How does Chinese Remainder theorem help prove $\phi(ab) = \phi(a)\phi(b)$?

Question

I understand that one to one correspondence refers to one element in set A, for example, mapping to only 1 distinct element in set B. How does this one to one correspondence arise after applying Chinese Remainder theorem to the referred system of equations in the image?

Answer 1

The order of $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is given by $\varphi(n)$. If $a$ and $b$ are coprime, we get $(\mathbb{Z}/ab\mathbb{Z})^{\times} \cong (\mathbb{Z}/a\mathbb{Z})^{\times} \times (\mathbb{Z}/b\mathbb{Z})^{\times}$ by the chinese remainder theorem, such that the orders coincide which means that $\varphi(ab) = \varphi(a)\varphi(b)$.

Answer 2

Let $\Phi(n)=\{k\in\{0,1,\ldots,n-1\}\mid (k,n)=1\}$. Then $\varphi(n)=\#\Phi(n)$. So,$$\varphi(ab)=\varphi(a)\varphi(b)\iff\#\Phi(ab)=\#\Phi(a)\times\#\Phi(b).$$And the Chinese Remainder Theorem is exctly what you need in order to define a bijection between $\Phi(ab)$ and $\Phi(a)\times\Phi(b)$, whose existence is equivalent to the equality $\#\Phi(ab)=\#\Phi(a)\times\#\Phi(b)$.