5.3: One-to-One Functions

We distinguish two special families of functions: one-to-one functions and onto functions. We shall discuss one-to-one functions in this section. Onto functions were introduced in section 5.2 and will be developed more in section 5.4.

One-to-One (Injective)

Recall that under a function each value in the domain has a unique image in the range.  For a one-to-one function, we add the requirement that each image in the range has a unique pre-image in the domain. 

Definition: One-to-One (Injection)

A function ${f}:{A}\to{B}$ is said to be one-to-one if

$$f(x_1) = f(x_2) \Rightarrow x_1=x_2$$

for all elements $x_1,x_2\in A$. A one-to-one function is also called an injection, and we call a function injective if it is one-to-one. A function that is not one-to-one is referred to as many-to-one.

The contrapositive of this definition is: A function ${f}:{A}\to{B}$ is  one-to-one  if $$x_1\neq x_2 \Rightarrow f(x_1)\neq f(x_2)$$

Any function is either one-to-one or many-to-one. A function cannot be one-to-many because no element can have multiple images. The difference between one-to-one and many-to-one functions is whether there exist distinct elements that share the same image. There are no repeated images in a one-to-one function.

 

Definition: Identity Function

The identity function on any nonempty set $A$ $${I_A}:{A}\to{A}, \qquad I_A(x)=x,$$ maps any element back to itself.

 It is clear that all identity functions are one-to-one.

Example $\PageIndex{1}\label{eg:oneonefcn-01}$

The function $h : {A}\to{A}$ defined by $h(x)=c$ for some fixed element $c\in A$, is an example of a constant function. It is a function with only one image. This is the exact opposite of an identity function. It is clearly not  one-to-one unless $|A|=1$.

For domains with a small number of elements, one can use inspection on the images to determine if the function is one-to-one. This becomes impossible if the domain contains a larger number of elements.

In practice, it is easier to use the contrapositive of the definition to test whether a function is one-to-one:

$$f(x_1) = f(x_2) \Rightarrow x_1 = x_2$$

To prove a function is One-to-One

To prove $f:A \rightarrow B$ is one-to-one:

• Assume $f(x_1)=f(x_2)$ 
• Show it must be true that $x_1=x_2$
• Conclude: we have shown if $f(x_1) = f(x_2)$ then $x_1 = x_2$, therefore $f$ is one-to-one, by definition of one-to-one.

Example $\PageIndex{2}$

Prove the function $f : {\mathbb{R}}\to{\mathbb{R}}$ defined by $f(x)=3x+2$ is one-to-one.

Solution

Assume $f(x_1)=f(x_2)$, which means $3x_1+2 = 3x_2+2.$
Thus $3x_1=3x_2$
so  $x_1=x_2$.
We have shown if $f(x_1)=f(x_2)$ then $x_1=x_2$. Therefore $f$ is one-to-one, by definition of one-to-one.

Hands-on exercise $\PageIndex{1}\label{he:oneonefcn-01}$

Prove the function $g : {\mathbb{R}}\to{\mathbb{R}}$ defined by $g(x)=5-7x$ is one-to-one.

Hands-on exercise $\PageIndex{2}\label{he:oneonefcn-02}$

Determine whether the function $h : {[2,\infty)}\to{\mathbb{R}}$ defined by $h(x)=\sqrt{x-2}$ is one-to-one.

Interestingly, sometimes we can use calculus to determine if a real function is one-to-one. A real function $f$ is increasing if $$x_1 < x_2 \Rightarrow f(x_1) < f(x_2),$$ and decreasing if $$x_1 < x_2 \Rightarrow f(x_1) > f(x_2).$$ Obviously, both increasing and decreasing functions are one-to-one. From calculus, we know that

• A function is increasing over an open interval $(a,b)$ if $f'(x)>0$ for all $x\in(a,b)$.
• A function is decreasing over an open interval $(a,b)$ if $f'(x)<0$ for all $x\in(a,b)$.

Therefore, if the derivative of a function is always positive, or always negative, then the function must be one-to-one.

Example $\PageIndex{4}\label{eg:oneonefcn-04}$

The function $p : {\mathbb{R}}\to{\mathbb{R}}$ defined by $$p(x) = 2x^3-5$$ is one-to-one, because $p'(x)=6x^2>0$ for any $x\in\mathbb{R}^*$. Likewise, the function $q:{\big(-\frac{\pi}{2},\frac{\pi}{2}\big)}\to{\mathbb{R}}$ defined by $$q(x) = \tan x$$ is also one-to-one, because $q'(x) = \sec^2x > 0$ for any $x\in \big(-\frac{\pi}{2},\frac{\pi}{2}\big)$.

