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:
| \(\ds g \circ f\) | \(=\) | \(\ds I_S\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds h \circ f\) |
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.
$\blacksquare$
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.