If and are two non-empty sets where , is function is defined such that
for any and , then
for any and , then
- A.
- B., only if
- C., only if
- D.
Show answer and solution
Correct option: C
Solution
The key facts are:
1. f{f⁻¹(B)} = B ∩ f(X). So f{f⁻¹(B)} = B only when B ⊆ f(X), not in general. This makes option 3 correct.
2. For option 1/2: f⁻¹{f(A)} ⊇ A always, with equality only when f is injective — not just when f(X) = Y. So neither option 1 nor 2 is correctly stated.
The decisive step: if y ∈ B but y ∉ f(X), then no x maps to y, so f⁻¹(B) misses any preimage for y, and f{f⁻¹(B)} can only produce elements actually in f(X). Hence f{f⁻¹(B)} = B ∩ f(X), which equals B if and only if B ⊆ f(X).
1. f{f⁻¹(B)} = B ∩ f(X). So f{f⁻¹(B)} = B only when B ⊆ f(X), not in general. This makes option 3 correct.
2. For option 1/2: f⁻¹{f(A)} ⊇ A always, with equality only when f is injective — not just when f(X) = Y. So neither option 1 nor 2 is correctly stated.
The decisive step: if y ∈ B but y ∉ f(X), then no x maps to y, so f⁻¹(B) misses any preimage for y, and f{f⁻¹(B)} can only produce elements actually in f(X). Hence f{f⁻¹(B)} = B ∩ f(X), which equals B if and only if B ⊆ f(X).
[AIEEE 2005]