Problem 4 - A Book of Abstract Algebra

Question

Let \(f: A \rightarrow B\) and \(g: B \rightarrow A\) be functions. Suppose that \(y=f(x)\) iff \(x=g(y)\). Prove that \(f\) is bijective, and \(g=f^{-1}\)

Step-by-Step Solution

Answer

Function \(f\) is bijective and \(g = f^{-1}\).

Step 1: Define Bijective Function

A bijective function is one that is both injective (one-to-one) and surjective (onto). To show that \(f\) is bijective, we first need to prove that \(f\) is injective.

Step 2: Prove Injective Property of \(f\)

To prove \(f\) is injective, assume \(f(x_1) = f(x_2)\). By the given condition, this implies \(x_1 = g(f(x_1)) = g(f(x_2)) = x_2\). Therefore, \(f\) is injective.

Step 3: Prove Surjective Property of \(f\)

To prove \(f\) is surjective, we must show that for every \(b \in B\), there exists an \(a \in A\) such that \(f(a) = b\). Given \(y = f(x)\) iff \(x = g(y)\), for any \(b \in B\), let \(a = g(b)\). Then \(f(g(b)) = b\) which shows \(f\) is surjective.

Step 4: Show that \(g = f^{-1}\)

To show \(g = f^{-1}\), we need to verify that \(g(f(x)) = x\) for all \(x \in A\) and \(f(g(y)) = y\) for all \(y \in B\). By the given condition \(y = f(x)\) iff \(x = g(y)\), both conditions are satisfied. Hence, \(g = f^{-1}\).

Key Concepts

Injective Function

An injective function, also known as a one-to-one function, is one where every element in the domain is mapped to a unique element in the codomain. This means that no two different elements in the domain share the same image.

• Imagine having a set of keys (domain) and locks (codomain) where each key fits exactly one lock. If two keys could fit the same lock, the function would not be injective.
• Mathematically, if we assume that for a function \(f\), \(f(x_1) = f(x_2)\) implies that \(x_1 = x_2\), then the function is injective.

For the function \(f: A \rightarrow B\), the exercise demonstrates injectiveness by assuming \(f(x_1) = f(x_2)\) and proving that it results in \(x_1 = x_2\) using the condition \(y=f(x)\) iff \(x=g(y)\). This condition is crucial as it allows us to state that if two things point to the same place, they must have started from the same point.
This kind of logical reasoning showcases why injective functions are so precise and well-defined.

Surjective Function

A surjective function, or an onto function, is one where every element in the codomain is an image of at least one element from the domain. In simpler terms, this means that the function covers the entire codomain.

• Think of it as ensuring every locker (codomain) must have at least one key (domain) assigned to it.
• Mathematically, for a function \(f: A \rightarrow B\), surjectiveness means for every \(b \in B\), there exists an \(a \in A\) such that \(f(a) = b\).

To show that \(f\) is surjective, the problem uses the relationship \(y=f(x)\) iff \(x=g(y)\). For any \(b\) in \(B\), we can find an element \(a\) in \(A\) by taking \(a=g(b)\). Since the function maps \(g(b)\) to \(b\), all elements in \(B\) are covered by the mapping from \(A\), confirming the surjective property.

Inverse Function

An inverse function essentially reverses the roles of the domain and codomain. For a function to have an inverse, it must be bijective.

• The concept of a function having an inverse is like having a reversible key – not only can you unlock the door, but you can also lock it back up with the same key.
• Mathematically, for a function \(f\) to have an inverse \(f^{-1}\), it must hold that \(f(f^{-1}(y)) = y\) for every \(y\) in the codomain and \(f^{-1}(f(x)) = x\) for every \(x\) in the domain.

In the exercise, we demonstrate that \(g = f^{-1}\) by ensuring that \(g(f(x)) = x\) and \(f(g(y)) = y\) for all \(x \in A\) and \(y \in B\). The condition \(y=f(x)\) iff \(x=g(y)\) confirms these identities.
Thus, the function \(g\) not only acts as the inverse of \(f\) but also guarantees that each element can be traced back to its original in the domain or codomain.