## Section1.2Sets and Functions

### Subsection1.2.1Set Theory

A set is a well-defined collection of objects; that is, it is defined in such a manner that we can determine for any given object $x$ whether or not $x$ belongs to the set. The objects that belong to a set are called its elements or members. We will denote sets by capital letters, such as $A$ or $X\text{;}$ if $a$ is an element of the set $A\text{,}$ we write $a \in A\text{.}$

A set is usually specified either by listing all of its elements inside a pair of braces or by stating the property that determines whether or not an object $x$ belongs to the set. We might write

\begin{equation*} X = \{ x_1, x_2, \ldots, x_n \} \end{equation*}

for a set containing elements $x_1, x_2, \ldots, x_n$ or

\begin{equation*} X = \{ x :x \text{ satisfies }{\mathcal P}\} \end{equation*}

if each $x$ in $X$ satisfies a certain property ${\mathcal P}\text{.}$ For example, if $E$ is the set of even positive integers, we can describe $E$ by writing either

\begin{equation*} E = \{2, 4, 6, \ldots \} \quad \text{or} \quad E = \{ x : x \text{ is an even integer and } x \gt 0 \}\text{.} \end{equation*}

We write $2 \in E$ when we want to say that 2 is in the set $E\text{,}$ and $-3 \notin E$ to say that $-3$ is not in the set $E\text{.}$

Some of the more important sets that we will consider are the following:

\begin{gather*} {\mathbb N} = \{n: n \text{ is a natural number}\} = \{1, 2, 3, \ldots \};\\ {\mathbb Z} = \{n : n \text{ is an integer} \} = \{\ldots, -1, 0, 1, 2, \ldots \};\\ {\mathbb Q} = \{r : r \text{ is a rational number}\} = \{p/q : p, q \in {\mathbb Z} \text{ where } q \neq 0\};\\ {\mathbb R} = \{ x : x \text{ is a real number} \};\\ {\mathbb C} = \{z : z \text{ is a complex number}\}\text{.} \end{gather*}

We can find various relations between sets as well as perform operations on sets. A set $A$ is a subset of $B\text{,}$ written $A \subset B$ or $B \supset A\text{,}$ if every element of $A$ is also an element of $B\text{.}$ For example,

\begin{equation*} \{4,5,8\} \subset \{2, 3, 4, 5, 6, 7, 8, 9 \} \end{equation*}

and

\begin{equation*} {\mathbb N} \subset {\mathbb Z} \subset {\mathbb Q} \subset {\mathbb R} \subset {\mathbb C}\text{.} \end{equation*}

Trivially, every set is a subset of itself. A set $B$ is a proper subset of a set $A$ if $B \subset A$ but $B \neq A\text{.}$ If $A$ is not a subset of $B\text{,}$ we write $A \notsubset B\text{;}$ for example, $\{4, 7, 9\} \notsubset \{2, 4, 5, 8, 9 \}\text{.}$ Two sets are equal, written $A = B\text{,}$ if we can show that $A \subset B$ and $B \subset A\text{.}$

It is convenient to have a set with no elements in it. This set is called the empty set and is denoted by $\emptyset\text{.}$ Note that the empty set is a subset of every set.

To construct new sets out of old sets, we can perform certain operations: the union $A \cup B$ of two sets $A$ and $B$ is defined as

\begin{equation*} A \cup B = \{x : x \in A \text{ or } x \in B \}; \end{equation*}

the intersection of $A$ and $B$ is defined by

\begin{equation*} A \cap B = \{x : x \in A \text{ and } x \in B \}\text{.} \end{equation*}

If $A = \{1, 3, 5\}$ and $B = \{ 1, 2, 3, 9 \}\text{,}$ then

\begin{equation*} A \cup B = \{1, 2, 3, 5, 9 \} \quad \text{and} \quad A \cap B = \{ 1, 3 \}\text{.} \end{equation*}

We can consider the union and the intersection of more than two sets. In this case we write

\begin{equation*} \bigcup_{i = 1}^{n} A_{i} = A_{1} \cup \ldots \cup A_n \end{equation*}

and

\begin{equation*} \bigcap_{i = 1}^{n} A_{i} = A_{1} \cap \ldots \cap A_n \end{equation*}

for the union and intersection, respectively, of the sets $A_1, \ldots, A_n\text{.}$

