5.4 Inverse Functions

Objectives:

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Define the composition of two functions
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Connect our understanding of inverse functions with composition of functions

5.4.1 Arriving at Our Goal

Our penultimate step on our journey: it is time to work with inverse functions. We have sneakily already defined them in a previous page, but now that we are in Section 5.4 we can give a formal definition:

Definition 5.4.1.

Let $X$ and $Y$ be sets, and let $f:X\rightarrow Y$ . We say that $f$ is invertible if there exists a function $g:Y\rightarrow X$ such that for all $x\in X$ and for all $y\in Y$ ,

y=f(x)\iff x=g(y).

We say that such a function $g$ is an inverse function of $f$ .

Example 5.4.2.

Show that $f:\mathbb{R}\rightarrow\mathbb{R}$ given by $f(x)=3x-1$ is an invertible function by finding its inverse function. Then show that, if $y=3x-1$ , then $x=g(y)$ .

As we did on the last page, define the identity function $\iota:X\rightarrow X$ to be the function $f(x)=x$ . If it is unclear from context what the shared domain and codomain of the function $\iota$ are, we will clarify by adding the domain as a subscript, as we do in the theorem below.

Theorem 5.4.3.

Let $X$ and $Y$ be sets, and let $f:X\rightarrow Y$ and $g:Y\rightarrow X$ . Then $f$ is invertible and $g$ is an inverse function of $f$ iff $g\circ f=\iota_{X}$ and $f\circ g=\iota_{Y}$ .

We briefly review why we need the function $f:X\rightarrow Y$ to be bijective in order for there to be an inverse function:

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If a graph is not surjective, the reflected graph will have a gap in the domain because the original function has a gap in the codomain.
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If a graph is not injective, it fails the vertical line test. Hence the graph’s reflection across the line $y=x$ fails the vertical line test and hence is not a function.

5.4.2 Bijectivity Iff Existence of Inverse

We now show that bijectivity is exactly what we need for an inverse function to exist.

Theorem 5.4.4.

Let $f:X\rightarrow Y$ be a function. Then $f$ is invertible if and only if $f$ is a bijection.

(More space for Theorem 5.4.4.)

Example 5.4.5.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by the rule $f(x)=x^{3}+1$ for all $x\in\mathbb{R}$ . Show that $f$ is a bijection and find $f^{-1}$ .

Theorem 5.4.6.

If $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ are both bijections, show that $g\circ f$ is also a bijection and that $(g\circ f)^{-1}=f^{-1}\circ g^{-1}$ .