Inverse Mapping is Unique

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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