Hands-on exercise $\PageIndex{3}\label{he:oneonefcn-03}$

Use both methods to show that the function $k:{(0,\infty)}\to{\mathbb{R}}$ defined by $k(x) = \ln x$ is one-to-one.

To prove a function is NOT one-to-one

To prove $f:A \rightarrow B$ is NOT one-to-one:

• Exhibit one case (a counterexample) where  $x_1\neq x_2$ and $f(x_1)=f(x_2).$
• Conclude: we have shown there is a case where  $x_1\neq x_2$ and $f(x_1)=f(x_2)$, therefore $f$ is NOT one-to-one.

Example $\PageIndex{5}\label{eg:oneonefcn-05}$

Prove the function $h : {\mathbb{R}}\to{\mathbb{R}}$ given by $h(x)=x^2$ is not one-to-one.
 

Solution

Consider $a=3$ and $b=-3$. Clearly $a \neq b$.  However, $h(3)=9$ and $h(-3)=9$ so $h(a)=h(b).$
we have shown there is a case where  $a \neq b$ and $h(a)=h(b)$, therefore $h$ is NOT one-to-one.

Example $\PageIndex{6}\label{eg:oneonefcn-06}$

The function $f : {\mathbb{Z}}\to{\mathbb{Z}}$ defined by $$f(n) = \cases{ \frac{n}{2} & if $n$ is even \cr \frac{n+1}{2} & if $n$ is odd \cr}$$ is not one-to-one, because, for example, $f(0)=f(-1)=0$. The function $g : {\mathbb{Z}}\to{\mathbb{Z}}$ defined by $$g(n) = 2n$$ is one-to-one, because if $g(n_1)=g(n_2)$, then $2n_1=2n_2$ implies that $n_1=n_2$.

hands-on exercise $\PageIndex{4}\label{he:oneonefcn-04}$

Show that the function $h : {\mathbb{Z}}\to{\mathbb{N}}$ defined by $$h(n) = \cases{ 2n+1 & if $n\geq0$, \cr -2n & if $n < 0$, \cr}$$ is one-to-one.

Example $\PageIndex{7}\label{eg:oneonefcn-7}$

Let $A$ be the set of all married individuals from a monogamous community who are neither divorced nor widowed. Then the function $s:{A}\to{A}$ defined by $$s(x) = \mbox{ spouse of } x$$ is one-to-one. The reason is, it is impossible to have $x_1\neq x_2$ and yet $s(x_1)=s(x_2)$.

Summary and Review

• A function $f$ is said to be one-to-one if $f(x_1) = f(x_2) \Rightarrow x_1=x_2$.
• No two images of a one-to-one function are the same.
• Know how to write a proof to show a function is one-to-one.
• To show that a function $f$ is not one-to-one, all we need is to find two different $x$-values that produce the same image; that is, find $x_1\neq x_2$ such that $f(x_1)=f(x_2)$.

Exercises 

Exercise $\PageIndex{1}\label{ex:oneonefcn-01}$

Which of the following functions are one-to-one? Explain.

(a) $f : {\mathbb{R}}\to{\mathbb{R}}$, $f(x)=x^3-2x^2+1$.

(b) $g : {[\,2,\infty)}\to{\mathbb{R}}$, $f(x)=x^3-2x^2+1$.
 

Solution

(a) No. For example, $f(0)=f(2)=1$
(b) Yes, since $g'(x)=3x^2-4x=x(3x-4)>0$ for $x>2$.

Exercise $\PageIndex{2}\label{ex:oneonefcn-02}$

Decide if this function is one-to-one or not.  Then prove your conclusion.

$p :{\mathbb{R}}\to{\mathbb{R}}$, $p(x)=|1-3x|$.

Exercise $\PageIndex{3}\label{ex:oneonefcn-03}$

Decide if this function is one-to-one or not.  Then prove your conclusion.

$q :{\mathbb{R}}\to{\mathbb{R}}$, $q(x)=x^4$.
 

Solution

 No. For example, $2 \neq -2$, but $q(2)=16$ and  $q(-2)=16$. 
We have shown a case where $x_1 \neq x_2$ and $q(x_1)=q(x_2)$, so $q$ is NOT one-to-one.

Exercise $\PageIndex{4}\label{ex:oneonefcn-04}$

Decide if this function is one-to-one or not.  Then prove your conclusion.

$f :{\mathbb{R}}\to{\mathbb{R}}$, $f(x)=6-5x$.

Exercise $\PageIndex{5}\label{ex:oneonefcn-05}$

Determine which of the following are one-to-one functions.

