CRT as a ring isomorphism and the multiplicativity of $\varphi$

First, $\psi$ is well-defined: if $x\equiv x'\pmod{m_1 m_2}$ then $m_1 m_2\mid(x-x')$, so both $m_1$ and $m_2$ divide $x-x'$, meaning $x\equiv x'\pmod{m_1}$ and $x\equiv x'\pmod{m_2}$.

Second, $\psi$ respects addition and multiplication in each coordinate because reduction modulo $m_i$ does (proved in the first topic on congruences), so $\psi$ is a ring homomorphism.

Third, Theorem 1 above says precisely that for every pair $(a_1,a_2)$ there is an $x$ mapped to it (so $\psi$ is surjective) and that $x$ is unique modulo $m_1 m_2$ (so $\psi$ is injective). Hence $\psi$ is a bijection and therefore a ring isomorphism.

Finally, in a product ring $R_1\times R_2$, an element $(u_1,u_2)$ has a multiplicative inverse $(v_1,v_2)$ iff $u_1 v_1=1$ in $R_1$ and $u_2 v_2=1$ in $R_2$ — that is, iff both coordinates are invertible. Since a ring isomorphism preserves invertibility, the invertible elements of $\mathbb{Z}/(m_1 m_2)\mathbb{Z}$ (of which there are $\varphi(m_1 m_2)$) correspond bijectively to pairs of invertible elements in $(\mathbb{Z}/m_1\mathbb{Z})\times(\mathbb{Z}/m_2\mathbb{Z})$ (of which there are $\varphi(m_1)\,\varphi(m_2)$). Counting both sides gives $\varphi(m_1 m_2)=\varphi(m_1)\,\varphi(m_2)$. Combined with $\varphi(p^r)=p^r-p^{r-1}$ for a prime power (where the non-coprime numbers are just the $p^{r-1}$ multiples of $p$), this immediately yields the general product formula for $\varphi(n)$ used in the previous topic. $\blacksquare$