GATE 2025 CS1 Q.17: solution and answer

GATE 2025 · CS1Q.17 · 1 mark · MCQ

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

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

(A)
$g (\cdot)$ must be an onto (surjective) function.
(B)
$g (\cdot)$ must be a one-to-one (injective) function.
(C)
$g (\cdot)$ must be a bijective function, that is, both one-to-one and onto.
(D)
$g (\cdot)$ is not required to be a one-to-one or onto function.

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Answer and solution

ANSWER

Option (D): $g$ is not required to be one-to-one or onto.

Official answer: DSource: Official answer key, GATE 2025 CS1 (IIT Roorkee)

Given: $f : B \to C$ is onto and $f \circ g : A \to C$ is onto. We ask what is forced on $g : A \to B$ .
The condition on $f$ is already given, so the only other condition is that $f \circ g$ is onto. Whether $g$ must be onto or one-to-one can be settled with small counterexamples that satisfy both given conditions.
Test whether $g$ must be onto. Take $A = {1}$ , $B = {1, 2}$ , $C = {1}$ , with $g (1) = 1$ and $f (1) = f (2) = 1$ . Then $f$ is onto $C$ , and $(f \circ g) (1) = 1$ so $f \circ g$ is onto $C$ . But $2 \in B$ is not in the range of $g$ , so $g$ is not onto. This rules out (A) and also (C), since a bijective function must be onto.
Test whether $g$ must be one-to-one. Take $A = {1, 2}$ , $B = {1}$ , $C = {1}$ , with $g (1) = g (2) = 1$ and $f (1) = 1$ . Then $f$ is onto $C$ , and $f \circ g$ maps both elements to $1$ , so it is onto $C$ . But $g (1) = g (2)$ with $1 \neq = 2$ , so $g$ is not one-to-one. This rules out (B).
Both properties can fail while the hypotheses hold, so $g$ is not required to be one-to-one or onto. Option (D) is true.

KEY IDEA

Knowing that $f$ and $f \circ g$ are onto says nothing about injectivity or surjectivity of $g$ . To refute a "must" claim, one small counterexample that meets all the hypotheses is enough.

WATCH OUT FOR

Thinking that because $f \circ g$ is onto, $g$ must be onto. The true fact is that $f \circ g$ onto forces $f$ to be onto, not $g$ . Likewise, $f \circ g$ one-to-one would force $g$ to be one-to-one, but nothing here gives that.

From the GATE 2025 CS1 paper. Every question of this paper · Download the full paper · All 8 Sets & relations questions, with notes

GATE 2025 CS1 Q.17: solution and answer

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