6.5: Properties of Functions

In this section, we will study some properties of functions. To facilitate our discussion, we need to introduce some notations. Some students may find them confusing and difficult to use. Besides memorizing the definitions, try to understand what they really mean.

Definition: image of under

Given a function \(f :{A}\to{B}\), and \(C\subseteq A\), the image of under is defined as \[f(C) = \{ f(x) \mid x\in C \}.\] In words, \(f(C)\) is the set of all the images of the elements of \(C\).

A few remarks about the definition:

1. It is about the image of a subset \(C\) of the domain of \(A\). Do not confuse it with the image of an element \(x\) from \(A\).
2. Therefore, do not merely say “the image.” Be specific: the image of an element, or the image of a subset.
3. Better yet: include the notation \(f(x)\) or \(f(C)\) in the discussion.
4. While \(f(x)\) is an element in the codomain, \(f(C)\) is a subset of the codomain.
5. Perhaps, the most important thing to remember is:

If \(y\in f(C)\), then \(y\in B\), and there exists an \(x\in C\)
such that \(f(x)=y\).

This key observation is often what we need to start a proof with.

Definition: image

Let \(f :{A}{B}\) be a function. The image or range of \(f\), denoted \(\mathrm{ im }{f}\), is defined as the set \(f(A)\). Hence, \(\mathrm{ im }{f}\) is the set of all possible images that \(f\) can assume.

The definition implies that a function \(f :{A}\to{B}\) is onto if \(\mathrm{ im }{f} = B\). Unfortunately, this observation is of limited use, because it is not always easy to find \(\mathrm{ im }{f}\).

Example \(\PageIndex{1}\label{eg:propfcn-01}\)

For the function \(f :{\mathbb{R}}\to{\mathbb{R}}\) defined by

\[f(x) = x^2, \nonumber \]

we find \(\mathrm{ im }{f}=[0,\infty)\). We also have, for example, \(f\big([\,2,\infty)\big) = [4,\infty)\). It is clear that \(f\) is neither one-to-one nor onto.

Example \(\PageIndex{2}\label{eg:propfcn-02}\)

For the function \(g :{\mathbb{Z}}\to{\mathbb{Z}}\) defined by \[g(n) = n+3,\nonumber\] we find \(\mathrm{ im }{g}=\mathbb{Z}\), and \(g(\mathbb{N})=\{4,5,6,\ldots\}\). The function \(g\) is both one-to-one and onto.

Exercise \(\PageIndex{1}\label{he:propfcn-01}\)

The function \(p :{\mathbb{R}}\to{\mathbb{R}}\) is defined as \(p(x)=3x+11\). Find \(p(\mathbb{R}^+)\) and \(\mathrm{ im }{p}\).

Exercise \(\PageIndex{2}\label{he:propfcn-02}\)

The function \(q :{\mathbb{R}}\to{\mathbb{R}}\) is defined as \(q(x)=x^2-x-7\). Find \(\mathrm{ im }{q}\).

Example \(\PageIndex{3}\label{eg:propfcn-03}\)

The function \(h :{\mathbb{Z}_{15}}\to{\mathbb{Z}_{15}}\) is defined by

\[h(x) \equiv 5x-11 \pmod{15}.\nonumber\]

From the tabulated data

\[\begin{array}{|c||*{9}{c|}} \hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & \cdots & 14 \\ \hline f(x) & 4 & 9 & 14 & 4 & 9 & 14 & 4 & \cdots & 14 \\ \hline \end{array} \nonumber\]

it becomes clear that the images repeat the pattern 4, 9, 14 five times. Therefore, we determine that \(\mathrm{ im }{h}=\{4,9,14\}\).

Exercise \(\PageIndex{3}\label{he:propfcn-03}\)

Determine \(h(\{0,3,4\})\), where \(h\) is defined in Example 6.5.3.

Example \(\PageIndex{4}\label{eg:propfcn-04}\)

Determine \(f(\{(0,2), (1,3)\})\), where the function \(f :{\{0,1,2\} \times\{0,1,2,3\}}{\mathbb{Z}}\) is defined according to

\[f(a,b) = a+b. \nonumber\]

Remark: Strictly speaking, we should write \(f((a,b))\) because the argument is an ordered pair of the form \((a,b)\). However, we often write \(f(a,b)\), because \(f\) can be viewed as a two-variable function. The first variable comes from \(\{0,1,2\}\), the second comes from \(\{0,1,2,3\}\), and we add them to form the image.

