Is gof^-1 = fog?
If f: A → B, g: B → C are functions then define gof and fog.
I am so confused about fog, I want to know the domain and co-domain of fog.

To clarify the relationship between the compositions of functions, let's break down what \( g \circ f \) (read as "g of f") and \( f \circ g \) (read as "f of g") mean, and then we'll address your specific question about whether \( g \circ f = f \circ g \). This is a common area of confusion, but once we define the terms and their domains and codomains, it should become clearer.

Understanding Function Composition

When we have two functions, \( f: A \to B \) and \( g: B \to C \), we can compose them in two different ways:

• Composition \( g \circ f \): This means we first apply \( f \) and then apply \( g \). The result is a function that takes an input from set \( A \) and produces an output in set \( C \).
• Composition \( f \circ g \): This means we first apply \( g \) and then apply \( f \). However, this composition is only valid if the output of \( g \) is in the domain of \( f \).

Defining the Compositions

Let's define these compositions more formally:

• For \( g \circ f \): The domain is \( A \) (the domain of \( f \)), and the codomain is \( C \) (the codomain of \( g \)). So, \( g \circ f: A \to C \).
• For \( f \circ g \): The domain is \( B \) (the domain of \( g \)), and the codomain is \( B \) (the codomain of \( f \)). Thus, \( f \circ g: B \to B \) is only defined if \( g \) maps to \( B \).

Domain and Codomain of \( f \circ g \)

Now, focusing specifically on \( f \circ g \), we need to ensure that the output of \( g \) falls within the domain of \( f \). If \( g: B \to C \) and \( f: A \to B \), then:

• The domain of \( f \circ g \) is \( B \), which is the set of all possible outputs from \( g \).
• The codomain of \( f \circ g \) is \( B \), as \( f \) maps from \( B \) to \( B \).

Are \( g \circ f \) and \( f \circ g \) Equal?

In general, \( g \circ f \) is not equal to \( f \circ g \). The order of function composition matters significantly. To illustrate this, consider a simple example:

• Let \( f(x) = x + 1 \) (where \( f: \mathbb{R} \to \mathbb{R} \)) and \( g(y) = 2y \) (where \( g: \mathbb{R} \to \mathbb{R} \)).
• Then, \( g \circ f(x) = g(f(x)) = g(x + 1) = 2(x + 1) = 2x + 2 \).
• On the other hand, \( f \circ g(y) = f(g(y)) = f(2y) = 2y + 1 \).

As you can see, \( g \circ f(x) = 2x + 2 \) and \( f \circ g(y) = 2y + 1 \) yield different results, confirming that \( g \circ f \) is not equal to \( f \circ g \).

Wrapping Up

In summary, \( g \circ f \) and \( f \circ g \) are distinct compositions of functions, each with its own domain and codomain. Understanding the order of operations is crucial in function composition, and it’s essential to ensure that the output of one function fits the input requirements of the next. If you have any more questions or need further clarification, feel free to ask!