## Section7.4Identity Functions

For every set $$A$$ there is a special function under which the image of each element is the element itself. Because the image of an element is identical to the element, it is called the identity function.

### Definition7.4.1.

For any set $$A\text{,}$$ the function $$\id_A:A\to A$$ given by $$\id_A(b)=b$$ for all $$b \in A$$ is the identity function on $$A\text{.}$$

In the video in Figure 7.4.2 we introduce identity functions and give examples.

The identity function on $$\Z_3=\{0,1,2\}$$ is the function $$\id_{\Z_3}:\Z_3\to\Z_3$$ given by:

\begin{align*} \id_{\Z_3}(0)\amp=0\\ \id_{\Z_3}(1)\amp=1\\ \id_{\Z_3}(2)\amp=2 \end{align*}

The behavior of the identity function with respect to composition is described in the following theorem.

Let $$A$$ and $$B$$ be sets, and let $$f:A\to B$$ be a function. Let $$a\in A\text{.}$$ Then,

\begin{equation*} (f\circ \id_A)(a)=f(\id_A(a))=f(a)\text{,} \end{equation*}

implying that $$f\circ \id_A=f\text{.}$$ Furthermore, $$\id_B(b)=b$$ for all $$b\in B\text{.}$$ So in particular,

\begin{equation*} (\id_B\circ f)(a)=\id_B(f(a))=f(a)\text{,} \end{equation*}

implying that $$\id_B\circ f=f\text{.}$$

Now complete the definition of the identity function on a set and the statement of the theorem.

Complete the following.

Let A be a nonempty set. The identity function on A is the function idA: A->A given by idA(x) =

• 0

• 1

• -1

• x

• -x

• 1/x

• -1/x

.

Let B be a set and f:A->B and g:B->A be functions. Then

(f $$\circ$$ idA)(x) =

• 0

• 1

• -1

• x

• -x

• f(x)

• -f(x)

• g(x)

• 1/g(x)

• 1/x

• -1/x

and (idA $$\circ$$ g)(x) =
• 0

• 1

• -1

• x

• -x

• f(x)

• -f(x)

• g(x)

• 1/g(x)

• 1/x

• -1/x

.

$$\text{x}$$

$$\text{f(x)}$$

$$\text{g(x)}$$

To determine whether a function $$f$$ is the identity function we can evaluate it at all elements of its domain to see whether $$f(a)=a$$ for all element $$a$$ of its domain.

Decide whether the function $$f:\Z_5\to\Z_5$$ given by $$f(x)=(x^5)\fmod 5$$ is the identity function on $$\Z_5\text{.}$$

Solution.

We evaluate the function $$f$$ at all elements of its domain $$\Z_5=\{0,1,2,3,4\}\text{.}$$ If $$f(x)=x$$ for all $$x\in\Z_5$$ then $$f$$ is the identity function on $$\Z_5\text{.}$$

\begin{align*} f(0)\amp =(0^5)\fmod 5 = 0\fmod 5= 0\\ f(1)\amp =(1^5)\fmod 5 = 1\fmod 5= 1\\ f(2)\amp =(2^5)\fmod 5 = 32\fmod 5=2\\ f(3)\amp =(3^5)\fmod 5 = 243\fmod 3= 3\\ f(4)\amp =(4^5)\fmod 5 = 1024\fmod 5= 4 \end{align*}

As $$f(x)=x$$ for all $$x\in\Z_5$$ it is the identity function on $$\Z_5\text{.}$$

Decide whether the function $$f:\Z_4\to\Z_4$$ given by $$f(x)=(x^2)\fmod 4$$ is the identity function on $$\Z_4\text{.}$$

Solution.

