# Inverse Mapping is Unique/Proof 1

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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

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.

$\blacksquare$

## Sources

- 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $2$: Maps and relations on sets: Proposition $2.14$

- For a video presentation of the contents of this page, visit the Khan Academy.