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Chapter 3: Problem 42

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

Short Answer

Expert verified
To show that \(g \circ f\) is bijective, we must prove that it is injective and surjective. We prove injectivity by assuming \(g(f(x_1)) = g(f(x_2))\) and showing that \(x_1 = x_2\). We prove surjectivity by finding an element \(x\) in the domain of \(f\) such that \(g(f(x)) = z\) for every element \(z\) in the codomain of \(g \circ f\). Thus, \(g \circ f\) is bijective.

Step by step solution

01

Definitions of Injective and Surjective Functions

A function is injective if every input has a unique output, and it is surjective if for every element in the codomain, there exists an input that maps to it. In other words, a function is injective if no two inputs have the same output, and it is surjective if the function covers the entire range of its codomain.
02

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

To prove that \(g \circ f\) is injective, assume that \(g(f(x_1)) = g(f(x_2))\), where \(x_1\) and \(x_2\) are elements in the domain of \(f\). Since \(g\) is bijective, it is injective, which means that if \(g(f(x_1)) = g(f(x_2))\), then \(f(x_1) = f(x_2)\). As \(f\) is also injective, we can further deduce that \(x_1 = x_2\). Therefore, \(g \circ f\) is injective.
03

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

To prove that \(g \circ f\) is surjective, let \(z\) be an element in the codomain of \(g \circ f\). Since \(g\) is surjective, there exists an element \(y\) in the domain of \(g\) such that \(g(y) = z\). Similarly, since \(f\) is surjective as well, there exists an element \(x\) in the domain of \(f\) such that \(f(x) = y\). Now, we can write the composition \(g \circ f\) as \(g(f(x)) = g(y) = z\). This demonstrates that for every element \(z\) in the codomain of \(g \circ f\), there exists an input \(x\) that maps to it, proving that \(g \circ f\) is surjective.
04

Conclusion

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 Functions
Injective functions, also known as one-to-one functions, have a unique characteristic: each element in the domain maps to a distinct and separate element in the codomain. In simpler terms, two different inputs will never produce the same output. This property ensures that every output from the function relates back to a single, unique input.

For a function \( f \) to be injective, it must satisfy the condition: if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \). This means there are no duplicate outputs for different inputs. An injective function ensures no overlapping in the mappings, which is crucial in exercises involving function composition.
  • Injective functions maintain distinctiveness
  • The uniqueness of mapping helps in proving bijective properties when functions are combined
Surjective Functions
A surjective function, or onto function, has a different kind of uniqueness - it makes sure that every element in the codomain has a preimage in the domain. In other words, the function covers every possible output in its codomain, leaving no gaps.

For a function to be surjective, for every element \( y \) in the codomain, there must exist at least one element \( x \) in the domain such that \( f(x) = y \). This ensures that all targets in the codomain have at least one arrow pointing to them from the domain.
  • Surjective functions ensure that the entire codomain is utilized
  • This coverage is essential when proving the surjectivity of composite functions
Function Composition
Function composition involves creating a new function by combining two functions, often noted as \( g \circ f \), where \( f \) is applied first and then \( g \). This operation allows us to use one function as the input for another, essentially chaining their behaviors together.

When we deal with bijective functions, both injective and surjective properties are maintained through the composition. If both \( f \) and \( g \) are bijective, \( g \circ f \) will also be bijective:
  • Injective Aspect: Since both functions don鈥檛 duplicate outputs, their composition also won鈥檛.
  • Surjective Aspect: Both functions cover their entire codomains, meaning the composition will cover its complete codomain as well.
This is crucial when solving problems like the one in this exercise, where showing that \( g \circ f \) is bijective confirms its complete one-to-one and onto nature.

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Most popular questions from this chapter

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) defined by \(f(x)=a x+b,\) where \(a, b \in \mathbb{R}\) and \(a \neq 0 .\) Show that \(f\) is surjective; that is, find a real number \(x\) such that \(f(x)=c.\)

Just as \(\sum\) is used to denote sums, the product \(a_{k} a_{k+1} \ldots a_{m}\) is denoted by \(\prod_{i=k}^{m} \mathrm{a}_{i} .\) The product symbol \(\Pi\) is the Greek capital letter \(p i .\) For example, \(n !=\prod_{i=1}^{n} i .\) Evaluate each product. $$\prod_{j=-5}^{50} 1$$

Rewrite each sum using the summation notation. $$1+3+5+\cdots+23$$

Prove. Any subset of a countable set is countable.

Evaluate each sum. $$\sum_{j=-2}^{2} j(j-2)$$
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