Question:

g(periodcentered) is a function from A to B, f(periodcentered) is a function from B to C, and their composition defined as f(g(periodcentered)) is a mapping from A to C. If f(periodcentered) and f(g(periodcentered)) are onto (surjective) functions, which ONE of the following is TRUE about the function g(periodcentered)?

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For composition to be surjective, the second function \( f \) must cover the entire codomain. The first function \( g \) does not necessarily need to be surjective or injective.
Updated On: Apr 4, 2025
  • g(periodcentered) must be an onto (surjective) function.
  • g(periodcentered) must be a one-to-one (injective) function.
  • g(periodcentered) must be a bijective function, that is, both one-to-one and onto.
  • g(periodcentered) is not required to be a one-to-one or onto function.
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The Correct Option is D

Solution and Explanation

We are given that both \( f \) and \( f(g) \) are surjective (onto) functions.

\( f(g(x)) \) represents the composition of functions \( f \) and \( g \), where \( g \) maps from \( A \) to \( B \), and \( f \) maps from \( B \) to \( C \). The composition \( f(g(x)) \) maps from \( A \) to \( C \).

Let's analyze the problem:

- The fact that \( f(g(x)) \) is surjective means that for every element in \( C \), there is an element in \( A \) that maps to it through the composition of \( f \) and \( g \).

- For \( f(g(x)) \) to be surjective, the image of \( g(x) \) (which lies in \( B \)) must cover all the elements in the domain of \( f \), i.e., \( f \) must cover the entire set \( C \).

- However, \( g(x) \) does not need to be surjective (onto) because it is not required to cover the entire set \( C \), just the relevant part that \( f \) needs to map onto \( C \). Therefore, \( g(x) \) does not need to be injective or surjective for the composition to be surjective.

As a result, \( g(x) \) is not required to be either one-to-one (injective) or onto (surjective).

Thus, the correct answer is \( \boxed{D} \).
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