Solution

Because \[f(0,2)=0+2=2, \qquad\mbox{and}\qquad f(1,3)=1+3=4,\] we determine that \(f(\{(0,2),(1,3)\}) = \{2,4\}\).

Exercise \(\PageIndex{4}\label{he:propfcn-04}\)

Find \(\mathrm{ im }{f}\), where \(f\) is defined in Example 6.5.4.

We are now ready to present the first collection of properties of functions.

Theorem \(\PageIndex{1}\label{thm:propfcn-01}\)

Given \(f :{A}\to{B}\), the following properties hold for any \(C_1,C_2\subseteq A\).

• \(f(C_1\cup C_2) = f(C_1)\cup f(C_2)\)
• \(f(C_1\cap C_2) \subseteq f(C_1)\cap f(C_2)\)
• \(f(C_1 - C_2) \supseteq f(C_1) - f(C_2)\)
• \(C_1\subseteq C_2 \Rightarrow f(C_1) \subseteq f(C_2)\)

Remark

These results provide excellent opportunities to learn how to write mathematical proofs. We only provide the proof of (a) below, and leave the proofs of (b)–(d) as exercises. In (a), we want to establish the equality of two sets. One way to prove that \(S=T\) is to show that \(S\subseteq T\), and \(T\subseteq S\). Now, in order to prove that \(S\subseteq T\), we need to show that \(z\in S\) implies \(z\in\)T; to show that \(T\subseteq S\), we want to prove that \(z\in T\) implies \(z\in s\).

Proof of (a)

First, we want to show that \(f(C_1\cup C_2) \subseteq f(C_1)\cup f(C_2)\). Let \(y\in f(C_1\cup C_2)\), then there exists \(x\in C_1\cup C_2\) such that \(f(x)=y\). Having \(x\in C_1\cup C_2\) means either \(x\in C_1\) or \(x\in C_2\), so we have to consider two cases.

If \(x\in C_1\), then \(f(x)\in f(C_1)\).

If \(x\in C_2\), then \(f(x)\in f(C_2)\).

Thus, \(y=f(x)\) belongs to either \(f(C_1)\) or \(f(C_2)\), which means \(y=f(x) \in f(C_1)\cup f(C_2)\). This proves that \(f(C_1\cup C_2) \subseteq f(C_1)\cup f(C_2)\).

Next, we want to show that \(f(C_1)\cup f(C_2) \subseteq f(C_1\cup C_2)\). Let \(y\in f(C_1)\cup f(C_2)\), then \(y\) belongs to either \(f(C_1)\) or \(f(C_2)\).

If \(y\in f(C_1)\), then there exists \(x_1\in C_1\) such that \(f(x_1)=y\).

If \(y\in f(C_2)\), then there exists \(x_2\in C_2\) such that \(f(x_2)=y\).

These two possibilities together imply that there exists an element \(x\) belonging to either \(C_1\) or \(C_2\), that is, \(x\in C_1\cup C_2\), such that \(f(x)=y\). This means \(f(x)\in f(C_1\cup C_2)\). This proves that \(f(C_1)\cup f(C_2) \subseteq f(C_1\cup C_2)\). This concludes the proof of \(f(C_1\cup C_2) = f(C_1)\cup f(C_2)\).

Exercise \(\PageIndex{5}\label{he:propfcn-05}\)

Prove part (b) of Theorem 6.5.1.

Remark

Part (b) of Theorem 6.5.2 only gives a subset relationship. The reason is: having \(y\in f(C_1)\) and \(y\in f(C_2)\) does not necessarily mean that \(y\) is the image of the same element. Since \(f\) can be many-to-one, it is possible to have \(x_1\in C_1-C_2\) and \(x_2\in C_2-C_1\) such that \(f(x_1)=f(x_2)=y\). Consider \(f :{\{1,2,3\}}\to{\{a,b\}}\) defined by \[f(1)=f(3)=a, \qquad\mbox{and}\qquad f(2)=b.\] If \(C_1=\{1,2\}\) and \(C_2=\{2,3\}\), then \(f(C_1)=f(C_2)=\{a,b\}\), and \[f(C_1\cap C_2) = f(\{2\}) = \{b\} \subset \{a,b\} = f(C_1)\cap f(C_2).\] Therefore, we can only conclude that \(y\in f(C_1\cap C_2) \Rightarrow y\in f(C_1) \cap f(C_2)\).

