Let f: A → B be an onto (surjective) function, where A and B are nonempty…
2023
Let f: A → B be an onto (surjective) function, where A and B are nonempty sets. Define an equivalence relation ∼ on A by
a1 ∼ a2 if f(a1) = f(a2), where a1, a2 ∈ A.
Let ε = {[x] : x ∈ A} be the set of all equivalence classes under ∼. Define F: ε → B by F([x]) = f(x) for every [x] ∈ ε.
Which of the following statements is/are TRUE?
Answer: B. F is an onto (or surjective) function.; C. F is a one-to-one (or injective) function.; D. F is a bijective function. — ConceptA quotient set groups elements according to an equivalence relation. A map defined on equivalence classes is well-defined only when its value is…
- 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 124 students.
Show answer & explanation
Correct answer: B, C, D
Concept
A quotient set groups elements according to an equivalence relation. A map defined on equivalence classes is well-defined only when its value is independent of the representative chosen from each class.
For the kernel relation x ∼ y exactly when f(x) = f(y), each equivalence class is a fiber of f. Collapsing each fiber removes repeated preimages while preserving the image set.
Application
Representative independence: if [x] = [y], then x ∼ y, so f(x) = f(y). Therefore the rule F([x]) = f(x) assigns the same value whichever representative is used.
One-to-one property: if F([x]) = F([y]), then f(x) = f(y). Hence x ∼ y and therefore [x] = [y].
Onto property: for any b ∈ B, surjectivity of f gives some a ∈ A with f(a) = b. The class [a] belongs to ε and F([a]) = b.
Combining the previous two properties shows that F is both injective and surjective, so it is bijective.
Cross-check and result
Define G: B → ε by choosing any a with f(a) = b and setting G(b) = [a]. If another representative has the same image b, it belongs to the same equivalence class, so G is well-defined. Then F(G(b)) = b and G(F([x])) = [x], confirming that F and G are inverses.
Thus the statements that F is surjective, injective, and bijective are true; the statement that F is not well-defined is false.