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6. Suppose that the function f: X→ Y is a one to one and onto, with inverse ¹Y → X. Show that fof=1, and fo f¹ = ly-

6. Suppose that the function f: X→ Y is a one to one and onto, with inverse ¹Y → X. Show that fof=1, and fo f¹ = ly-

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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**6. Suppose that the function \( f: X \rightarrow Y \) is a one-to-one and onto function, with inverse \( f^{-1}: Y \rightarrow X \). Show that \( f^{-1} \circ f = I_X \) and \( f \circ f^{-1} = I_Y \).** This problem deals with the properties of bijective functions and their inverses. A function \( f \) is considered bijective if it is both injective (one-to-one) and surjective (onto). The presence of an inverse function \( f^{-1} \) signifies that every element in set \( Y \) corresponds to precisely one element in set \( X \). **Key Concepts:** - **Injective (One-to-One):** Each element of the domain maps to a unique element of the codomain. - **Surjective (Onto):** Every element of the codomain is mapped to by at least one element of the domain. - **Inverse Function:** The function \( f^{-1} \) reverses the mapping of \( f \), taking elements from the codomain back to the domain. **Objective:** - **Identity Function:** The identity function \( I_X \) on set \( X \) is defined so that \( I_X(x) = x \) for all \( x \in X \). Similarly, \( I_Y(y) = y \) for all \( y \in Y \). To prove these conditions, we need to show that applying \( f \) and then \( f^{-1} \) (or vice versa) results in the original input, meaning that the composite of these functions behaves as the identity function on their respective sets.
Transcribed Image Text:**6. Suppose that the function \( f: X \rightarrow Y \) is a one-to-one and onto function, with inverse \( f^{-1}: Y \rightarrow X \). Show that \( f^{-1} \circ f = I_X \) and \( f \circ f^{-1} = I_Y \).** This problem deals with the properties of bijective functions and their inverses. A function \( f \) is considered bijective if it is both injective (one-to-one) and surjective (onto). The presence of an inverse function \( f^{-1} \) signifies that every element in set \( Y \) corresponds to precisely one element in set \( X \). **Key Concepts:** - **Injective (One-to-One):** Each element of the domain maps to a unique element of the codomain. - **Surjective (Onto):** Every element of the codomain is mapped to by at least one element of the domain. - **Inverse Function:** The function \( f^{-1} \) reverses the mapping of \( f \), taking elements from the codomain back to the domain. **Objective:** - **Identity Function:** The identity function \( I_X \) on set \( X \) is defined so that \( I_X(x) = x \) for all \( x \in X \). Similarly, \( I_Y(y) = y \) for all \( y \in Y \). To prove these conditions, we need to show that applying \( f \) and then \( f^{-1} \) (or vice versa) results in the original input, meaning that the composite of these functions behaves as the identity function on their respective sets.
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