For a mapping %:A -> B, let == denote the kernel equivalence of %, and let *:A -> A== denote the natural mapping. Define $:A== -> B by $([a]) = %(a) for every equivalence class [a] in A==.

1. Show that $ is well defined and one-to-one, and that $ is onto if % is onto. Furthermore, show that % = $*, so that % is the composite of an onto mapping followed by a one-to-one mapping.

2. If %(A) is a finite set, show that the set A== of equivalence classes is also finite and that the cardinality of A== is equal to the cardinality of %(A) (called the Bijection Theorem).

3. In each of the following two cases, find the cardinality of A== for the given set A and the given mapping %:

i. A = U X U, where U = {1, 2, 3, 4, 6, 12}, and %:A -> Q is defined by %(n, m) = n/m, where Q is the set of all rational numbers.

ii. A = {n: n is an element of Z and 1 <= n <= 99}, where Z is the set of all integers, and %:A -> N is defined by %(n) = the sum of the digits of n, where N is the set of all natural numbers.