Definition: preimage of under

Given a function \(f :{A}\to{B}\), and \(D\subseteq B\), the preimage of under is defined as \[f^{-1}(D) = \{ x\in A \mid f(x) \in D \}.\] Hence, \(f^{-1}(D)\) is the set of elements in the domain whose images are in \(C\). The symbol \(f^{-1}(D)\) is also pronounced as “\(f\) inverse of \(D\).”

Some remarks about the definition:

1. The preimage of \(D\) is a subset of the domain \(A\).

2. In particular, the preimage of \(B\) is always \(A\).

3. The key thing to remember is:

   If \(x\in f^{-1}(D)\), then \(x\in A\), and \(f(x)\in D\).

4. It is possible that \(f^{-1}(D)=\emptyset\) for some subset \(D\). If this happens, \(f\) is not onto.

5. Therefore, \(f\) is onto if and only if \(f^{-1}(\{b\})\neq \emptyset\) for every \(b\in B\).

Example \(\PageIndex{5}\label{eg:propfcn-05}\)

If \(t :{\mathbb{R}\to}{\mathbb{R}}\) is defined by \(t(x)=x^2-5x+5\), find \(t^{-1}(\{-1\})\).

Solution

We want to find \(x\) such that \(t(x)=x^2-5x+5=-1\). Hence, we have to solve the equation \[0 = x^2-5x+6 = (x-2)(x-3).\nonumber\] The solutions are \(x=2\) and \(x=3\). Therefore, \(t^{-1}(\{-1\}) = \{2,3\}\).

Exercise \(\PageIndex{6}\label{he:propfcn-06}\)

If \(k :{\mathbb{Q}}\to{\mathbb{R}}\) is defined by \(k(x)=x^2-x-7\), find \(k^{-1}(\{3\})\).

Example \(\PageIndex{6}\label{eg:propfcn-06}\)

For the function \(f :{\{0,1,2\}\times\{0,1,2,3\}}\to{\mathbb{Z}}\) defined by \[f(a,b) = a+b,\] we find \[\begin{aligned} f^{-1}(\{3\}) &=& \{(0,3), (1,2), (2,1)\}, \\ f^{-1}(\{4\}) &=& \{(1,3), (2,2)\}. \end{aligned}\] Since preimages are sets, we need to write the answers in set notation.

Exercise \(\PageIndex{7}\label{he:propfcn-07}\)

Find \(h^{-1}(\{4\})\) and \(h^{-1}(\{2\})\), where the function \(h\) is defined in Example 6.5.3.

Theorem \(\PageIndex{2}\label{thm:propfcn-02}\)

Given \(f :{A}\to{B}\), and \(D_1,D_2\subseteq B\), the following properties hold.

• \(f^{-1}(D_1\cup D_2) = f^{-1}(D_1)\cup f^{-1}(D_2)\)
• \(f^{-1}(D_1\cap D_2) = f^{-1}(D_1)\cap f^{-1}(D_2)\)
• \(f^{-1}(D_1 - D_2) = f^{-1}(D_1) - f^{-1}(D_2)\)
• \(D_1\subseteq D_2 \Rightarrow f^{-1}(D_1)\subseteq f^{-1}(D_2)\)

Proof of (a)

First, we want to prove that \(f^{-1}(D_1\cup D_2) \subseteq f^{-1}(D_1) \cup f^{-1}(D_2)\). Let \(x\in f^{-1}(D_1\cup D_2)\), then \(f(x)\in D_1\cup D_2\). This means either \(f(x)\in D_1\) or \(f(x)\in D_2\).

If \(f(x)\in D_1\), then \(x\in f^{-1}(D_1)\).

If \(f(x)\in D_2\), then \(x\in f^{-1}(D_2)\).

Since \(x\) belongs to either \(f^{-1}(D_1)\) or \(f^{-1}(D_2)\), we determine that \(x\in f^{-1}(D_1)\cup f^{-1}(D_2)\). Therefore, \(f^{-1}(D_1\cup D_2) \subseteq f^{-1}(D_1)\cup f^{-1}(D_2)\).