When two sets have no elements in common, they are said to be disjoint; for example, if $E$ is the set of even integers and $O$ is the set of odd integers, then $E$ and $O$ are disjoint. Two sets $A$ and $B$ are disjoint exactly when $A \cap B = \emptyset\text{.}$

Sometimes we will work within one fixed set $U\text{,}$ called the universal set. For any set $A \subset U\text{,}$ we define the complement of $A\text{,}$ denoted by $A'\text{,}$ to be the set

\begin{equation*} A' = \{ x : x \in U \text{ and } x \notin A \}\text{.} \end{equation*}

We define the difference of two sets $A$ and $B$ to be

\begin{equation*} A \setminus B = A \cap B' = \{ x : x \in A \text{ and } x \notin B \}\text{.} \end{equation*}
###### Example1.1.

Let ${\mathbb R}$ be the universal set and suppose that

\begin{equation*} A = \{ x \in {\mathbb R} : 0 \lt x \leq 3 \} \quad \text{and} \quad B = \{ x \in {\mathbb R} : 2 \leq x \lt 4 \}\text{.} \end{equation*}

Then

\begin{align*} A \cap B & = \{ x \in {\mathbb R} : 2 \leq x \leq 3 \}\\ A \cup B & = \{ x \in {\mathbb R} : 0 \lt x \lt 4 \}\\ A \setminus B & = \{ x \in {\mathbb R} : 0 \lt x \lt 2 \}\\ A' & = \{ x \in {\mathbb R} : x \leq 0 \text{ or } x \gt 3 \}\text{.} \end{align*}
###### Proof.

We will prove (1) and (3) and leave the remaining results to be proven in the exercises.

(1) Observe that

\begin{align*} A \cup A & = \{ x : x \in A \text{ or } x \in A \}\\ & = \{ x : x \in A \}\\ & = A \end{align*}

and

\begin{align*} A \cap A & = \{ x : x \in A \text{ and } x \in A \}\\ & = \{ x : x \in A \}\\ & = A\text{.} \end{align*}

Also, $A \setminus A = A \cap A' = \emptyset\text{.}$

(3) For sets $A\text{,}$ $B\text{,}$ and $C\text{,}$

\begin{align*} A \cup (B \cup C) & = A \cup \{ x : x \in B \text{ or } x \in C \}\\ & = \{ x : x \in A \text{ or } x \in B, \text{ or } x \in C \}\\ & = \{ x : x \in A \text{ or } x \in B \} \cup C\\ & = (A \cup B) \cup C\text{.} \end{align*}

A similar argument proves that $A \cap (B \cap C) = (A \cap B) \cap C\text{.}$

###### Proof.

(1) If $A \cup B = \emptyset\text{,}$ then the theorem follows immediately since both $A$ and $B$ are the empty set. Otherwise, we must show that $(A \cup B)' \subset A' \cap B'$ and $(A \cup B)' \supset A' \cap B'\text{.}$ Let $x \in (A \cup B)'\text{.}$ Then $x \notin A \cup B\text{.}$ So $x$ is neither in $A$ nor in $B\text{,}$ by the definition of the union of sets. By the definition of the complement, $x \in A'$ and $x \in B'\text{.}$ Therefore, $x \in A' \cap B'$ and we have $(A \cup B)' \subset A' \cap B'\text{.}$

To show the reverse inclusion, suppose that $x \in A' \cap B'\text{.}$ Then $x \in A'$ and $x \in B'\text{,}$ and so $x \notin A$ and $x \notin B\text{.}$ Thus $x \notin A \cup B$ and so $x \in (A \cup B)'\text{.}$ Hence, $(A \cup B)' \supset A' \cap B'$ and so $(A \cup B)' = A' \cap B'\text{.}$

The proof of (2) is left as an exercise.

