GATE CS 2025 Set 1 Q.17: Sets, Relations, Functions, Partial Orders and Lattices

$g(.)$ 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 $g(.)$ ?

$g(.)$ must be an onto (surjective) function.

$g(.)$ must be a one-to-one (injective) function.

$g(.)$ must be a bijective function, that is, both one-to-one and onto.

$g(.)$ is not required to be a one-to-one or onto function.

Official answer and solution

Official answer: D
From: Official GATE 2025 answer key

Solution

Test whether $f$ and $f \circ g$ being surjective forces $g$ to be injective or surjective using a small counterexample.

When $f \circ g: A \to C$ is onto, $f$ is guaranteed to be onto, but $g: A \to B$ does not need to be either one-to-one or onto.

Counterexample: Let $A = \{1, 2\}$ , $B = \{1, 2\}$ , and $C = \{1\}$ . Define:

$g(1) = 1, \; g(2) = 1$
$f(1) = 1, \; f(2) = 1$

Here both $f$ and $f \circ g$ map onto $C = \{1\}$ , yet $g$ is neither one-to-one ( $g(1) = g(2)$ ) nor onto ( $2 \in B$ has no preimage). Thus $g(.)$ is not required to be a one-to-one or onto function.

Final answer

Option (D)

Common trap

Confusing the theorem ' $f \circ g$ onto $\implies f$ onto' with ' $f \circ g$ onto $\implies g$ onto'.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2025 Computer Science and Information Technology - Session 1 (CS1)
Question: 17 of 65
Section: Computer Science and Information Technology

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