GATE CS 2025 Set 1 | Question 17

MCQ · 1 mark · −0.33

\(g(.)\) is a function from \(A\) to \(B\), \(f(.)\) is a function from \(B\) to \(C\), and their composition defined as \(f(g(.))\) is a mapping from \(A\) to \(C\).

If \(f(.)\) and \(f(g(.))\) are onto (surjective) functions, which ONE of the following is TRUE about the function \(g(.)\)?

  1. Option A: \(g(.)\) must be an onto (surjective) function.
  2. Option B: \(g(.)\) must be a one-to-one (injective) function.
  3. Option C: \(g(.)\) must be a bijective function, that is, both one-to-one and onto.
  4. Option D: \(g(.)\) is not required to be a one-to-one or onto function. (correct)

Answer

(D)

Explanation

For \(f(g(.))\) to be onto, the values produced by \(g\) only have to be enough for \(f\) to reach every element of \(C\). That puts no one-to-one or onto condition on \(g\) itself.

Counterexample: \(A = \{1, 2\}\), \(B = \{p, q\}\), \(C = \{z\}\), with \(g(1) = g(2) = p\) and \(f(p) = f(q) = z\). Both \(f\) and \(f(g(.))\) are onto, yet \(g\) is not one-to-one (1 and 2 both go to \(p\)) and not onto (\(q\) is never reached). So (A), (B) and (C) are false.

Answer: D