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Let f: A → B be an onto (or surjective) function, where A and B are nonempty sets. Define an equivalence relation ~ on the set A as a1 ~ a2 if f(a1) = f(a2), where a1, a2 ∈ A. Let E = {[x]: x ∈ A} be the set of all the equivalence classes under ~. Define a new mapping F: E → B as F([x]) = f(x), for all the equivalence classes [x] in E. Which of the following statements is/are TRUE?

1. F is NOT well-defined.
2. F is an onto (or surjective) function.
3. F is a one-to-one (or injective) function.
4. F is a bijective function.

Correct answer: F is a bijective function.

Solution

F is a bijective function because it is both onto and one-to-one. Since f is onto, every element in B has a pre-image in A, ensuring that F covers all of B. Additionally, because the equivalence relation groups elements of A that map to the same element in B, F is injective, as different equivalence classes map to different elements in B.

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