Next, we want to prove that \(f^{-1}(D_1)\cup f^{-1}(D_2) \subseteq f^{-1}(D_1\cup D_2)\). Let \(x\in f^{-1}(D_1)\cup f^{-1}(D_2)\). Then \(x\) belongs to either \(f^{-1}(D_1)\) or \(x\in f^{-1}(D_2)\).

If \(x\in f^{-1}(D_1)\), then \(f(x)\in D_1\).

If \(x\in f^{-1}(D_2)\), then \(f(x)\in D_2\).

Hence, \(f(x)\) belongs to either \(D_1\) or \(D_2\), which means \(f(x)\in D_1\cup D_2\). Thus, \(x\in f^{-1}(D_1\cup D_2)\). We have proved that \(f^{-1}(D_1)\cup f^{-1}(D_2) \subseteq f^{-1}(D_1\cup D_2)\). Together with \(f^{-1}(D_1\cup D_2) \subseteq f^{-1}(D_1)\cup f^{-1}(D_2)\), we conclude that \(f^{-1}(D_1\cup D_2) = f^{-1}(D_1)\cup f^{-1}(D_2)\).

Exercise \(\PageIndex{8}\label{he:propfcn-08}\)

Prove part (b) of Theorem 6.5.2.

Whether a function \(f :{A}{B}\) is one-to-one or onto can be determined by the cardinality of the preimages.

• \(f\) is one-to-one if and only if \(|f^{-1}(\{b\})|\leq1\) for every \(b\in B\).
• \(f\) is onto if and only if \(|f^{-1}(\{b\})|\geq1\) for every \(b\in B\).

If \(A\) and \(B\) are finite sets, then

• \(|A|\leq|B|\) if \(f\) is one-to-one, and
• \(|A|\geq|B|\) if \(f\) is onto.

In particular, if \(f\) is one-to-one and onto, we have \(|A|=|B|\).

Example \(\PageIndex{7}\label{eg:propfcn-08}\)

A function \(f :{\mathbb{Z}_{14}}\to{\mathbb{Z}_{10}}\) cannot be one-to-one because in order for it to be one-to-one, we need 14 distinct images. Since the codomain has only 10 elements, it is impossible for it to come up with 14 different images.

Likewise, a function \(g :{\mathbb{Z}_{23}}\to{\mathbb{Z}_{57}}\) cannot be onto because the domain has 23 elements, hence, we can have at most 23 different images. But the codomain has 57 elements, therefore, some of its elements must be left unused.

Example \(\PageIndex{8}\label{eg:propfcn-07}\)

Consider the function \(h :{\mathbb{Z}_{23}}\to{\mathbb{Z}_{57}}\) defined by \[h(x) \equiv 43x \pmod{57}.\] If \(y\equiv43x\) (mod 57), then, since \(43^{-1}\equiv4\) (mod 57), we find, in \(\mathbb{Z}_{23}\), \[x = 43^{-1}y = 4y.\] Since we can also express \(x\) in terms of \(y\), we declare that \(f\) is onto. Yet, we have learned from the previous example that \(f\) cannot be onto. Is there any contradiction?

Solution

There is an error in the argument. We should have said \[x \equiv 43^{-1}y \equiv 4y \pmod{57}.\] Since \(x\) is reduced modulo 57, its value may exceed 23. If this happens, \(x\notin\mathbb{Z}_{23}\). For example, if \(y=11\), we would have \(x=44\notin\mathbb{Z}_{23}\). Even if we reduce 44 modulo 23, we obtain \(x\equiv21\) (mod 23), we would have \[43\cdot21 \equiv 48\not\equiv 11 \pmod{57}.\] So it is still not the correct preimage. This example again illustrates the importance of taking caution when a function involves different moduli in its domain and codomain.

Summary and Review

• Given a function \(f :{A}\to{B}\), the image of \(C\subseteq A\) is defined as \(f(C) = \{f(x) \mid x\in C\}\).
• If \(y\in f(C)\), then \(y\in B\), and there exists an \(x\in C\) such that \(f(x)=y\).
• See Theorem 6.5.1 for a list of properties of the image of a set.
• The preimage of \(D\subseteq B\) is defined as \(f^{-1}(D) = \{x\in A \mid f(x)\in D\}\).
• If \(x\in f^{-1}(D)\), then \(x\in A\), and \(f(x)\in D\).
• See Theorem 6.5.2 for a list of properties of the preimage of a set.

