Problem 47 - Discrete Mathematics with Applications

Question

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

Step-by-Step Solution

Answer

Since \((f^{-1} \circ f)(x) = f^{-1}(f(x))\) and \(f^{-1}(f(y)) = y\) for any \(y\) in \(Y\), we can substitute \(y\) with \(f(x)\) to get \((f^{-1} \circ f)(x) = x\). This is the definition of the identity function \(1_X(x)\), so \(f^{-1} \circ f = 1_{X}\).

Step 1: Recall the definition of the composition of functions

The composition of two functions, denoted \(\circ\), is the operation of applying one function after another. In our case, we want to find the composition \(f^{-1} \circ f\). This means that we first apply the function \(f\) and then apply the function \(f^{-1}\) to the result. More formally, for any \(x\) in \(X\), \((f^{-1} \circ f)(x) = f^{-1}(f(x))\).

Step 2: Use the definition of inverse functions

The given functions \(f\) and \(g\) are invertible, which means they have inverses, denoted as \(f^{-1}\) and \(g^{-1}\). The inverse of a function "undoes" the action of the original function. In other words, for any \(x\) in \(X\), \(f(f^{-1}(x)) = x\), and for any \(y\) in \(Y\), \(f^{-1}(f(y)) = y\). With this definition, we can now simplify the expression \((f^{-1} \circ f)(x)\).

Step 3: Simplify the expression for \((f^{-1} \circ f)(x)\)

We want to simplify the expression \((f^{-1} \circ f)(x)\), which is defined as \(f^{-1}(f(x))\). From Step 2, we know that \(f^{-1}(f(y)) = y\) for any \(y\) in \(Y\). Here, \(f(x)\) is an element of \(Y\), so if we replace \(y\) with \(f(x)\) in this equation, we get: $$f^{-1}(f(f(x))) = f(x)$$ Now, let's substitute this result back into our expression for \((f^{-1} \circ f)(x)\): $$(f^{-1} \circ f)(x) = f^{-1}(f(x)) = x$$

Step 4: Compare the composition to the identity function

In the previous step, we found that \((f^{-1} \circ f)(x) = x\) for any \(x\) in \(X\). This is precisely the definition of the identity function \(1_X\). So, \((f^{-1} \circ f)(x) = 1_X(x)\) for all \(x\) in \(X\).

Step 5: Conclusion

Since \((f^{-1} \circ f)(x) = 1_X(x)\) for all \(x\) in \(X\), we have shown that \(f^{-1} \circ f = 1_{X}\).

Key Concepts

Composition of Functions

When we talk about functions, composition is like stacking two functions together. Imagine you have two functions, say \( f \) and \( g \). You can combine these into a single function by passing the output of \( f \) into \( g \). This operation is called the composition of functions and is symbolized by \( \circ \). So, if you have \( f : X \rightarrow Y \) and \( g : Y \rightarrow Z \), the composition \( g \circ f \) means that for each input \( x \), you apply \( f \) first to get \( f(x) \), and then \( g \) to the result, resulting in \( g(f(x)) \).
In more simple words, it's like following a path of functions, one step at a time.

• Apply one function, then another.
• Write as \( g \circ f \) - "\( g \) after \( f \)".
• Order matters - changing the order generally changes the result.

Using composition is very helpful when dealing with multiple transformations or operations in one go.

Inverse Functions

Inverse functions essentially "reverse" what the original function does. If \( f \) is a function that turns \( X \) into \( Y \), then its inverse \( f^{-1} \) will take \( Y \) back to \( X \). It's important to remember that not all functions have inverses. For a function to have an inverse, it must be one-to-one and onto. This means each input corresponds to one unique output and covers the entire output range, respectively.
This is how they work:

• Function \( f \) maps \( x \) to \( y \).
• Inverse \( f^{-1} \) maps \( y \) back to \( x \).
• When composed, \( f^{-1} \circ f \) or \( f \circ f^{-1} \) should return you to the start.

For instance, you might have \( f(x) = 2x + 3 \); then \( f^{-1}(x) = \frac{x-3}{2} \) works inversely.

Identity Function

An identity function is the simplest type of function because it does absolutely nothing to the input! It's symbolically represented as \( 1_X(x) = x \), meaning each element maps to itself. It's like a mirror for function inputs - no transformation at all. The role of an identity function becomes crucial in proving properties of compositions and inverses.
Here's why it's important:

• Acts as a "do nothing" function - \( x \) stays \( x \).
• In function operations, \( f^{-1} \circ f \) should end up as \( 1_X \), proving \( f \) undoes \( f^{-1} \).
• It preserves the element structure while showing functional integrity.

Understanding the identity function helps you recognize when you've returned to the starting value, making it a handy reference in the study of function theory.