# Inverse Mapping is Unique/Proof 1

## 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

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.

$\blacksquare$