GATE2025_CS1_Q17

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(.)?

✅ Final Answer:

The correct option is (D) g(.) is not required to be a one-to-one or onto function.

🔹 Short Answer:

The property of function composition states that if the composite function f(g(x)) is surjective (onto), then the outer function f(x) must also be surjective. However, this property does not impose any necessary conditions on the inner function g(x). It is possible for f and f(g) to be surjective while g is neither surjective nor injective. Therefore, g(.) is not required to have either property.

🔸 Long Answer:

This question tests our understanding of function properties, specifically surjectivity (onto) and composition. Let's analyze the given information and the properties of function composition.

Key Definitions:

• Function Mappings: We have g: A → B and f: B → C. The composition is f(g): A → C.
• Surjective (Onto) Function: A function h: X → Y is surjective if its range is equal to its codomain. This means for every element y in Y, there is at least one element x in X such that h(x) = y.

Given Conditions:

1. f: B → C is surjective.
2. f(g): A → C is surjective.

A fundamental theorem in discrete mathematics states: If a composite function f(g(x)) is surjective, then the outer function f(x) must be surjective. The question gives us that both are surjective, which is consistent. The question asks what this implies about the inner function, g(x).

Analysis with a Counterexample

Let's construct a counterexample to test if g must be surjective or injective. We define sets A, B, and C and functions g and f that meet the given criteria.

• Let Set A = {1, 2}
• Let Set B = {x, y, z}
• Let Set C = {p, q}

Now, define the functions:

1. Function g: A → B

g(1) = x
g(2) = y

Is g surjective? No, because the element z in the codomain B is not in the range of g. The range is {x, y}.
Is g injective? Yes, because distinct elements in A map to distinct elements in B.

2. Function f: B → C

f(x) = p
f(y) = q
f(z) = q

Is f surjective? Yes, because every element in C ({p, q}) is in the range of f.

3. Composite Function f(g): A → C

f(g(1)) = f(x) = p
f(g(2)) = f(y) = q

Is f(g) surjective? Yes, because its range is {p, q}, which is equal to the codomain C.

Conclusion from the Counterexample

• The conditions are met: f is surjective, and f(g) is surjective.
• But, the function g is not surjective (it misses 'z').
• (In this specific example, g is injective, but we could easily make it non-injective too. For instance, if A = {1, 2, 3} and g(3)=y, g would no longer be injective, but the conditions on f and f(g) would still hold).

Since we found a valid case where g is not surjective, option (A) is false. Since we can also construct a case where g is not injective, option (B) is false. Consequently, option (C) is also false. This leaves only option (D) as correct: there are no required properties for g based on the given information.

Bibliography:
- Rosen, K. H. (2018). Discrete Mathematics and Its Applications. McGraw-Hill Education. (Chapter 2: Basic Structures: Sets, Functions, Sequences, Sums, and Matrices). [1]
- "Composition of Functions." Wikipedia, Wikimedia Foundation.

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