Section 3.2 – Inverses and Compositions¶

Example

$$\begin{aligned} A &= \{ 1,2,3,4 \} \newline B &= \{ w,x,y,z \} \newline C &= \{ a,b,c \} \newline \end{aligned}$$

Let $f: A \mapsto B$,

$$\begin{aligned} f = \{ (1,x),(2,y),(3,z),(4,w) \} \end{aligned}$$

and,

$$\begin{aligned} g = \{ (1,a),(2,b),(3,c),(4,c) \} \end{aligned}$$

Solution

$f$ is one-to-one and onto since every element in $B$ is related to exactly one element in $A$, it is therefore a bijection. $g$ is not one-to-one since both $3$ and $4$ map to $c$ but it is onto since every element in $C$ is mapped to.

Reversing the order of each pair in $f$ gives its inverse $f^{-1}: B \mapsto A$, where,

$$\begin{aligned} f^{-1} = \{ (x,1),(y,2),(z,3),(w,4) \} \end{aligned}$$

$g$ does not have an inverse since it wouldn't be a function.

Inverses¶

definition: A function $f: A \mapsto B$ has an inverse,

$$f^{-1}: B \mapsto A$$

if and only if the relation obtained by reversing $f$'s ordered pairs is a function and is defined by,

$$f^{-1} = \{(y,x): (x,y) \in f \}.$$

To find the inverse,

1. Switch $x$ and $y$,
2. Solve for $y$.

Example

Find $f^{-1}: \mathbb R \setminus { 1} \mapsto \mathbb R \setminus {2}$ where,

$$f(x) = y =\frac{2x}{x-1}$$

Solution

$$\begin{aligned} \frac{2y}{y-1} &= x \newline 2y &= x (y-1) \newline 2y - xy &= -x \newline y(2-x) &= -x \newline y &= \frac{-x}{2-x} \newline y &= \frac{x}{x-2} \newline \end{aligned}$$

Proposition¶

A function $f$ has an inverse if and only if $f$ is a bijection (it's one-to-one and onto).

Fact¶

$$(f^{-1})^{-1} = f$$

So every inverse is a bijection as well.

Compositions¶

definition: The composition of $g: B \mapsto C$ with $f: A \mapsto B$ is,

$$g \circ f = g(f(x)), \forall x \in A$$

Example

$$\begin{aligned} A &= \{ 1,2,3,4 \} \newline B &= \{ w,x,y,z \} \newline C &= \{ a,b,c \} \newline \end{aligned}$$

$$\begin{aligned} f&: A \mapsto B &\text{ where } f = \{ (1,w),(2,z),(3,x),(4,w) \} \newline g&: B \mapsto C &\text{ where } g = \{ (w,a),(x,c),(y,b),(z,a) \} \newline \end{aligned}$$

Solution

Potatoes.

$$\begin{aligned} g \circ f = \{ (1,a),(2,a),(3,c)(4,a) \} \end{aligned}$$ $$\begin{aligned} f \circ g = \text{ not a function} \end{aligned}$$ $$\begin{aligned} (g \circ f)^{-1} = \text{ not a function} \end{aligned}$$

Notation¶

• $f^2 = f \circ f$
• $f^n = f \circ f \circ ... \circ f$

Proposition¶

if $f: A \mapsto B$ and $g: B \mapsto C$ are bijections, then $g \circ f$ is also a bijection.

Proof¶

(one-to-one) Supposed that $g \circ f(x_1) = g \circ f(x_2)$, for some $x_1, x_2 \in A$, or in another notation,

$$%Aligned \begin{aligned} g(f(x_1)) = g(f(x_2)) \end{aligned}$$

Since $g$ is one-to-one and $g(f(x_1)) = g(f(x_2))$,

$$%Aligned \begin{aligned} f(x_1) = f(x_2) \end{aligned}$$

then since, $f$ is one-to-one and $f(x_1) = f(x_2)$,

$$%Aligned \begin{aligned} x_1 = x_2 \end{aligned}$$

So $g \circ f$ is one-to-one.

(onto) Consider $c \in C$, since $g$ is onto,

$$%Aligned \begin{aligned} \exists b \in B \text{ such that } g(b) = c \end{aligned}$$

and since $f$ is onto,

$$%Aligned \begin{aligned} \exists a \in A \text{ such that } f(a) = b \end{aligned}$$

$$%Aligned \begin{aligned} \therefore g(b) = g(f(a)) = c \end{aligned}$$

and $g \circ f$ is onto.

Since $g \circ f$ is one-to-one and onto it is a bijection.

Inverses and Identities¶

Proposition¶

Given two functions $f: A \mapsto B$ and $g: B \mapsto A$ are inverses if and only is their compositions are identities.

$$%Aligned \begin{aligned} g \circ f &= i_A \newline f \circ g &= i_B \newline \end{aligned}$$

Example

$$%Aligned \begin{aligned} f(x) &= \frac{2x}{x-1} \newline g(x) &= \frac{x}{x-2} \newline \end{aligned}$$

Solution

$$%Aligned \begin{aligned} g \circ f &= \frac{\frac{2x}{x-1}}{\frac{2x}{x-1}-2} \newline &= \frac{\frac{2x}{x-1}}{\frac{2x}{x-1}-2} \cdot \frac{x-1}{x-1} \newline &= \frac{2x}{2x-2x+2} \newline &= \frac{2x}{2} \newline &= x \newline \end{aligned}$$

$$%Aligned \begin{aligned} f \circ g &= \frac{2\left( \frac{x}{x-2} \right)}{\left( \frac{x}{x-2} \right)-1} \newline &= \frac{2\left( \frac{x}{x-2} \right)}{\left( \frac{x}{x-2} \right)-1} \cdot \frac{x-2}{x-2} \newline &= \frac{2x}{2} \newline &= x \newline \end{aligned}$$