###### Example1.4.

Other relations between sets often hold true. For example,

\begin{equation*} ( A \setminus B) \cap (B \setminus A) = \emptyset\text{.} \end{equation*}

To see that this is true, observe that

\begin{align*} ( A \setminus B) \cap (B \setminus A) & = ( A \cap B') \cap (B \cap A')\\ & = A \cap A' \cap B \cap B'\\ & = \emptyset\text{.} \end{align*}

### Subsection1.2.2Cartesian Products and Mappings

Given sets $A$ and $B\text{,}$ we can define a new set $A \times B\text{,}$ called the Cartesian product of $A$ and $B\text{,}$ as a set of ordered pairs. That is,

\begin{equation*} A \times B = \{ (a,b) : a \in A \text{ and } b \in B \}\text{.} \end{equation*}
###### Example1.5.

If $A = \{ x, y \}\text{,}$ $B = \{ 1, 2, 3 \}\text{,}$ and $C = \emptyset\text{,}$ then $A \times B$ is the set

\begin{equation*} \{ (x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3) \} \end{equation*}

and

\begin{equation*} A \times C = \emptyset\text{.} \end{equation*}

We define the Cartesian product of $n$ sets to be

\begin{equation*} A_1 \times \cdots \times A_n = \{ (a_1, \ldots, a_n): a_i \in A_i \text{ for } i = 1, \ldots, n \}\text{.} \end{equation*}

If $A = A_1 = A_2 = \cdots = A_n\text{,}$ we often write $A^n$ for $A \times \cdots \times A$ (where $A$ would be written $n$ times). For example, the set ${\mathbb R}^3$ consists of all of 3-tuples of real numbers.

Subsets of $A \times B$ are called relations. We will define a mapping or function $f \subset A \times B$ from a set $A$ to a set $B$ to be the special type of relation where each element $a \in A$ has a unique element $b \in B$ such that $(a, b) \in f\text{.}$ Another way of saying this is that for every element in $A\text{,}$ $f$ assigns a unique element in $B\text{.}$ We usually write $f:A \rightarrow B$ or $A \stackrel{f}{\rightarrow} B\text{.}$ Instead of writing down ordered pairs $(a,b) \in A \times B\text{,}$ we write $f(a) = b$ or $f : a \mapsto b\text{.}$ The set $A$ is called the domain of $f$ and

\begin{equation*} f(A) = \{ f(a) : a \in A \} \subset B \end{equation*}

is called the range or image of $f\text{.}$ We can think of the elements in the function's domain as input values and the elements in the function's range as output values.

###### Example1.6.

Suppose $A = \{1, 2, 3 \}$ and $B = \{a, b, c \}\text{.}$ In Figure 1.7 we define relations $f$ and $g$ from $A$ to $B\text{.}$ The relation $f$ is a mapping, but $g$ is not because $1 \in A$ is not assigned to a unique element in $B\text{;}$ that is, $g(1) = a$ and $g(1) = b\text{.}$

Given a function $f : A \rightarrow B\text{,}$ it is often possible to write a list describing what the function does to each specific element in the domain. However, not all functions can be described in this manner. For example, the function $f: {\mathbb R} \rightarrow {\mathbb R}$ that sends each real number to its cube is a mapping that must be described by writing $f(x) = x^3$ or $f:x \mapsto x^3\text{.}$

Consider the relation $f : {\mathbb Q} \rightarrow {\mathbb Z}$ given by $f(p/q) = p\text{.}$ We know that $1/2 = 2/4\text{,}$ but is $f(1/2) = 1$ or $2\text{?}$ This relation cannot be a mapping because it is not well-defined. A relation is well-defined if each element in the domain is assigned to a unique element in the range.

If $f:A \rightarrow B$ is a map and the image of $f$ is $B\text{,}$ i.e., $f(A) = B\text{,}$ then $f$ is said to be onto or surjective. In other words, if there exists an $a \in A$ for each $b \in B$ such that $f(a) = b\text{,}$ then $f$ is onto. A map is one-to-one or injective if $a_1 \neq a_2$ implies $f(a_1) \neq f(a_2)\text{.}$ Equivalently, a function is one-to-one if $f(a_1) = f(a_2)$ implies $a_1 = a_2\text{.}$ A map that is both one-to-one and onto is called bijective.

