Chinese remainder theorem for a finite pairwise-coprime list: simultaneous residues determine one class modulo the product, and the resulting bijection preserves addition and multiplication

For $r=0$, $N=P_0=1$. The quotient $\mathbb Z/1$ has one class because congruence modulo $1$ is universal, and the empty Cartesian product has its single empty tuple; hence $\Phi$ is a bijection and all componentwise preservation assertions are vacuous.

Assume the simultaneous-residue assertion for the first $k$ moduli. Given classes modulo the first $k+1$ moduli, the induction hypothesis supplies a unique class $[y]_{P_k}$ having the first $k$ residues. Since $\gcd(P_k,n_k)=1$ by [L1], [L2] supplies a unique class $[x]_{P_kn_k}=[x]_{P_{k+1}}$ reducing to $[y]_{P_k}$ and to the prescribed class modulo $n_k$. It therefore has all prescribed residues.

If $x\equiv y\pmod N$, every $n_i$ divides $N$ and hence divides $x-y$, so $\Phi$ is well defined. For representatives $x,y$, its $i$th coordinate sends $[x+y]_N$ to $[x+y]_{n_i}=[x]_{n_i}+[y]_{n_i}$ and sends $[xy]_N$ to $[xy]_{n_i}=[x]_{n_i}[y]_{n_i}$; it similarly sends $[0]_N,[1]_N$ to the componentwise identities.

If two classes modulo $P_{k+1}$ have the same first $k+1$ residues, their representatives have difference divisible by every $n_i$ for $i<k+1$, hence by $P_{k+1}$ using [L1]; the two classes are equal. Thus the class constructed in step 1.2 is unique.

Induction gives existence and uniqueness for every finite length $r$. Therefore $\Phi$ is surjective and injective, hence bijective by [L5].

Steps 3.1 and 1.3 prove all assertions, including the empty-list boundary case from step 1.1.