Inverse of Bijection is Bijection

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

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

It follows that $$f^{-1}$$ is itself a bijection.

Proof
First we establish the definition of $f^{-1}$ in this context.

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

Then $$f^{-1} \subseteq T \times S$$ is a relation defined as:
 * $$f^{-1} = \left\{{\left({t, s}\right): t = f \left({s}\right)}\right\}$$

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

As $$f$$ is a bijection, it is by definition an injection.

So, by Inverse of Injection is Functional Relation‎, $$f^{-1}$$ is functional.

Also, as $$f$$ is a bijection, it is by definition a surjection.

From Surjection iff Image equals Codomain, $$\operatorname{Im} \left({f}\right) = T$$.

Thus from the definition of inverse relation, the domain of $$\operatorname {Dom} \left({f^{-1}}\right) = T$$.

Thus we have established that:


 * $$\forall y \in T: \exists x \in S: f^{-1} \left({y}\right) = x$$ as $$\operatorname {Dom} \left({f^{-1}}\right) = T$$;
 * $$\forall y_1, y_2 \in T: f^{-1} \left({y_1}\right) \ne f^{-1} \left({y_2}\right) \implies y_1 = y_2$$ as $$f^{-1}$$ is functional.

Hence by definition $$f^{-1}$$ is a mapping.

Necessary Condition
Now suppose $$f^{-1}$$ is a mapping.

Then by definition:
 * $$\forall y_1, y_2 \in T: f^{-1} \left({y_1}\right) \ne f^{-1} \left({y_2}\right) \implies y_1 = y_2$$

which implies that:
 * $$\forall y_1, y_2 \in T: f \left({x_1}\right) = f \left({x_2}\right) \implies x_1 = x_2$$

and so $$f$$ is an injection.

From Preimage of Mapping equals Domain we have that:
 * $$\operatorname{Im}^{-1} \left({f^{-1}}\right) = \operatorname{Dom} \left({f^{-1}}\right)$$.

from the definition of inverse relation, the domain of $$\operatorname {Dom} \left({f^{-1}}\right) = T$$.

From Surjection iff Image equals Codomain, $$\operatorname{Im} \left({f}\right) = T$$.

So $$f$$ is a surjection.

Being both an injection and a surjection, it follows by definition that $$f$$ is a bijection.

Now, from Inverse of Inverse Relation, we have $$\left({f^{-1}}\right)^{-1} = f$$.

So we have that the inverse of $f^{-1}$ is $$f$$.

But then $$f$$, being a bijection, is by definition a mapping.

So if the inverse of $$f^{-1}$$ is a mapping, then $$f^{-1}$$ must be a bijection.

Also see
It also follows, from Bijection Composite with Inverse, that $$f^{-1}$$ is the two-sided inverse of $$f$$, i.e.: where $$I_S$$ and $$I_T$$ are the identity mappings on $$S$$ and $$T$$.
 * $$f \circ f^{-1} = I_S$$
 * $$f^{-1} \circ f = I_T$$