Codomain of Bijection is Domain of Inverse
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Theorem
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a bijection.
Let $f^{-1}: T \to S$ be the inverse of $f$.
Then the domain of $f^{-1}$ equals the codomain of $f$.
Proof
Follows directly from the definition of domain and codomain:
- $\Dom f = S$ and $\Cdm f = T$
- $\Dom {f^{-1} } = T$ and $\Cdm {f^{-1} } = S$
That is:
- $\Dom {f^{-1} } = T = \Cdm f$
$\blacksquare$
Also see
Sources
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $3$
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries