Problem 12 - Applied Discrete Structures

Question

Let \(f\) and \(g\) be functions whose inverses exist. Prove that \((f \circ g)^{-1}=\) \(g^{-1} \circ f^{-1}\)

Step-by-Step Solution

Answer

The inverse of a composition \((f \circ g)^{-1}\) is \(g^{-1} \circ f^{-1}\), proven by reversing the order of applying inverse functions.

Step 1: Understanding the notation

First, let's break down the notation. The expression \((f \circ g)^{-1}\) refers to the inverse of the composition of functions \(f\) and \(g\). We want to prove that this is equal to \(g^{-1} \circ f^{-1}\), which is the composition of the inverses of \(g\) and \(f\), but in reverse order.

Step 2: Recall the definition of inverse functions

Recall that if a function \(h\) has an inverse \(h^{-1}\), then \(h(h^{-1}(x)) = x\) and \(h^{-1}(h(x)) = x\) for all \(x\) in the domains of \(h^{-1}\) and \(h\) respectively.

Step 3: Applying composition of functions

Consider the composition \((f \circ g)(x) = f(g(x))\). We need to find a function that, when composed with \((f \circ g)(x)\), returns \(x\).

Step 4: Using inverses to find \((f \circ g)^{-1}\)

To find \((f \circ g)^{-1}\), we apply the inverse functions in the reverse order. Start with \(y = (f \circ g)(x) = f(g(x))\). We need \(x\) back, so first apply \(f^{-1}\): \(f^{-1}(y) = g(x)\). Then apply \(g^{-1}\): \(g^{-1}(f^{-1}(y)) = x\).

Step 5: Conclusion

From Step 4, applying \(g^{-1}\) after \(f^{-1}\) reverses the effect of \(f \circ g\). Hence \((f \circ g)^{-1} = g^{-1} \circ f^{-1}\). We've proven that the inverse of the composition is the composition of the inverses in reverse order.

Key Concepts

Composition of Functions

Function composition is an essential concept in mathematics. It's the operation of applying one function to the results of another, creating a new function. Imagine you have two functions, \( f \) and \( g \). When you compose these, you're essentially feeding the output of \( g \) into the function \( f \). In function notation, this is referred to as \( (f \circ g)(x) = f(g(x)) \).
To put it simply:

• Start with an input \( x \).
• First, put \( x \) into \( g \) to produce \( g(x) \).
• Then, take \( g(x) \) and put it into \( f \), resulting in \( f(g(x)) \).

This operation is useful as it allows us to create complex functions built on simpler ones, helping to model more intricate real-world scenarios. Remember, order matters with composition. \( f(g(x)) \) usually isn't the same as \( g(f(x)) \).

Function Notation

Function notation is a way to represent functions in a clear and concise manner, enabling mathematicians to work with functions effortlessly. It's like a shorthand that tells us the process needed to get outputs from inputs.

• For a function \( f \), you would write \( f(x) \) to denote the result of \( f \) when you put \( x \) into it.
• The letters used (like \( f \), \( g \), \( h \)) are simply placeholders. They can be almost anything you wish, as long as they are consistent.
• A special note about inverses: when you write \( f^{-1}(x) \), this represents the inverse function of \( f \), which, if it exists, allows us to reverse \( f \)'s operation to find the original input \( x \).

This clarity makes it easier to perform operations like composition or finding inverses. By understanding function notation, you'll be better equipped to not only deal with straightforward functions but also more complex combinations of them.

Proof by Composition

Proof by composition is a method used to demonstrate that two functions are inverses of each other. This technique uses the properties stored within the functions

• To show two functions, \( f \) and \( g \), are inverses, you must prove that \( (f \circ g)(x) = x \) and \( (g \circ f)(x) = x \) for all \( x \) in the domains of \( f \) and \( g \).
• The core idea is that when you "undo" each function with its inverse, you're left with just the original input.
• In the exercise, this means demonstrating that \( (f \circ g)^{-1} = g^{-1} \circ f^{-1} \) using these proof concepts.

This proof structure not only solidifies your understanding of function inverses but also reinforces why order is so critical with compositional inverses. The subtle nature of proof by composition underscores the beauty and intricacy of mathematical operations.