Problem 50 - Discrete Mathematics with Applications

Question

Let \(f: X \rightarrow Y\) and \(g: Y \rightarrow Z\) be invertible functions. Prove each. $$\left(f^{-1}\right)^{-1}=f$$

Step-by-Step Solution

Answer

Since we've shown that \(\left(f^{-1}\right)^{-1} \circ f^{-1} = Id_Y\) and \(f^{-1} \circ \left(f^{-1}\right)^{-1} = Id_X\), we can conclude that \(\left(f^{-1}\right)^{-1} = f\).

Step 1: Invertibility and Inverse Functions

An invertible function is a function that has an inverse, meaning there exists another function that "reverses" the action of our original function. If \(f:X \rightarrow Y\) is invertible, then there exists a function \(f^{-1}:Y \rightarrow X\) such that: 1. \(f \circ f^{-1} = Id_Y\) (applying them in sequence is an identity function on \(Y\)) 2. \(f^{-1} \circ f = Id_X\) (applying them in sequence is an identity function on \(X\))

Step 2: Composition of Inverse Functions

Now, let's consider the inverse function of \(f^{-1}\), denoted by \(\left(f^{-1}\right)^{-1}\). To prove that \(\left(f^{-1}\right)^{-1} = f\), we need to show that: 1. \(\left(f^{-1}\right)^{-1} \circ f^{-1} = Id_Y\) 2. \(f^{-1} \circ \left(f^{-1}\right)^{-1} = Id_X\)

Step 3: Verifying the First Condition

To show that \(\left(f^{-1}\right)^{-1} \circ f^{-1} = Id_Y\), we need to apply both functions in sequence. Following the definition of composition of functions, we have: \((\left(f^{-1}\right)^{-1}) \circ f^{-1}(y) = (\left(f^{-1}\right)^{-1})(f^{-1}(y))\), where \(y \in Y\). We know that \(f \circ f^{-1} = Id_Y\), so \(f(f^{-1}(y)) = y\). Replacing \(f(f^{-1}(y))\) by \(y\), we get: \((\left(f^{-1}\right)^{-1})(y)=y\), for all \(y\in Y\). This satisfies the condition, giving us: \(\left(f^{-1}\right)^{-1} \circ f^{-1} = Id_Y\)

Step 4: Verifying the Second Condition

To show \(f^{-1} \circ \left(f^{-1}\right)^{-1} = Id_X\), we need to apply both functions in sequence. Following the definition of composition of functions, we have: \(f^{-1} \circ (\left(f^{-1}\right)^{-1})(x) = f^{-1}((\left(f^{-1}\right)^{-1})(x))\), where \(x \in X\). We know that \(f^{-1} \circ f = Id_X\), so \(f^{-1}(f(x)) = x\). Replacing \(\left(f^{-1}\right)^{-1}\) by \(f\), we get: \(f^{-1}(f(x))=x\), for all \(x\in X\). This satisfies the condition, giving us: \(f^{-1} \circ \left(f^{-1}\right)^{-1} = Id_X\)

Step 5: Conclusion

Since we've shown that both conditions for the inverse function of \(f^{-1}\) are satisfied, we can conclude that \(\left(f^{-1}\right)^{-1} = f\).

Key Concepts

Function Composition

Understanding how functions can be combined is a cornerstone of algebra and higher mathematics. Function composition refers to applying one function to the results of another. The notation for composition is \( (f \circ g)(x) \), which means you first apply \( g(x) \) and then apply \( f \) to the result. When \( f : X \rightarrow Y \) and \( g : Y \rightarrow Z \) are both invertible, their composition \( f \circ g \) is also invertible, with the inverse given by \( g^{-1} \circ f^{-1} \).

This property is incredibly useful because it allows us to 'chain' together transformations and then 'unchain' them by applying the inverse transformations in reverse order. Using composition, complex operations can be broken down into simpler, more manageable parts. Moreover, if a function and its inverse are composed, they simplify to the identity function, which leads us directly into our next concept.

Identity Function

The identity function, often denoted as \( Id_X \), is a function that always returns its input. So for every element \( x \) in the domain \( X \) of \( Id_X \), \( Id_X(x) = x \). It's like a mathematical mirror reflecting whatever number looks into it.

In the realm of invertible functions, the identity function plays the role of a neutral element. If \( f \) is an invertible function with inverse \( f^{-1} \), then \( f \circ f^{-1} = Id_Y \) and \( f^{-1} \circ f = Id_X \). This concept is pivotal when demonstrating that a function is indeed invertible, as it must satisfy the conditions that the composition of the function and its inverse result in identity functions for their respective sets.

Inverse of a Function

The inverse of a function, denoted as \( f^{-1} \) for a function \( f \), reverses the effect of \( f \). It's the mathematical equivalent of hitting the 'undo' button. For \( f \) to have an inverse \( f^{-1} \), it must be bijective, meaning it's both injective (one-to-one) and surjective (onto).

Whenever we compose \( f \) and its inverse \( f^{-1} \) we should get the identity function. This property ensures that for every \( y \) in the codomain \( Y \), there is a unique \( x \) in the domain \( X \) such that \( f^{-1}(f(x)) = x \) and \( f(f^{-1}(y)) = y \). The concept of inverse functions is fundamental in mathematics because it provides a way to 'reverse' functions, making it possible to 'solve' equations and undo operations.

Proofs in Discrete Mathematics

In discrete mathematics, proofs are logical arguments that establish the truth of mathematical statements. There are many proof techniques, including direct proof, proof by contradiction, and mathematical induction, to name a few. When working with functions in discrete mathematics, a common proof strategy is to show that certain properties hold, such as the existence of an identity function when functions are composed.

For invertible functions, we often use direct proofs to show properties like \( \left(f^{-1}\right)^{-1} = f \). We demonstrate such claims by using function compositions and the definitions of the identity function and function inverses. By establishing both \( \left(f^{-1}\right)^{-1} \circ f^{-1} = Id_Y \) and \( f^{-1} \circ \left(f^{-1}\right)^{-1} = Id_X \), we have a clear, step-by-step verification of the original statement. Proofs are the language through which mathematicians communicate certainty and provide the foundation upon which all mathematical theory is built.