# 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:

 $\ds g \circ f$ $=$ $\ds I_S$ $\ds$ $=$ $\ds h \circ f$

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.