Problem 9 - Applied Discrete Structures

Question

Let \(A\) be a nonempty set. Prove that if \(f\) is a bijection on \(A\) and \(f \circ f=f\), then \(f\) is the identity function, \(i\) Hint. You have seen a similar proof in matrix algebra.

Step-by-Step Solution

Answer

The function \(f\) is the identity function on \(A\).

Step 1: Understand the Given Information

The exercise states that we have a function \(f\) which is a bijection, meaning it is both injective (one-to-one) and surjective (onto), and it is defined as \(f: A \to A\). We are also given that \(f \circ f = f\). Our goal is to prove that \(f\) is the identity function \(i\), meaning \(f(x) = x\) for all \(x \in A\).

Step 2: Interpret the Condition f ∘ f = f

The condition \(f \circ f = f\) implies that applying \(f\) twice is the same as applying \(f\) once. Specifically, for any element \(x \in A\), we have \(f(f(x)) = f(x)\). This repetitive application means that each element behaves like a fixed point under \(f\).

Step 3: Analyze Injectivity

Since \(f\) is injective, if \(f(x) = f(y)\), then \(x = y\). By using the condition \(f(f(x)) = f(x)\), injectivity implies that \(f(x) = x\). This is because if \(f\) maps \(f(x)\) and \(x\) to the same value which is \(f(x)\), then \(x = f(x)\) must hold.

Step 4: Verify Surjectivity

Given that \(f\) is surjective and \(f(x) = x\) for each \(x \in A\) (shown by injectivity), \(f\) maps to every element in \(A\). Therefore, for every element in \(A\), there is an element \(x\) such that \(f(x) = x\). This is consistent and further confirms \(f\) is the identity function.

Step 5: Conclude That f is the Identity Function

With \(f(x) = x\) for every element \(x\), \(f(x)\) behaves exactly like the identity function \(i(x) = x\) for all elements in \(A\). Therefore, \(f\) must be the identity function on \(A\).

Key Concepts

Bijection

A bijection is a type of function that creates a perfect "pairing" between elements of two sets. When a function is a bijection, it means there is a one-to-one correspondence between the elements of the domain and the codomain. This is because the function is both injective and surjective.

• Injective: Also known as "one-to-one." This guarantees that no two different elements in the domain map to the same element in the codomain.
• Surjective: Also called "onto," this ensures every element in the codomain is an output of the function from some element of the domain.

A bijective function is a really nice structure to work with, because every output value is hit exactly once, with a unique input being mapped to it. For example, if you think of function like a machine, a bijection ensures that for every output, there’s an exact input that produces it, no more and no less. This characteristic makes bijections essential in understanding equivalences and setting up a direct comparison between the structure of two sets.

Injective Function

An injective function is like a precise arrow; each element of the domain is aiming at a unique target in the codomain without overlapping. In simpler terms, if you picture an injective function, imagine that if you were labeling things, no two arrows can point to the same labeled spot.

The formal definition: If a function \( f: A \to B \) is injective, for all pairs of distinct elements \( x \) and \( y \) in \( A \), \( f(x) eq f(y) \).

Injectivity can be checked by ensuring that every element in the input gets a unique output. If \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), then \( f \) is injective. This condition was crucial in proving that the function in our exercise is indeed the identity function, as it demonstrated that applying the function twice left each element unchanged, which showed \( f(x) = x \) for all \( x \).

Surjective Function

A surjective function ensures that every possible output value is covered. It’s like making sure every seat in a theater is filled with a person; no empty spots.

The formal definition of surjective is: If \( f: A \to B \) is surjective, for every \( y \) in \( B \), there is at least one \( x \) in \( A \) such that \( f(x) = y \).

This trait was particularly important in our solution because it means our function \( f \) can "reach" every element in the set \( A \). When combined with injectivity in our problem, it reassured us that not only does \( f \) map elements distinctly, but it also doesn’t leave any part of \( A \) uncovered. This exhaustive coverage paired with distinct mapping implies that \( f(x) = x \) for each \( x \) in set \( A \), which confirms our function acts as the identity function.