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3. Let f: A → B and g: B → C be functions. Recall that the composition of g and f is the function go f : A → C defined by (gof)(a) = g(ƒ(a)) for each a € A. (a) Prove that if g and f are injective, then go f is injective. (b) Prove that if g and f are surjective, then go f is surjective. (c) Prove that if g and f are bijective, then go f is bijective. (Hint: This is all about the definitions. Start by writing out the hypotheses and the desired conclusion in terms of the definitions. Use the logical structure of the relevant definition (e.g., that of injective for (a)) to create an outline for your argument.) For any set A, the identity function iA: A → A is the function defined by i(a) = a for all a € A. An inverse for a function f : A → B is a function g: B → A such that go f = i and fog=ig. A function f can admit at most one inverse, and if it does, it is sometimes ¹Recall that, for a set S, P(S) denotes the power set of S, which is, by definition, the set consisting of all subsets of S. The notation (R) - {0} indicates the power set of R with the empty set removed.

3. Let f: A → B and g: B → C be functions. Recall that the composition of g and f is the function go f : A → C defined by (gof)(a) = g(ƒ(a)) for each a € A. (a) Prove that if g and f are injective, then go f is injective. (b) Prove that if g and f are surjective, then go f is surjective. (c) Prove that if g and f are bijective, then go f is bijective. (Hint: This is all about the definitions. Start by writing out the hypotheses and the desired conclusion in terms of the definitions. Use the logical structure of the relevant definition (e.g., that of injective for (a)) to create an outline for your argument.) For any set A, the identity function iA: A → A is the function defined by i(a) = a for all a € A. An inverse for a function f : A → B is a function g: B → A such that go f = i and fog=ig. A function f can admit at most one inverse, and if it does, it is sometimes ¹Recall that, for a set S, P(S) denotes the power set of S, which is, by definition, the set consisting of all subsets of S. The notation (R) - {0} indicates the power set of R with the empty set removed.

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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### Composition and Properties of Functions 3. Consider \( f : A \to B \) and \( g : B \to C \) as functions. Recall that the **composition** of \( g \) and \( f \) is the function \( g \circ f : A \to C \) defined by \( (g \circ f)(a) = g(f(a)) \) for each \( a \in A \). a. **Injectivity**: Prove that if \( g \) and \( f \) are injective, then \( g \circ f \) is injective. b. **Surjectivity**: Prove that if \( g \) and \( f \) are surjective, then \( g \circ f \) is surjective. c. **Bijectivity**: Prove that if \( g \) and \( f \) are bijective, then \( g \circ f \) is bijective. **Hint**: This problem is based on the definition. Start by writing out the hypotheses and the desired conclusion in terms of the definitions. Use the logical structure of the relevant definition (e.g., that of injectivity for (a)) to create an outline for your argument. For any set \( A \), the **identity function** \( i_A : A \to A \) is the function defined by \( i_A(a) = a \) for all \( a \in A \). An **inverse** for a function \( f : A \to B \) is a function \( g : B \to A \) such that \( g \circ f = i_A \) and \( f \circ g = i_B \). A function \( f \) can admit at most one inverse, and if it does, it is sometimes denoted \( f^{-1} \); it is clear from the definition that \( f \) is also the inverse of \( f^{-1} \) in this case. For example, the functions \( \exp : \mathbb{R} \to (0, \infty) \) and \( \log : (0, \infty) \to \mathbb{R} \) with the \(\mathbb{R} \) (the natural) exponential and logarithm functions) familiar from calculus are inverse of one another. --- **Footnotes:** 1. Recall
Transcribed Image Text:### Composition and Properties of Functions 3. Consider \( f : A \to B \) and \( g : B \to C \) as functions. Recall that the **composition** of \( g \) and \( f \) is the function \( g \circ f : A \to C \) defined by \( (g \circ f)(a) = g(f(a)) \) for each \( a \in A \). a. **Injectivity**: Prove that if \( g \) and \( f \) are injective, then \( g \circ f \) is injective. b. **Surjectivity**: Prove that if \( g \) and \( f \) are surjective, then \( g \circ f \) is surjective. c. **Bijectivity**: Prove that if \( g \) and \( f \) are bijective, then \( g \circ f \) is bijective. **Hint**: This problem is based on the definition. Start by writing out the hypotheses and the desired conclusion in terms of the definitions. Use the logical structure of the relevant definition (e.g., that of injectivity for (a)) to create an outline for your argument. For any set \( A \), the **identity function** \( i_A : A \to A \) is the function defined by \( i_A(a) = a \) for all \( a \in A \). An **inverse** for a function \( f : A \to B \) is a function \( g : B \to A \) such that \( g \circ f = i_A \) and \( f \circ g = i_B \). A function \( f \) can admit at most one inverse, and if it does, it is sometimes denoted \( f^{-1} \); it is clear from the definition that \( f \) is also the inverse of \( f^{-1} \) in this case. For example, the functions \( \exp : \mathbb{R} \to (0, \infty) \) and \( \log : (0, \infty) \to \mathbb{R} \) with the \(\mathbb{R} \) (the natural) exponential and logarithm functions) familiar from calculus are inverse of one another. --- **Footnotes:** 1. Recall
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