###### Example1.8.

Let $f:{\mathbb Z} \rightarrow {\mathbb Q}$ be defined by $f(n) = n/1\text{.}$ Then $f$ is one-to-one but not onto. Define $g : {\mathbb Q} \rightarrow {\mathbb Z}$ by $g(p/q) = p$ where $p/q$ is a rational number expressed in its lowest terms with a positive denominator. The function $g$ is onto but not one-to-one.

Given two functions, we can construct a new function by using the range of the first function as the domain of the second function. Let $f : A \rightarrow B$ and $g : B \rightarrow C$ be mappings. Define a new map, the composition of $f$ and $g$ from $A$ to $C\text{,}$ by $(g \circ f)(x) = g(f(x))\text{.}$

###### Example1.10.

Consider the functions $f: A \rightarrow B$ and $g: B \rightarrow C$ that are defined in Figure 1.9 (top). The composition of these functions, $g \circ f: A \rightarrow C\text{,}$ is defined in Figure 1.9 (bottom).

###### Example1.11.

Let $f(x) = x^2$ and $g(x) = 2x + 5\text{.}$ Then

\begin{equation*} (f \circ g)(x) = f(g(x)) = (2x + 5)^2 = 4x^2 + 20x + 25 \end{equation*}

and

\begin{equation*} (g \circ f)(x) = g(f(x)) = 2x^2 + 5\text{.} \end{equation*}

In general, order makes a difference; that is, in most cases $f \circ g \neq g \circ f\text{.}$

###### Example1.12.

Sometimes it is the case that $f \circ g= g \circ f\text{.}$ Let $f(x) = x^3$ and $g(x) = \sqrt[3]{x}\text{.}$ Then

\begin{equation*} (f \circ g )(x) = f(g(x)) = f( \sqrt[3]{x}\, ) = (\sqrt[3]{x}\, )^3 = x \end{equation*}

and

\begin{equation*} (g \circ f )(x) = g(f(x)) = g( x^3) = \sqrt[3]{ x^3} = x\text{.} \end{equation*}
###### Example1.13.

Given a $2 \times 2$ matrix

\begin{equation*} A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\text{,} \end{equation*}

we can define a map $T_A : {\mathbb R}^2 \rightarrow {\mathbb R}^2$ by

\begin{equation*} T_A (x,y) = (ax + by, cx +dy) \end{equation*}

for $(x,y)$ in ${\mathbb R}^2\text{.}$ This is actually matrix multiplication; that is,

\begin{equation*} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx +dy \end{pmatrix}\text{.} \end{equation*}

Maps from ${\mathbb R}^n$ to ${\mathbb R}^m$ given by matrices are called linear maps or linear transformations.

###### Example1.14.

Suppose that $S = \{ 1,2,3 \}\text{.}$ Define a map $\pi :S\rightarrow S$ by

\begin{equation*} \pi( 1 ) = 2, \qquad \pi( 2 ) = 1, \qquad \pi( 3 ) = 3\text{.} \end{equation*}

This is a bijective map. An alternative way to write $\pi$ is

\begin{equation*} \begin{pmatrix} 1 & 2 & 3 \\ \pi(1) & \pi(2) & \pi(3) \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix}\text{.} \end{equation*}

For any set $S\text{,}$ a one-to-one and onto mapping $\pi : S \rightarrow S$ is called a permutation of $S\text{.}$

###### Proof.

We will prove (1) and (3). Part (2) is left as an exercise. Part (4) follows directly from (2) and (3).

(1) We must show that

\begin{equation*} h \circ (g \circ f) = (h \circ g) \circ f\text{.} \end{equation*}

For $a \in A$ we have

\begin{align*} (h \circ (g \circ f))(a) & = h((g \circ f)(a))\\ & = h(g(f(a)))\\ & = (h \circ g)(f(a))\\ & = ((h \circ g) \circ f)(a)\text{.} \end{align*}

