Any function induces a surjection by restricting its codomain to its range. This

Question

Any function induces a surjection by restricting its codomain to its range. This should be obvious. What may or may not be obvious is that any onto function induces a bijection defined on a quotient of its domain. Let f: A ? B be an onto function. Let Q be the set of all equivalence classes of domain A under the equivalence relation x ~ y if and only if f(x) = f(y). (Equivalently, Q is the set of all pre-images under f). 2. Let p be the function that sends every element of A into its equivalence class; p(x) ? [x]. The induced map g: Q ? B given by g([x]) = f(x) is a well defined. Show that g is a bijection and that f = g ? p. Comment on the importance of such a result.

Explanation / Answer

Let g([x]) = g([y]). Then f(x) = f(y), meaning that x ~ y. Thus x and y are in the same equivalence class, i.e., [x] = [y]. Hence g is injective. Now let y be in B. Then there exists x in A such that f(x) = y since f is a surjection. Thus g([x]) = y and g is a surjection as well. Therefore g is a bijection. Since p : A -> Q and g : Q -> B, then g o p : A -> B, just like f. All we need to show is that they send the same elements to the same place. So let x be in A. Then g(p(x)) = g([x]) = f(x). Hence f(x) = g o p (x). Therefore f = g o p. A comment on this result? I don't know, I'll let you answer that one since this is the easy part :P

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