Exercise \(\PageIndex{1}\label{ex:propfcn-01}\)

For each of the following functions, find the image of \(C\), and the preimage of \(D\).

• \({f_1}:{\{1,2,3,4,5\}}\to{\{a,b,c,d\}}\); \(f_1(1)=b\), \(f_1(2)=c\), \(f_1(3)=a\), \(f_1(4)=a\), \(f_1(5)=c\);\(C=\{1,3\}\), \(D=\{a.c\}\).
• \({f_2}:{\{1,2,3,4\}}\to{\{a,b,c,d,e\}}\); \(f_2(1)=c\), \(f_2(2)=b\), \(f_2(3)=a\), \(f_2(4)=d\);\(C=\{1,3\}\), \(D=\{b,d\}\).
• \({f_3}:{\{1,2,3,4,5\}}\to{\{a,b,c,d,e\}}\); \(f_3(1)=b\), \(f_3(2)=b\), \(f_3(3)=b\), \(f_3(4)=a\), \(f_3(5)=d\);\(C=\{1,3,5\}\), \(D=\{c\}\).
• \({f_4}:{\{1,2,3,4,5\}}\to{\{a,b,c,d,e\}}\); \(f_4(1)=d\), \(f_4(2)=b\), \(f_4(3)=e\), \(f_4(4)=a\), \(f_4(5)=c\);\(C=\{3\}\), \(D=\{c\}\).

Exercise \(\PageIndex{2}\label{ex:propfcn-02}\)

For each of the following functions, find the image of \(C\), and the preimage of \(D\).

1. \({f_5}:{\mathbb{Z}}\to{\mathbb{Z}}\); \(f_5(n)=-n\); \(C=2\mathbb{Z}\), \(D=\mathbb{N}\).
2. \({f_6}:{\mathbb{Z}}\to{\mathbb{Z}}\); \(f_6(n) = \cases{\hfill 2n & if\)n < 0\(, \cr \hfill -3n & if\)n0\(; \cr}\) \(C=\mathbb{N}\), \(D=2\mathbb{Z}\).
3. \({f_7}:{\mathbb{N}}\to{\mathbb{N}}\); \(f_7(n) = \cases{(n+1)/2 & if\)n\(is odd, \cr n/2 & if\)n\(is even; \cr}\) \(C=D=2\mathbb{N}\).
4. \({f_8}:{\mathbb{N}}\to{\mathbb{N}}\); \(f_8(n) = \cases{n+1 & if\)n\(is odd, \cr n-1 & if\)n\(is even; \cr}\) \(C=D=2\mathbb{N}\).

Exercise \(\PageIndex{3}\label{ex:propfcn-03}\)

The function \(s :{\mathbb{Z}_{12}}\to{\mathbb{Z}_{12}}\) is defined as \[s(x) \equiv 4x+7 \pmod{12}.\]

1. Find \(s(\{2,5,7\})\).
2. Find \(s^{-1}(\{2,5,7\})\).
3. Find \(\mathrm{ im }{s}\).

Exercise \(\PageIndex{4}\label{ex:propfcn-04}\)

The function \(t :{\mathbb{Z}_{15}}\to{\mathbb{Z}_{15}}\) is defined as \[t(x) \equiv 3x^2-5 \pmod{15}.\]

1. Find \(t(\{2,3,5,13\})\).
2. Find \(t^{-1}(\{1,5,7\})\).
3. Find \(\mathrm{ im }{t}\).

Exercise \(\PageIndex{5}\label{ex:propfcn-05}\)

The function \(u :{\mathbb{R}}\to{\mathbb{R}}\) is defined as \(u(x)=3x+11\), and the function \(v :{\mathbb{Z}}\to{\mathbb{R}}\) is defined as \(v(x)=3x+11\).