We evaluate the function $$f$$ at all elements of its domain $$\Z_4=\{0,1,2,3\}\text{.}$$ If $$f(x)=x$$ for all $$x\in\Z_4$$ then $$f$$ is the identity function on $$\Z_4\text{.}$$

\begin{align*} f(0)\amp =(0^2)\fmod 4 = 0\fmod 4= 0\\ f(1)\amp =(1^2)\fmod 4 = 1\fmod 4= 1\\ f(2)\amp =(2^2)\fmod 4 = 4\fmod 4= 0 \end{align*}

We have found that $$f(2)=0\ne 2\text{.}$$ So $$f$$ is not the identity function on $$\Z_4\text{.}$$

In Checkpoint 7.4.8 follow the methods from the two problems.

Let $$f:\mathbb{Z}_{5}\to\mathbb{Z}_{5},\;$$ $$f(x) = \left(3x^{4}+x^{3}\right) \bmod 5\text{.}$$

Evaluate $$f$$ at all elements of the domain:

$$f(0) =$$

$$f(1) =$$

$$f(2) =$$

$$f(3) =$$

$$f(4) =$$

Now conclude whether $$f$$ is equal to the identity function on $${\mathbb{Z}_{5}}\text{.}$$

The function $$f$$

• select

• is the identity function

• is NOT the identity function

on $$\mathbb{Z}_{5}\text{.}$$

$$0$$

$$4$$

$$1$$

$$0$$

$$2$$

$$\text{is NOT the identity function}$$

In Example 7.4.9 we consider the composite of two functions and find that it is an identity function.

Consider the two functions

\begin{equation*} f:\Z_5\to \Z_5\text{ given by }f(x)=3\cdot x \end{equation*}

and

\begin{equation*} g:\Z_5\to \Z_5\text{ given by }g(x)=2\cdot x. \end{equation*}

Then $$(g\circ f)$$ is a function with domain $$\Z_5$$ and codomain $$\Z_5\text{.}$$ We show that $$(g\circ f)$$ is the identity function on $$\Z_5\text{.}$$

Since $$\Z_5=\{0,1,2,3,4\}$$ does not have too many elements we can do this by exhausting all elements of $$\Z_5\text{.}$$ We get

 $$x$$ $$f(x)$$ $$g(f(x))$$ $$\mathbf{0}$$ $$f(0)=(3\cdot 0)\fmod 5=0$$ $$g(f(0))=(2\cdot f(0))\fmod 5=\mathbf{0}$$ $$\mathbf{1}$$ $$f(1)=(3\cdot 1)\fmod 5 =3$$ $$g(f(1))=(2\cdot f(1))\fmod 5=(2\cdot 3)\mod 5 = \mathbf{1}$$ $$\mathbf{2}$$ $$f(2)=(3\cdot 2)\fmod 5 = 1$$ $$g(f(1))=(2\cdot f(2))\fmod 5= 2\fmod 5 = \mathbf{2}$$ $$\mathbf{3}$$ $$f(3)=(3\cdot 3)\fmod 5 =4$$ $$g(f(1))=(2\cdot f(3))\fmod 5=(2\cdot 9)\mod 5=\mathbf{3}$$ $$\mathbf{4}$$ $$f(4)=(3\cdot 4)\fmod 5 =2$$ $$g(f(4))=(2\cdot f(4))\fmod 5=(2\cdot 2)\mod 5=\mathbf{4}$$

We see that in each row of the table the columns for $$x$$ and for $$g(f(x))$$ have the same entries, that is $$g(f(x))=x$$ for all $$x\in\Z_5\text{.}$$ Thus $$(g\circ f)=\id_{\Z_5}\text{.}$$

Even the composite of a function with itself can be the identity function.

Consider the function

\begin{equation*} f:\Z\to \Z\text{ given by }f(x)=(-x). \end{equation*}

We show that $$(f\circ f)$$ is the identity function on $$\Z\text{.}$$

Clearly both the domain and codomain of $$(f\circ f)$$ are the set of integers $$\Z\text{.}$$ Let $$x\in\Z\text{.}$$ We have

\begin{equation*} (f\circ f)(x)=f(f(x))=f(-x)=-(-x)=x. \end{equation*}

Thus, by Definition 7.4.1, the composite $$(f\circ f)$$ is the identity function on $$\Z\text{.}$$