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Chapter 3: Problem 46

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

Short Answer

Expert verified
If \(g \circ f\) is bijective, then for any \(x_1, x_2\) in the domain of f with \(f(x_1) = f(x_2)\), it follows that \(x_1 = x_2\), proving that f is injective. Moreover, for any \(y\) in the codomain of g, there exists an \(x\) in the domain of g such that \(g(x) = y\), proving that g is surjective. Therefore, if \(g \circ f\) is bijective, then f is injective and g is surjective.

Step by step solution

01

Recalling Definitions

- A function is called injective (or one-to-one) if every element of the domain maps to a unique element of the codomain, meaning no two elements in the domain of the function have the same image. - A function is called surjective (or onto) if every element in the codomain has at least one element in the domain that maps to it. - A function is called bijective if it is both injective and surjective. Now let's move on to prove the given statement using these definitions.
02

Proof of f being injective

Assume that \(g \circ f\) is bijective. In order to prove that f is injective, we need to show that for any \(x_1, x_2\) in the domain of f such that \(f(x_1) = f(x_2)\), it follows that \(x_1 = x_2\). Let's start by assuming that \(f(x_1) = f(x_2)\) for some \(x_1, x_2\). We need to show that this implies \(x_1 = x_2\). Since \(g \circ f\) is bijective, it is also injective. So, \(g(f(x_1)) \neq g(f(x_2))\) whenever \(x_1 \neq x_2\). However, we have assumed that \(f(x_1) = f(x_2)\), so \(g(f(x_1)) = g(f(x_2))\). This can only happen if \(x_1 = x_2\). Thus, we have proved that f is injective.
03

Proof of g being surjective

Now we need to prove that g is surjective. In order to prove this, we need to show that for any \(y\) in the codomain of g, there exists an \(x\) in the domain of g such that \(g(x) = y\). Since \(g \circ f\) is bijective, it is also surjective. This means that for every \(y\) in the codomain of \(g \circ f\), there exists an \(x\) in the domain of f (let's call it \(x'\)) such that \(g(f(x')) = y\). Now, consider any \(y\) in the codomain of g. We have, \(g(f(x')) = y\). This means that there exists an \(x = f(x')\) in the domain of g such that \(g(x) = y\). Thus, we have proved that g is surjective. In conclusion, if the function composition \(g \circ f\) is bijective, then the function f is injective and the function g is surjective.

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Key Concepts

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

Injective Function
Understanding the concept of an injective function, commonly referred to as a 'one-to-one' function, is crucial in the realm of mathematics, particularly in the topic of function analysis. An injective function holds a distinctive property, where each element of the domain pairs uniquely with an element in the codomain. In other words, no two different elements in the domain map to the same value in the codomain.

For example, consider a function f that assigns each student in a class a unique student ID number. This function is injective because no two students will ever have the same ID number. It's this property that makes injective functions important for showing uniqueness and building various mathematical constructs.

To determine if a function is injective, we check if assuming two inputs, say, a and b, produce the same output (that is, f(a) = f(b)), necessitates that a must equal b. This is a pivotal test and lies at the heart of understanding injective functions.
Surjective Function
Moving on, the concept of a surjective or 'onto' function is another fundamental piece of the puzzle in function theory. A function is surjective if every element in the codomain is the image of at least one element from the domain. Essentially, this ensures that the function 'covers' the entire codomain.

An everyday metaphor could be a factory assembly line. If the factory represents the function, the raw materials are elements from the domain, and the finished products are elements of the codomain. The factory is surjective if every type of finished product has some raw material from which it's made.

To verify surjectivity, we must show that for every element y in the codomain, there exists an element x in the domain so that f(x) = y. This attribute makes surjective functions integral in demonstrating that every possible output is attainable from some input within the domain.
Function Composition
Function composition is a fascinating process in mathematics, serving as the cornerstone for building complex relationships between sets. It involves the combination of two functions, f and g, to create a new function, expressed as g(f(x)). This new function is constructed by taking the output of function f, and using this output as the input for function g.

Imagine a relay race where the first runner (function f) hands off the baton (the output) to the second runner (function g) who completes the race. The final result of the race (the output of g(f(x))) depends on both runners' performances.

To understand function composition more deeply, consider that the resulting composition g(f(x)) will exhibit properties related to the two original functions, f and g. Specifically, if the resulting function g(f(x)) is bijective, meaning both injective and surjective, then we can infer that the first function f by necessity is injective and the second function g is surjective. This dependency is part of the magic that function composition unfolds within the realm of mathematics.

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Most popular questions from this chapter

Prove. The open interval \((a, b)\) is uncountable. [Hint: Find a suitable bijection from \((0,1)\) to \((a, b) . ]\)

Expand each. $$\sum_{j=1}^{2} a_{i j}$$

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Prove each, where \(x \in \mathbb{R}\) and \(n \in \mathbf{Z}.\) \(\left\lfloor\frac{n^{2}}{4}\right\rfloor=\frac{n^{2}-1}{4}\) if \(n\) is odd

Let \(A\) and \(B\) be any two sets, and \(U\) the universe. Let \(f_{S}\) denote the characteristic function of a subset \(S\) of \(U\) and \(x\) an arbitrary element in \(U .\) Prove each. $$f_{A \oplus B}(x)=f_{A}(x)+f_{B}(x)-2 f_{A \cap B}(x)$$
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