Inverse Mapping is Unique

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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


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

then $g = h$.

Proof 1

By the definition of inverse mapping:

\(\ds g \circ f\) \(=\) \(\ds I_S\)
\(\ds \) \(=\) \(\ds h \circ f\)


\(\ds f \circ g\) \(=\) \(\ds I_T\)
\(\ds \) \(=\) \(\ds f \circ h\)


\(\ds h\) \(=\) \(\ds h \circ I_T\)
\(\ds \) \(=\) \(\ds h \circ \paren {f \circ g}\)
\(\ds \) \(=\) \(\ds \paren {h \circ f} \circ g\) Composition of Mappings is Associative
\(\ds \) \(=\) \(\ds I_S \circ g\)
\(\ds \) \(=\) \(\ds g\)

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


Proof 2

We need to show that:

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


\(\ds \map f {\map g t}\) \(=\) \(\ds t\) 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 $\forall s \in S: \map h {\map f s} = s$


Hence the result.