Domain of Bijection is Codomain of Inverse

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 codomain of $f^{-1}$ equals the domain 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:

$\Cdm {f^{-1} } = S = \Dom f$

$\blacksquare$

Also see

• Codomain of Bijection is Domain of Inverse

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