Let $m$ and $n$ be positive integers with $\gcd(m, n) = 1$.
Let $A = \{1, 2, \dots, m\}$, $B = \{1, 2, \dots , n\}$, $X = \{1, 2, \dots, mn\}$, and let $Y = A \times B$.
a) Show that for every integer $x \in X$, there exists a unique ordered pair $(a, b)$ in $Y$ such that $x \equiv a \mod m $ and $x \equiv b \mod n$.
b) Show that this defines a one-to-one correspondence between the sets $X$ and $Y$.
My attempt:
a) By the Chinese Remainder theorem, we know that there is a unique solution $x$ to the system of congruences modulo $mn$. Assume for some $x$ there also exists a pair $(a',b')$ which satisfies the system of congruences.
Then,
$x \equiv a \equiv a' \mod m$
$x \equiv b \equiv b' \mod n$
I want to show that $(a,b) = (a',b')$ but I get stuck. Any tips as to where to go from here? Is showing that $a \equiv a' \mod m$ enough to show that $a=a'$ ?
Since
b) For this bit, doesn't it naturally follow that since for each $x$, there is a unique $(a,b)$ such that that $F : X \to Y$ is one-to-one?