(3) Assume that $f$ and $g$ are both onto functions. Given $c \in C\text{,}$ we must show that there exists an $a \in A$ such that $(g \circ f)(a) = g(f(a)) = c\text{.}$ However, since $g$ is onto, there is an element $b \in B$ such that $g(b) = c\text{.}$ Similarly, there is an $a \in A$ such that $f(a) = b\text{.}$ Accordingly,

\begin{equation*} (g \circ f)(a) = g(f(a)) = g(b) = c\text{.} \end{equation*}

If $S$ is any set, we will use $id_S$ or $id$ to denote the identity mapping from $S$ to itself. Define this map by $id(s) = s$ for all $s \in S\text{.}$ A map $g: B \rightarrow A$ is an inverse mapping of $f: A \rightarrow B$ if $g \circ f = id_A$ and $f \circ g = id_B\text{;}$ in other words, the inverse function of a function simply “undoes” the function. A map is said to be invertible if it has an inverse. We usually write $f^{-1}$ for the inverse of $f\text{.}$

###### Example1.16.

The function $f(x) = x^3$ has inverse $f^{-1}(x) = \sqrt[3]{x}$ by Example 1.12.

###### Example1.17.

The natural logarithm and the exponential functions, $f(x) = \ln x$ and $f^{-1}(x) = e^x\text{,}$ are inverses of each other provided that we are careful about choosing domains. Observe that

\begin{equation*} f(f^{-1}(x)) = f(e^x) = \ln e^x = x \end{equation*}

and

\begin{equation*} f^{-1}(f(x)) = f^{-1}(\ln x) = e^{\ln x} = x \end{equation*}

whenever composition makes sense.

###### Example1.18.

Suppose that

\begin{equation*} A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\text{.} \end{equation*}

Then $A$ defines a map from ${\mathbb R}^2$ to ${\mathbb R}^2$ by

\begin{equation*} T_A (x,y) = (3x + y, 5x + 2y)\text{.} \end{equation*}

We can find an inverse map of $T_A$ by simply inverting the matrix $A\text{;}$ that is, $T_A^{-1} = T_{A^{-1}}\text{.}$ In this example,

\begin{equation*} A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}; \end{equation*}

hence, the inverse map is given by

\begin{equation*} T_A^{-1} (x,y) = (2x - y, -5x + 3y)\text{.} \end{equation*}

It is easy to check that

\begin{equation*} T^{-1}_A \circ T_A (x,y) = T_A \circ T_A^{-1} (x,y) = (x,y)\text{.} \end{equation*}

Not every map has an inverse. If we consider the map

\begin{equation*} T_B (x,y) = (3x , 0 ) \end{equation*}

given by the matrix

\begin{equation*} B = \begin{pmatrix} 3 & 0 \\ 0 & 0 \end{pmatrix}\text{,} \end{equation*}

then an inverse map would have to be of the form

\begin{equation*} T_B^{-1} (x,y) = (ax + by, cx +dy) \end{equation*}

and

\begin{equation*} (x,y) = T_B \circ T_B^{-1} (x,y) = (3ax + 3by, 0) \end{equation*}

for all $x$ and $y\text{.}$ Clearly this is impossible because $y$ might not be $0\text{.}$

###### Example1.19.

Given the permutation

\begin{equation*} \pi = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} \end{equation*}

on $S = \{ 1,2,3 \}\text{,}$ it is easy to see that the permutation defined by

\begin{equation*} \pi^{-1} = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} \end{equation*}

is the inverse of $\pi\text{.}$ In fact, any bijective mapping possesses an inverse, as we will see in the next theorem.

###### Proof.

Suppose first that $f:A \rightarrow B$ is invertible with inverse $g: B \rightarrow A\text{.}$ Then $g \circ f = id_A$ is the identity map; that is, $g(f(a)) = a\text{.}$ If $a_1, a_2 \in A$ with $f(a_1) = f(a_2)\text{,}$ then $a_1 = g(f(a_1)) = g(f(a_2)) = a_2\text{.}$ Consequently, $f$ is one-to-one. Now suppose that $b \in B\text{.}$ To show that $f$ is onto, it is necessary to find an $a \in A$ such that $f(a) = b\text{,}$ but $f(g(b)) = b$ with $g(b) \in A\text{.}$ Let $a = g(b)\text{.}$

Conversely, let $f$ be bijective and let $b \in B\text{.}$ Since $f$ is onto, there exists an $a \in A$ such that $f(a) = b\text{.}$ Because $f$ is one-to-one, $a$ must be unique. Define $g$ by letting $g(b) = a\text{.}$ We have now constructed the inverse of $f\text{.}$

###### 1.

Consider the sets $A = \{x \st x-3 \in \N\}$ and $B = \{x \st \frac{x}{2} \in \N\}\text{.}$ Find (describe) the sets $A \cap B$ and $A \cup B\text{.}$

Note that you can type such as $A \cap B = \{1,2,3\}$(although this is not the right answer). If you right-click on any math in this book and select “Show math as -> TeX commands”, you can see the commands that produce that math.

