I am trying to prove that if $r$ and $s$ are relatively prime, then $\phi(rs) = \phi(r) \phi(s)$, where $\phi$ is Euler's Phi function. I will use $U(m)$ to denote the set of units modulo $m$.
Define the map $f: U(rs) \to U(r) \times U(s)$ by the rule that $x \mapsto (x^*, x^{**})$, where $x^*$ is the remainder of $x$, modulo $r$, and $x^{**}$ is the remainder of $x$, modulo $s$.
It is necessary to prove that $f$ defines a bijection of sets, which I believe has to follow in some way from the fact that $\gcd(r,s) = 1$, and since $r,s \in U_m$, $\gcd(r,m) = 1$ and $\gcd(s,m) = 1$, so $\gcd(rs,m) = 1$. Though I know I need to in some way use these facts, I wasn't successful in demonstrating the definition of one-to-one or onto.
Assuming that we have demonstrated that $f$ is a bijection of sets, we have $|U(rs)| = |U(r) \times U(s)|$. Since $U(r)$ and $U(s)$ are finite sets, we know $|U(r) \times U(s)| = |U(r)||U(s)|$. We have: $$|U(rs)| = |U(r) \times U(s)| = |U(r)| |U(s)|.$$ The set of units modulo $rs$ is the set of integers $1, 2, \ldots, rs-1$, so $|U(rs)| = \phi(rs)$. Similarly, $|U(r)| = \phi(r)$ and $|U(s)| = \phi(s)$, so we have $$\phi(rs) = \phi(r) \phi(s).$$
Any help on this would be appreciated.