## SectionProject: More on Inverses, Composition and Relations

###### Problem2.32.
1. Is Problem 1.28 valid if we define $(a, b) = \{\{a, 1\}, \{2, b\}\}\text{?}$

2. How would one formally define an ordered triple? An ordered quadruple?

###### Problem2.33.

Show that the composition of one-to-one functions is one-to-one by letting $g:\B \to \C$ and $h:\A \to \B$ be one-to-one functions and showing that $g\circ h:\A \to \C$ is one-to-one.

###### Problem2.34.

Show that the composition of onto functions is onto.

The word identity is used in a number of different contexts in mathematics. The function $f: \R \to \R$ defined by $f(x) = x$ is called the identity function because it identifies every number with itself. Similarly, $0$ is called an additive identity because adding $0$ to a number does not change the number, so $0+x$ is identified with $x\text{.}$ By the same reasoning, $1$ is called the multiplicative identity.

###### Problem2.35.

Let $\mathbf S$ denote the set of all bijections from $\A$ to itself. According to Problems 2.33 and Problem 2.34, $\circ$ is a binary operation on $\mathbf S\text{,}$ that is, $f\circ g \in \mathbf S$ whenever $f$ and $g$ are in $\mathbf S\text{.}$ We define a special bijection $i:\A \to \A$ called the identity function on $\A$ as

\begin{equation*} i(x) = x \mbox{ for all } x\in \A\text{.} \end{equation*}

Show that $i$ is an identity for $\mathbf S\text{,}$ that is,

1. $f\circ i = f = i\circ f$ and

2. $\displaystyle f\circ f^{-1} = i = f^{-1}\circ f$

for all $f \in \mathbf S\text{.}$ This shows that $i$ serves as an identity for composition in the same way that 0, 1, $\emptyset$ work as identities for $+\text{,}$ × and $\oplus\text{,}$ respectively.

###### Problem2.36.

Consider the list below of all 6 different bijections from $\{0,1,2\}$ onto itself. For example $p$ maps $0 \to 1\text{,}$ $1 \to 2$ and $2 \to 0\text{.}$

Make a composition table for these 6 bijections, illustrating Problems 2.33, Problem 2.34 and Problem 2.35 above.

###### Problem2.38.

A function $f:\B \to \C$ is said to be left cancelative if, for all sets $\A$ and all functions $g,h:\A \to \B\text{,}$

\begin{equation*} f \circ g = f \circ h \ \mbox{ implies } \ \ g = h\text{.} \end{equation*}

Show that $f$ is left cancelative if and only if $f$ is one-to-one.

###### Problem2.39.

A function $f:\B \to \C$ is said to be right cancelative if, for all sets $\D$ and all functions $g,h:\C \to \D\text{,}$

\begin{equation*} g \circ f = h \circ f, \ \ \mbox{ implies } \ \ g = h\text{.} \end{equation*}

Show that $f$ is right cancelative if and only if $f$ is onto $\C\text{.}$