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3. Suppose f: A → B is injective and g: B→ A satisies f g = wherei is the identity function on B. Prove that f is surjective and 8=f-1.

3. Suppose f: A → B is injective and g: B→ A satisies f g = wherei is the identity function on B. Prove that f is surjective and 8=f-1.

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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**Mathematics Problem: Function Injectivity and Surjectivity** **Problem Statement:** 3. Suppose \( f: A \to B \) is injective and \( g: B \to A \) satisfies \( f \circ g = I_B \), where \( I_B \) is the identity function on \( B \). Prove that \( f \) is surjective and \( g = f^{-1} \). **Solution Outline:** 1. **Injectivity of Function \( f \):** - A function \( f \) is said to be injective (or one-to-one) if for any \( a_1, a_2 \in A \), whenever \( f(a_1) = f(a_2) \), it must follow that \( a_1 = a_2 \). 2. **Identity Function \( I_B \):** - The identity function \( I_B \) on the set \( B \) is defined such that for every \( b \in B \), \( I_B(b) = b \). 3. **Composition of Functions:** - The composition \( f \circ g = I_B \) means that for every \( b \in B \), \( (f \circ g)(b) = f(g(b)) = b \). 4. **Surjectivity Proof:** - To prove that \( f \) is surjective (onto), we need to show that for every element \( b \in B \), there exists an element \( a \in A \) such that \( f(a) = b \). - Given \( f(g(b)) = b \) for every \( b \in B \), we see that each \( b \in B \) can be mapped from some \( a \in A \), implying \( f \) is surjective. 5. **Inverse Function:** - If \( f \) is both injective and surjective, then it is bijective. - The function \( g \) acts as the inverse of \( f \), so \( g = f^{-1} \). By following these steps, we can prove that the function \( f \) is surjective and that \( g \) is indeed the inverse of \( f \). This exercise helps illustrate fundamental concepts in function theory, including injective, surjective, and bijective mappings, as well as the
Transcribed Image Text:**Mathematics Problem: Function Injectivity and Surjectivity** **Problem Statement:** 3. Suppose \( f: A \to B \) is injective and \( g: B \to A \) satisfies \( f \circ g = I_B \), where \( I_B \) is the identity function on \( B \). Prove that \( f \) is surjective and \( g = f^{-1} \). **Solution Outline:** 1. **Injectivity of Function \( f \):** - A function \( f \) is said to be injective (or one-to-one) if for any \( a_1, a_2 \in A \), whenever \( f(a_1) = f(a_2) \), it must follow that \( a_1 = a_2 \). 2. **Identity Function \( I_B \):** - The identity function \( I_B \) on the set \( B \) is defined such that for every \( b \in B \), \( I_B(b) = b \). 3. **Composition of Functions:** - The composition \( f \circ g = I_B \) means that for every \( b \in B \), \( (f \circ g)(b) = f(g(b)) = b \). 4. **Surjectivity Proof:** - To prove that \( f \) is surjective (onto), we need to show that for every element \( b \in B \), there exists an element \( a \in A \) such that \( f(a) = b \). - Given \( f(g(b)) = b \) for every \( b \in B \), we see that each \( b \in B \) can be mapped from some \( a \in A \), implying \( f \) is surjective. 5. **Inverse Function:** - If \( f \) is both injective and surjective, then it is bijective. - The function \( g \) acts as the inverse of \( f \), so \( g = f^{-1} \). By following these steps, we can prove that the function \( f \) is surjective and that \( g \) is indeed the inverse of \( f \). This exercise helps illustrate fundamental concepts in function theory, including injective, surjective, and bijective mappings, as well as the
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