My professor gave us a statement for the "Chinese Remainder Theorem" different from that stated in David M. Burton, which say:
If $n_1,\dots,n_k$ are coprime positive integers, then there exist a ring isomorphism $g:\mathbb Z_{n_1\cdots n_k}\to\mathbb Z_{n_1}\times\cdots\times\mathbb Z_{n_k}$.
And he defined it as: $$g(a\bmod n_1\cdots n_k)=(a\bmod n_1,\dots,a\bmod n_k)$$.
Then he proved that this function is well-defined, which I did not understand it, could anyone show me this proof please?
Let's call $m=n_1\cdots n_k$. If this function $g$ is to be well-defined, then $g(a+\ell m)$ must be componentwise equivalent to $g(a)$. That is, for each $1\le i \le k$, we must have $$a+\ell m \equiv a\pmod{n_i}.$$ However, $n_i\mid m$ so that $\ell m\equiv 0 \pmod{n_i}$ for each $i$. Thus, indeed $a+\ell m \equiv a \pmod{n_i}$. This shows well-definedness.
Maybe its also a good idea also to provide an explanation for why one must actually check that functions are "well-defined" at all. We know that $\mathbb{Q}$ consists of quotients of the form $\frac{a}{b}$ for $a,b\in \mathbb{Z}$. However, the elements of $\mathbb{Q}$ are actually equivalence classes of pairs $(a,b)\in \mathbb{Z}^2$ subject to $(a,b)\sim(c,d)$ if and only if $ad-bc=0$. (This is the notion of equivalent fractions taught in primary school.) Suppose now we define a "function" $f:\mathbb{Q}\to \mathbb{Z}$ by $\frac{a}{b}\mapsto a$. To ensure that this function is well-defined, we need to be sure that it is independent of the choice of representation of $x=\frac{a}{b}$. However, $x=\frac{2a}{2b}$ also, so that per our recipe $x\mapsto 2a$ also. Thus, this function is not well-defined. So a naïvely defined "function" on $\mathbb{Q}$ may not be a function at all.