Question

$$ \begin{aligned} &\text { Prove that if } f \text { and } g \text { are one-to-one functions, then }\\\ &(f \circ g)^{-1}(x)=\left(g^{-1} \circ f^{-1}\right)(x). \end{aligned} $$

Short answer

The inverse of the composition of two one-to-one functions, \(f\) and \(g\), equals the composition of the inverse of the two functions. This has been proven by using the basic property of inverse functions that the composition of a function and its inverse gives the original function back, i.e., an identity function.

Step-by-step solution

Understanding Inverse Functions

Let's begin with the concept of inverse functions. For a function \(g\) to have an inverse, it must be one-to-one (injective). This means that for every \(y\) in the codomain, there is exactly one \(x\) in the domain of \(g\) such that \(g(x) = y\). The inverse function, denoted as \(g^{-1}\), takes a \(y\) from the codomain of \(g\) and outputs the corresponding \(x\) in the domain of \(g\). In other words, \(g^{-1}(g(x)) = x\).

Applying the property of Inverse Functions

Now let's apply this understanding to solve the exercise. We need to prove that \((f \circ g)^{-1}(x) = (g^{-1} \circ f^{-1})(x)\). To prove this equality, we can take the composition of each side with \(f \circ g\), and show that this results in \(x\).

Proving the Equality

Let's try to take the composition of \((f \circ g)^{-1}(x)\) with \(f \circ g\). Using the property of inverse functions, \((f \circ g)( (f \circ g)^{-1}(x)) = x\). Now, let's do the same thing for the other side, \((g^{-1} \circ f^{-1})( f \circ g)(x)\). Since the composition of a function with its inverse gives the identity, this simplifies to \(x\) as well.