Let f : A → B be an onto (or surjective) function, where A and B are nonempty…
2023
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∈Aa1, a2∈A . Let ε={[x]:x∈A} ε={[x] : x∈A} be the set of all the equivalence classes under ∼. Define a new mapping F:ε→B as F([x])=f(x),F([x])=f(x) or all the equivalence classes [x] in ε
Which of the following statements is/are TRUE?
- A.
F is NOT well-defined.
- B.
F is an onto (or surjective) function.
- C.
F is a one-to-one (or injective) function.
- D.
F is a bijective function.
Attempted by 87 students.
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Correct answer: B, C, D
Key idea: The equivalence relation groups together all elements of A that share the same f-image, so each class [x] corresponds to the value f(x). Define F([x]) = f(x).
Well-defined: If [x] = [y], then f(x) = f(y) by definition of the relation, so F([x]) = f(x) = f(y) = F([y]).
Injective: If F([x]) = F([y]) then f(x) = f(y), hence [x] = [y]. Thus different equivalence classes map to different elements of B.
Surjective: Since f is onto, for any b in B there exists a in A with f(a) = b. Then F([a]) = b, so every element of B is hit by F.
Conclusion: F is well-defined, injective, and surjective, therefore F is a bijection between the set of equivalence classes and B.