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

For a ? A, f(a) = f(a). Thus a ~ a. This means ~ is reflexive. If a ~ b, then f(a) = f(b). By the symmetric property of = , f(b) = f(a). This means b ~ a. Therefore ~ is symmetric. If a ~ b and b ~ c, then f(a) = f(b) and f(b) = f(c). Since = is transitive, we have f(a) = f(c). This means a ~ c. Thus ~ is transitive. In conclusion, ~ is an equivalence relation. We want to show that g is a bijection (that is, one-to-one and onto). Onto is already given. Suppose g([x]) = g([y]). Then f(x) = f(y) This means x ~ y or [x] = [y] by equivalence. Thus g is a bijection. Next, suppose x ? A. g o p(x) = g(p(x)) = g([x]) = f(x). Thus g o p = f.

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