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Let f and g be bijections such that fog is defined. Prove that fog is bijective.

Let f and g be bijections such that fog is defined. Prove that fog is bijective.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Problem Statement:** Let \( f \) and \( g \) be bijections such that \( f \circ g \) is defined. Prove that \( f \circ g \) is bijective. **Explanation:** In the problem given, we are asked to demonstrate that the composition of two bijective functions, \( f \) and \( g \), results in a function \( f \circ g \) that is also bijective. This problem involves properties of function composition and the nature of bijections, which have both injective (one-to-one) and surjective (onto) properties. To prove this, you will show two things: 1. **Injectivity of \( f \circ g \):** Assume \( f \circ g(x_1) = f \circ g(x_2) \). Since \( f \) is injective, \( f(g(x_1)) = f(g(x_2)) \) implies \( g(x_1) = g(x_2) \). Since \( g \) is injective, \( x_1 = x_2 \). Therefore, \( f \circ g \) is injective. 2. **Surjectivity of \( f \circ g \):** For any \( y \) in the codomain of \( f \circ g \), since \( f \) is surjective, there exists a \( z \) such that \( f(z) = y \). Since \( g \) is surjective, there exists an \( x \) such that \( g(x) = z \). Thus, \( f \circ g(x) = y \), making \( f \circ g \) surjective. Together, these properties prove that \( f \circ g \) is bijective.
Transcribed Image Text:**Problem Statement:** Let \( f \) and \( g \) be bijections such that \( f \circ g \) is defined. Prove that \( f \circ g \) is bijective. **Explanation:** In the problem given, we are asked to demonstrate that the composition of two bijective functions, \( f \) and \( g \), results in a function \( f \circ g \) that is also bijective. This problem involves properties of function composition and the nature of bijections, which have both injective (one-to-one) and surjective (onto) properties. To prove this, you will show two things: 1. **Injectivity of \( f \circ g \):** Assume \( f \circ g(x_1) = f \circ g(x_2) \). Since \( f \) is injective, \( f(g(x_1)) = f(g(x_2)) \) implies \( g(x_1) = g(x_2) \). Since \( g \) is injective, \( x_1 = x_2 \). Therefore, \( f \circ g \) is injective. 2. **Surjectivity of \( f \circ g \):** For any \( y \) in the codomain of \( f \circ g \), since \( f \) is surjective, there exists a \( z \) such that \( f(z) = y \). Since \( g \) is surjective, there exists an \( x \) such that \( g(x) = z \). Thus, \( f \circ g(x) = y \), making \( f \circ g \) surjective. Together, these properties prove that \( f \circ g \) is bijective.
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