Question

If \( g \circ f \) is a bijective function, then:

• A. \( f \) is one-one and \( f \) is onto
• B. \( f \) is many-one and \( g \) is onto
• C. \( f \) is one-one and \( g \) is onto
• D. \( g \) is one-one and \( g \) is onto

Hint:

For the composition of functions \( g \circ f \) to be bijective, the function \( f \) must be injective (one-to-one) and the function \( g \) must be surjective (onto).

Answer 1

The Correct Option is C

We are given that the composition of functions \( g(f(x)) \) is bijective. This means that the composite function is both one-to-one (injective) and onto (surjective).

Now, let's break down what this means for the individual functions \( f \) and \( g \), and figure out the conditions on these functions for \( g \circ f \) to be bijective.

Step 1: What Does "One-to-One (Injective)" Mean?

- A function is one-to-one (injective) if different inputs produce different outputs. Mathematically, if \( f(x_1) = f(x_2) \), then we must have \( x_1 = x_2 \).

For the composition \( g \circ f \) to be one-to-one (injective), we need to look at the behavior of both \( f \) and \( g \):

• If \( f(x_1) = f(x_2) \), then for \( g \circ f \) to remain injective, it must hold that \( g(f(x_1)) = g(f(x_2)) \).
• For this to be true, \( f \) itself must be injective. If \( f \) weren’t injective, it could cause \( g(f(x_1)) = g(f(x_2)) \) without \( x_1 \) and \( x_2 \) being equal, which would break the injectivity of \( g \circ f \).

So, we conclude that for \( g \circ f \) to be injective, the function \( f \) must also be injective.

Step 2: What Does "Onto (Surjective)" Mean?

- A function is onto (surjective) if every element in the target set \( Y \) has at least one corresponding element in the domain \( X \) such that \( g(x) = y \). In other words, the function "covers" all the possible outputs in the target set.

For the composition \( g \circ f \) to be onto (surjective), we need to ensure that:

• For every element in the target set of \( g \), there is at least one element from the image of \( f \) such that \( g(f(x)) = y \). This ensures that \( g \circ f \) covers all possible elements in the target set.
• This can only happen if \( g \) itself is onto. If \( g \) isn’t onto, then there will be elements in the target set that \( g \circ f \) cannot reach.

So, we conclude that for \( g \circ f \) to be surjective, the function \( g \) must be surjective (onto).

Step 3: Conclusion – What Must Be True for \( g \circ f \) to be Bijective?

For the composition \( g \circ f \) to be bijective (both one-to-one and onto), the following must be true:

• Function \( f \) must be one-to-one (injective).
• Function \( g \) must be onto (surjective).

Final Answer:

Therefore, the correct answer is: Option 3: \( f \) is one-to-one and \( g \) is onto.