Inverse Mapping is Unique/Proof 2
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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
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.
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions