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Chapter 6: Problem 340

Explain why a function that has an inverse must be a bijection.

Short Answer

Expert verified
A function with an inverse must be bijective because it is both injective and surjective.

Step by step solution

01

Understand Inverse Function

A function has an inverse if there exists another function that 'reverses' the effect of the original function. Mathematically, if a function \( f \) has an inverse \( f^{-1} \), then for every \( x \) in the domain of \( f \), \( f^{-1}(f(x)) = x \) and for every \( y \) in the domain of \( f^{-1} \), \( f(f^{-1}(y)) = y \).
02

Define Bijection

A bijection is a function that is both injective and surjective. An injective (or one-to-one) function means no two different inputs map to the same output. A surjective (onto) function means every possible output is covered by the function.
03

Proving Injectivity

To show that a function with an inverse is injective, assume \( f(a) = f(b) \). Applying the inverse function to both sides gives \( f^{-1}(f(a)) = f^{-1}(f(b)) \), which simplifies to \( a = b \). Thus, no two different inputs can have the same output, proving injectivity.
04

Proving Surjectivity

To show that a function with an inverse is surjective, take any element \( y \) in the codomain. There exists an element \( x \) in the domain such that \( y = f(x) \), because \( f^{-1}(y) = x \). Hence, the function maps to every element in the codomain, proving surjectivity.
05

Conclusion

Since the function is both injective and surjective, it is bijective. Therefore, a function that has an inverse must be a bijection.

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Key Concepts

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

Understanding Bijective Functions
A function is bijective if and only if it is both injective and surjective. In simpler terms, a bijective function is a perfect matching between two sets. No element is left out on either side, and there are no duplicates. If you think about it, it's like assigning a unique ID to every student in a class: every student gets a distinct ID and every ID is assigned to exactly one student.

For a function to be bijective, it needs to satisfy two main properties. Let's break those down.

First is injectivity, which ensures no duplicates on the output side. Second is surjectivity, which ensures that every possible output is covered. If both properties hold, the function has a one-to-one correspondence with its outputs.
What is an Injective Function?
An injective function, also known as a one-to-one function, ensures that no two different inputs produce the same output. Imagine assigning a different unique locker to each student. No two students share the same locker. Mathematically, a function \( f \) is injective if \( f(a) = f(b) \) implies \( a = b \). If you find out that two students have been assigned the same locker, then the system is not injective.

This quality is crucial in creating a function that can be reversed. If a function is not injective, it cannot have an inverse because reversing the outputs would lead to ambiguity. You wouldn't know which input to match to the output since more than one input produces the same result.
Understanding Surjective Functions
A surjective function ensures that every possible output value is accounted for. It's like ensuring all slots in a bingo game have been called out at least once. Mathematically, a function \( f \) is surjective if for every element \( y \) in the codomain, there exists an element \( x \) in the domain such that \( f(x) = y \).

This quality ensures that there are no missed outputs, making the function 'onto'. If any output is missed, the function is not surjective. Surjectivity is important for a function to be reversible because it guarantees that every possible output has a corresponding input. Without surjectivity, some outputs would have no corresponding inputs, making it impossible to define an inverse.

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Most popular questions from this chapter

Another relation that you may have learned about in school, perhaps in the guise of "clock arithmetic," is the relation of equivalence modulo \(n\). For integers (positive, negative, or zero) \(a\) and \(b\), we write \(a \equiv b\) \((\bmod n)\) to mean that \(a-b\) is an integer multiple of \(n,\) and in this case, we say that \(a\) is congruent to \(b\) modulo \(n\) and write \(a \equiv b(\bmod n) . .\) Show that the relation of congruence modulo \(n\) is an equivalence relation.

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If \(f: S \rightarrow T, g: T \rightarrow X,\) and \(h: X \rightarrow Y,\) is \(h \circ(g \circ f)=\) \((h \circ g) \circ f ?\) What does this say about the status of the associative law $$ \rho \circ(\sigma \circ \varphi)=(\rho \circ \sigma) \circ \varphi $$ in a group of permutations?

Suppose a group acts on a set \(S\). Could an element of \(S\) be in two different orbits? (Say why or why not.) (h)
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