$\ds \map h {\map f {\map g t} }$

$\ds \leadsto \ \ $

$\ds \map h t$

$=$

$\ds \map g t$

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 $1.3: \ 12 \ \text{(a)}$
1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Exercise $\text{G}$
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