Inverse of Bijection is Bijection

Theorem
Let $$f: S \to T$$ be a mapping.

Then $$f$$ is a bijection iff $$f^{-1}$$ is a bijection from $$T$$ to $$S$$.

Proof

 * Let $$f: S \to T$$ be a bijection.

Then from Two-Sided Inverse, $$f^{-1}$$ exists and is a bijection.


 * Now let $$f: S \to T$$ have a Two-Sided Inverse $$f^{-1}$$ which is a bijection.

First we show $$f$$ is injective.

Let $$f \left({x_1}\right) = f \left({x_2}\right)$$:

$$ $$ $$ $$ $$ $$ $$

Thus $$f$$ is injective.

Now we show $$f$$ is surjective.

Let $$y \in T$$. Then:

$$ $$ $$ $$

Thus for any given $$y$$ in its range, there exists an $$x$$ in its domain, and so $$f$$ is surjective.

So $$f$$ is both injective and surjective, therefore bijective, and the proof is complete.