1. $f : {\mathbb{Z}}\to{\mathbb{Z}}$; $f(n)=n^3+1$
2. $g : {\mathbb{Q}}\to{\mathbb{Q}}$; $g(x)=n^2$
3. ${k} : {\mathbb{R}}\to{\mathbb{R}}$; $k(x)=5^x$

Solution

(a) One-to-one (b) Not one-to-one (c) One-to-one

Exercise $\PageIndex{6}\label{ex:oneonefcn-06}$

Determine which of the following are one-to-one functions and explain your answer.

1. $p :{\mathscr{P}(\{1,2,3,\ldots,n\})}\to{\{0,1,2,\ldots,n\}}$; $p(S)=|S|$
2. $q :{\mathscr{P}(\{1,2,3,\ldots,n\})}\to{\mathscr{P}(\{1,2,3,\ldots,n\})}$; $q(S)=\overline{S}$

Exercise $\PageIndex{7}\label{ex:oneonefcn-07}$

Determine which of the following functions are one-to-one.

1. ${f_1}:{\{1,2,3,4,5\}}\to{\{a,b,c,d\}}$; $f_1(1)=b$, $f_1(2)=c$, $f_1(3)=a$, $f_1(4)=a$, $f_1(5)=c$
2. ${f_2}:{\{1,2,3,4\}}\to{\{a,b,c,d,e\}}$; $f_2(1)=c$, $f_2(2)=b$, $f_2(3)=a$, $f_2(4)=d$

Solution

(a) Not one-to-one (b) One-to-one

Exercise $\PageIndex{8}\label{ex:oneonefcn-08}$

Determine which of the following functions are one-to-one.

1. ${g_1}:{\{1,2,3,4,5\}}\to{\{a,b,c,d,e\}}$; $g_1(1)=b$, $g_1(2)=b$, $g_1(3)=b$, $g_1(4)=a$, $g_1(5)=d$
2. ${g_2}:{\{1,2,3,4,5\}}\to{\{a,b,c,d,e\}}$; $g_2(1)=d$, $g_2(2)=b$, $g_2(3)=e$, $g_2(4)=a$, $g_2(5)=c$

Exercise $\PageIndex{9}\label{ex:oneonefcn-09}$

List all the one-to-one functions from $\{1,2\}$ to $\{a,b,c,d\}$.

Hint

List the images of each function.
 

Solution

There are twelve one-to-one functions from $\{1,2\}$ to $\{a,b,c,d\}$. The images of 1 and 2 under them are listed below. $$\begin{array}{|c||*{12}{c|}} \hline & f_1 & f_2 & f_3 & f_4 & f_5 & f_6 & f_7 & f_8 & f_9 & f_{10} & f_{11} & f_{12} \\ \hline\hline 1 & a & a & a & b & b & b & c & c & c & d & d & d \\ \hline 2 & b & c & d & a & c & d & a & b & d & a & b & c \\ \hline \end{array}$$

Exercise $\PageIndex{10}\label{ex:oneonefcn-10}$

Is it possible to find a one-to-one function from $\{1,2,3,4\}$ to $\{1,2\}$? Explain.

Exercise $\PageIndex{11}\label{ex:oneonefcn-11}$

Determine which of the following functions are one-to-one.

1. $f : {\mathbb{Z}_{10}}\to{\mathbb{Z}_{10}}$; $h(n)\equiv 3n$ (mod 10).
2. $g : {\mathbb{Z}_{10}}\to{\mathbb{Z}_{10}}$; $g(n)\equiv 5n$ (mod 10).
3. $h : {\mathbb{Z}_{36}}\to{\mathbb{Z}_{36}}$; $h(n)\equiv 3n$ (mod 36).

Solution

(a) One-to-one (b) Not one-to-one (c) Not one-to-one

Exercise $\PageIndex{12}\label{ex:oneonefcn-12}$

Decide if this function is one-to-one or not.  Then prove your conclusion.

 $k : {\mathbb{R}}\to{\mathbb{R}}$ defined by $k(x)=3x^2-5$ is one-to-one.

Exercise $\PageIndex{13}\label{ex:oneonefcn-13}$

Decide if this function is one-to-one or not.  Then prove your conclusion.

$f: {\mathbb{R}}\to{\mathbb{R}}$ defined by $f(x)=57$ is one-to-one.

Solution

 No. For example, $8 \neq 17$, but $f(8)=57$ and  $f(17)=57$. 
We have shown a case where $x_1 \neq x_2$ and $f(x_1)=f(x_2)$, so $f$ is NOT one-to-one.

Exercise $\PageIndex{14}\label{ex:oneonefcn-14}$

Give an example of a one-to-one function $f$ from $\mathbb{N}$ to $\mathbb{N}$ that is not the identity function.