Inverse of Bijection is Bijection

Theorem
Let $f: S \to T$ be a bijection in the sense that:
 * $(1): \quad f$ is an injection
 * $(2): \quad f$ is a surjection.

Then the inverse $f^{-1}$ of $f$ is itself a bijection by the same definition.

Proof
Let $f$ be both an injection and a surjection.

From Bijection iff Inverse is Mapping it follows that its inverse $f^{-1}$ is a mapping.

From Inverse of Inverse Relation:
 * $\left({f^{-1}}\right)^{-1} = f$

Thus the inverse of $f^{-1}$ is $f$.

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

So from Bijection iff Inverse is Mapping it follows that $f^{-1}$ is a bijection.

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

Some sources use this property as the definition of an inverse mapping.