Proving exponentiation in $\mathbb{Z}^{*}_{pq}$ is one to one

Question

We saw in our crypto class that in the group $\mathbb{Z}^{*}_{pq}$ where $p$ and $q$ are primes, that if for some $a$, $gcd(a, \phi(pq) = (p-1)(q-1))$ = 1 (where $\phi$ is Euler's totient function) then the function $f(x) = x^a$ is one to one, and he lecturer proved this by calculating the inverse of $a$ (the RSA trapdoor). What I have trouble understanding is why this proves the function is one to one. We computed an inverse function, that is showed that there is an inverse, but how does this show that there is no other inverse for an element?

Answer 1

If $f$ is an function and $g$ and $h$ are inverse functions for $f,$ then $g=h$ since $$g=g(fh)=(gf)h=h.$$

Answer 2

In general: if $G$ is a finite group and $k$ a positive integer. Then the map $f: G \rightarrow G$ defined by $f(g)=g^k$ is a bijection if and only if $gcd(k,|G|)=1$. (The proof of the "if" uses Bézout's Lemma, the "only if" leverages Cauchy's Theorem)