Problem 42 If \(f\) and \(g\) are bijective... [FREE SOLUTION]

Chapter 3: Problem 42

If \(f\) and \(g\) are bijective, then \(g\) of is bijective.

Short Answer

Expert verified

We can prove that the composite function \(g \circ f\) is bijective by showing it is both injective and surjective according to two steps. Step 1: If \(g(f(a)) = g(f(b))\), then since \(g\) is injective, \(f(a) = f(b)\). As \(f\) is also injective, this means \(a = b\), so \(g \circ f\) is injective. Step 2: For any element \(y\) in the codomain of \(g \circ f\), since \(g\) and \(f\) are surjective, we can find an element \(x\) in the domain of \(g \circ f\) such that \(g(f(x)) = y\). Thus, \(g \circ f\) is surjective. Therefore, \(g \circ f\) is bijective.

Step by step solution

Prove that \(g \circ f\) is injective

To prove that \(g \circ f\) is injective, we need to show that if \(g(f(a)) = g(f(b))\), then \(a = b\). Let's assume \(g(f(a)) = g(f(b))\). Since \(g\) is bijective, it is also injective, which means that if \(g(x) = g(y)\), then \(x = y\). In our case, we have \(g(f(a)) = g(f(b))\), so \(f(a) = f(b)\). Now, since \(f\) is also bijective, it is injective as well. This means that if \(f(a) = f(b)\), then \(a = b\). We already have \(f(a) = f(b)\), so we can conclude that indeed \(a = b\). Therefore, \(g \circ f\) is injective.

Prove that \(g \circ f\) is surjective

To prove that \(g \circ f\) is surjective, we need to show that for every element \(y\) in the codomain of \(g \circ f\), there exists an element \(x\) in the domain of \(g \circ f\) such that \(g(f(x)) = y\). We are given that \(g\) is bijective, which means it is also surjective. So for a given \(y\), there exists an element \(i\) in the domain of \(g\) such that \(g(i) = y\). Since \(f\) is also bijective, it is surjective as well, which means that for every element \(i\) in the codomain of \(f\), there exists an element \(x\) in the domain of \(f\) such that \(f(x) = i\). So we have found the element \(x\) in the domain of \(g \circ f\) such that \(g(f(x)) = g(i) = y\). Therefore, \(g \circ f\) is surjective. Since we have proven that \(g \circ f\) is both injective and surjective, we can conclude that \(g \circ f\) is bijective.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Injective Function

An injective function, or one-to-one function, ensures uniqueness in mapping. Specifically, if you have two distinct elements in the domain, they will be mapped to two distinct elements in the codomain.
Let's think of this in a real-world scenario for clarity. If each person has a unique ID number, then the function associating each person to their ID is injective. No two persons share the same number, just as in injective functions, no two different inputs result in the same output.

In the solution provided, we demonstrate injectivity by showing that if two function outputs are equal (that is, if two ID numbers are the same), then the inputs must have been the same as well (the ID numbers belong to the same person). This characteristic is crucial in understanding why the composition of two bijective functions is also injective since bijectivity includes injectivity as one of its components.

Surjective Function

In contrast, a surjective function, or onto function, focuses on covering the entire codomain so that every element in the codomain is the image of at least one element in the domain.
Imagine we have a set of job roles (codomain) and a group of employees (domain). If every job role is filled by at least one employee, the function from employees to job roles is surjective.

The solution shows surjectivity by ensuring that for any point you might pick in the codomain, you can always find a starting point from the domain that gets you there. When we combine two bijective functions, as in function composition, the resulting function inherits this 'job-filling' property, assuring us that the combination will be surjective as well.

Function Composition

Function composition involves creating a new function by combining two functions where the output of the first function becomes the input to the second function.
To better understand function composition, let's use the analogy of a multi-step process like making coffee. First, you grind the beans (function one), and then you brew the coffee (function two). The ground beans are passed from the first process directly into the second to end up with your coffee. The composition is the entire process from beans to your final cup of coffee.

In mathematical terms, given two functions, f and g, composing them (denoted as g 鈭� f) means you're applying f first and then applying g to the result of f. This can only be done if the output (codomain) of f fits the input (domain) of g. When both f and g are bijective, the composition g 鈭� f will also be bijective, combining the individual properties of injectivity and surjectivity from f and g, as detailed in the textbook solution.

91影视

If \(f\) and \(g\) are bijective, then \(g\) of is bijective.

Short Answer

Step by step solution

Prove that \(g \circ f\) is injective

Prove that \(g \circ f\) is surjective

Key Concepts

Injective Function

Surjective Function

Function Composition

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