Inversion Mapping is Permutation/Proof 2

Theorem
Let $\left({G, \circ}\right)$ be a group.

Let $\iota: G \to G$ be the inversion mapping on $G$.

Then $\iota$ is a permutation on $G$.

Proof of Surjection
Let $a \in G$.

By definition of $\iota$:


 * $\iota(a^{-1}) = \left({a^{-1}}\right)^{-1}$

By Inverse of Inverse:


 * $\left({a^{-1}}\right)^{-1} = a$

Hence $a$ has a preimage.

Since $a$ was arbitrary, $\iota$ is a surjection.

Proof of Injection
Suppose for some $a, b \in G$ that:


 * $\iota \left({a}\right) = \iota \left({b}\right)$

Then by the definition of $\iota$:


 * $a^{-1} = b^{-1}$

It follows from Inverse in Group is Unique that:


 * $a = b$

Hence $\iota$ is an injection.

Hence by definition $\iota$ is a bijection.

A bijection from a set to itself is by definition a permutation.