Chapter 1.5, Problem 39E

Solution

To Analyse: If with domain $D$ and range $R$ , $f$ is a one-to-one function, then with domain $R$ and range $D$ , $f^{-1}$ is a one-to-one function.

The given statement is true.

Given:

If $f$ is a one-to-one function with domain $D$ and range $R$ , then $f^{-1}$ is a one-to-one function with domain $R$ and range $D$ .

Explanation:

If $f$ is a one-to-one function, then if

  $D=\left\{a_1,a_2,a_3,\dots \right\}$

and

  $R=\left\{b_1,b_2,b_3,\dots \right\},$

If with the points $\left(a_1,b_1\right),\left(a_2,b_2\right),\dots$ the function $f$ has graphed, so that no point has the same $x-$ coordinate as another, and no point has the same $y-$ coordinate as another.

If $f^1$ defined as a relation, so, the graph is made of the points

  $\left(b_1,a_1\right),\left(b_2,a_2\right),\dots$

Where any point $\left(b_k,a_k\right)$ corresponds to a point $\left(a_k,b_k\right)$ in the graph of $f$ .

There is no other point in the graph of $f^1$ for any point $\left(b_k,a_k\right),$ where $b_k$ is the $x-$ coordinate. So, $f$ is a function.

There is no other point in the graph of $f^1$ for any point $\left(b_k,a_k\right)$ , where $a_k$ is the $y-$ coordinate is. So, $f$ is a one-to-one function.

So, it's clear that $f^{-1}$ is a one-to-one function with domain

  $D^{\prime }=\left\{b_1,b_2,b_3,\dots \right\}=R$

and range

  $R^{\prime }=\left\{a_1,a_2,a_3,\dots \right\}=D$

So, the statement is true.