###### 2.

What does it mean for two sets to be disjoint? Describe in words and in symbols, and give an example and non-example.

###### 3.

What is the difference between a function being one-to-one (injective) and being onto (surjective)?

###### 4.

After reading the section, what questions do you still have? Write at least one well formulated question (even if you think you understand everything).

### Exercises1.2.4Practice Problems

###### 1.

Same or different: $\{2^n \in \N \st n \in \N\}$ and $\{n \in \N \st 2^n \in \N\}\text{.}$

Different. The first set only has elements that are powers of 2, while the second set is all natural numbers.

###### 2.

Same or different: $\{2^n \st n \in \N\}$ and $\{n \st n = 2^x \text{ for some } x \in \N\}\text{.}$

Same.

###### 3.

Suppose that

\begin{align*} A & = \{ x : x \in \mathbb N \text{ and } x \text{ is even} \},\\ B & = \{x : x \in \mathbb N \text{ and } x \text{ is prime}\},\\ C & = \{ x : x \in \mathbb N \text{ and } x \text{ is a multiple of } 5\}\text{.} \end{align*}

Describe each of the following sets.

1. $\displaystyle A \cap B$

2. $\displaystyle B \cap C$

3. $\displaystyle A \cup B$

4. $\displaystyle A \cap (B \cup C)$

Hint

(a) $A \cap B = \{ 2 \}\text{;}$ (b) $B \cap C = \{ 5 \}\text{.}$

###### 4.

I'm trying to invent a function $f: \{1,2,3\} \to \{1,2,3,4\}\text{.}$ So far, I've decided $f(1) = 4$ and $f(2) = 1\text{.}$ What should $f(3)$ be ...

1. ... if I want $f$ to be one-to-one?

2. ... if I want $f$ to be onto?

To be one to one, I could take $f(3) = 2$ or $f(3) = 3\text{,}$ but the other two options would not work. There is no choice I can make to make $f$ onto.

###### 5.

I'm trying to invent a function $f: \{1,2,3\} \to \{1,2\}\text{.}$ So far, I've decided $f(1) = 2\text{.}$ What should $f(2)$ and $f(3)$ be ...

1. ... if I want $f$ to be one-to-one?

2. ... if I want $f$ to be onto?

There is no way to make $f$ one-to-one. To make it onto, I just need one of $f(2)$ or $f(3)$ to be 1, so every element in the codomain is the image of at least one element of the domain.

###### 6.

Consider the function $f:\Z \to \Z$ given by $f(n) = \begin{cases}n+1 \amp \text{ if }n\text{ is even} \\ n-3 \amp \text{ if }n\text{ is odd} . \end{cases}$

1. Is $f$ injective? Prove your answer.

2. Is $f$ surjective? Prove your answer.

### Exercises1.2.5Collected Homework

###### 1.

Consider the sets below.

\begin{equation*} A = \{2n \st n \in \Z\} \qquad \text{ and } \qquad B = \{x \in \Z \st x^2 \in A\}\text{.} \end{equation*}
1. Describe the sets $A$ and $B$ in words or by listing enough elements to see what is going on.

2. If $a \in A$ and $b \in A\text{,}$ must $a+b \in A\text{?}$ Explain.

3. If $a \in B$ and $b \in B\text{,}$ must $a+b \in B\text{?}$ Explain.

###### 2.

Let $f : A \rightarrow B$ and $g : B \rightarrow C$ be maps.

1. If $f$ and $g$ are both one-to-one functions, show that $g \circ f$ is one-to-one.

2. If $g \circ f$ is onto, show that $g$ is onto.

Hint

To show a function $h$ is one-to-one, you show that if $h(x) = h(y)$ then $x=y\text{.}$

To show a function $h$ is onto, show that for every element $c$ in the codomain, there is some element $a$ in the domain such that $h(a) = c\text{.}$