Let f : A → B be an injective (one-to-one) function. Define g : 2^A → 2^B as :…

2003

Let f : A → B be an injective (one-to-one) function. 

Define g : 2^A → 2^B as :
g(C) = {f(x) | x ∈ C}, for all subsets C of A.
Define h : 2^B → 2^A as :
h(D) = {x | x ∈ A, f(x) ∈ D}, for all subsets D of B.

Which of the following statements is always true ?

Answer: A. g(h(D)) ⊆ D — For any function f : A → B, define the image map g(C) = f(C) for C ⊆ A and the preimage map h(D) = f⁻¹(D) = {x ∈ A | f(x) ∈ D} for D ⊆ B. A standard…

  1. A.

    g(h(D)) ⊆ D

  2. B.

    g(h(D)) ⊇ D

  3. C.

    g(h(D)) ∩ D = ∅

  4. D.

    g(h(D)) ∩ (B - D) ≠ ∅

Attempted by 151 students.

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Correct answer: A

For any function f : A → B, define the image map g(C) = f(C) for C ⊆ A and the preimage map h(D) = f⁻¹(D) = {x ∈ A | f(x) ∈ D} for D ⊆ B. A standard set-theory identity governs the composite g(h(D)): applying f to its own preimage of D always lands back inside D, i.e. g(h(D)) = f(f⁻¹(D)) = D ∩ f(A) ⊆ D — and this holds for EVERY function f (injective or not, onto or not); equality g(h(D)) = D holds exactly when D ⊆ f(A) (in particular whenever f is surjective).

Applying this identity to the given g and h:

  1. Take any y ∈ g(h(D)).

  2. By definition of g, y = f(x) for some x ∈ h(D).

  3. By definition of h, f(x) ∈ D. Therefore y = f(x) ∈ D. This proves g(h(D)) ⊆ D always holds.

Checking the other options by counterexample (cross-check):

  1. The claim that g(h(D)) ⊇ D need not hold. Example: A = {1}, B = {a,b}, f(1)=a, and D={b}. Then h(D)=∅ and g(h(D))=∅, so g(h(D)) does not contain b.

  2. The claim that g(h(D)) ∩ D = ∅ need not hold. Example: A={1}, B={a}, f(1)=a, D={a}. Then h(D)={1}, g(h(D))={a}, and the intersection is {a}.

  3. The claim that g(h(D)) meets B − D (i.e., g(h(D)) ∩ (B − D) ≠ ∅) is impossible because g(h(D)) ⊆ D, so its intersection with B − D is always empty.

Answer: g(h(D)) ⊆ D is always true.

  • Equality remark: g(h(D)) = D ∩ f(A). Thus g(h(D)) equals D exactly when every element of D is in the image f(A).

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