1. Find \(u([\,3,5))\) and \(v(\{3,4,5\})\).
2. Find \(u^{-1}((2,7\,])\) and \(v^{-1}((2,7\,])\).

Exercise \(\PageIndex{6}\label{ex:propfcn-06}\)

Is the function \(h :{\mathbb{Z}}\to{\mathbb{Z}}\) defined by \[h(n) = \cases{ 2n & if $n\geq0$ \cr -n & if $n < 0$ \cr}\] one-to-one? Is it onto?

Exercise \(\PageIndex{7}\label{ex:propfcn-07}\)

Define the \(r :{\mathbb{Z}\times\mathbb{Z}}\to{\mathbb{Q}}\) according to \(r(m,n) = 3^m 5^n\).

1. Find \(r(\{1,2,3\}\times\{-1,0,1\})\).
2. Find \(r^{-1}\big(\big\{\frac{25}{27}\big\}\big)\).
3. Find \(r^{-1}(D)\), where \(D=\{3,9,27,81,\ldots\,\}\).

Exercise \(\PageIndex{8}\label{ex:propfcn-08}\)

Define the function \(p :{\mathbb{Z}\times\mathbb{Z}}\to{\mathbb{Z}}\) according to \(p(x,y) = 12x+15y\).

1. Find \(p^{-1}(\{18\})\). You may use the set-builder notation to describe your answer.
2. Find \(\mathrm{ im }{p}\).

Exercise \(\PageIndex{9}\label{ex:propfcn-09}\)

The sum of the entries in a particular row in a matrix is called a row sum, and the sum of the entries in a particular column is called a column sum. Discuss how can we use the row sums and column sums of the incidence matrix of a function to determine if the function is well-defined, one-to-one, and onto.

Exercise \(\PageIndex{10}\label{ex:propfcn-10}\)

Below is the incidence matrix of the function \(f :{\{a,b,c,d,e\}}\to{\{\alpha,\beta,\gamma,\delta,\epsilon\}}\): \[\begin{array}[t]{cc} & \begin{array}{ccccc} \alpha & \beta & \gamma & \delta &\epsilon \end{array} \\ \noalign{\medskip} \begin{array}{c} a \\ b \\ c \\ d \\ e \end{array} & \left(\begin{array}{ccccc} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{array}\right) \end{array}\]

1. Find \(f(\{a,d,e\})\).
2. Find \(f^{-1}(\{\alpha,\beta,\epsilon\})\).
3. Find \(\mathrm{ im }{f}\).

Exercise \(\PageIndex{11}\label{ex:propfcn-11}\)

Consider the function \(h_1\) defined in Problem 6.5.8a in Exercises 1.2. What is \(h_1^{-1}(\{m\})\), if \(m\) represents your mother?

Exercise \(\PageIndex{12}\label{ex:propfcn-12}\)

Let \(S\) denote the maternal family tree, that includes you, your mother, your maternal grandmother, your maternal great-grandmother, and so on. Define a function \({M}:{S}\to{S}\) by letting \(M(x)\) be the mother of \(x\). Determine \(\mathrm{ im }{M}\).

Exercise \(\PageIndex{13}\label{ex:propfcn-13}\)

Prove part (c) of Theorem 6.5.1.

Exercise \(\PageIndex{14}\label{ex:propfcn-14}\)

Prove part (c) of Theorem 6.5.2.

Exercise \(\PageIndex{15}\label{ex:propfcn-15}\)

1. (a) Prove part (d) of Theorem 6.5.1.
2. (b) Prove part (d) of Theorem 6.5.2.

Exercise \(\PageIndex{16}\label{ex:propfcn-16}\)

Construct an example of a function \(f :{A}{B}\), and \(C_1,C_2 \subseteq A\) such that \(f(C_1-C_2) \supsetneq f(C_1)-f(C_2)\). See part (c) of Theorem 6.5.1.

Exercise \(\PageIndex{17}\label{ex:propfcn-17}\)

Given a function \(f :{A}\to{B}\), and \(C\subset A\), since \(f(C)\) is a subset of \(B\), the preimage of this subset is indicated by the notation \(f^{-1}(f(C))\). Consider the function \(f :{\mathbb{Z}}\to{\mathbb{Z}}\) defined by \(f(x)=x^2\), and \(C=\{0,1,2,3\}\).

1. Find \(f(C)\).
2. Find \(f^{-1}(f(C))\).

Exercise \(\PageIndex{18}\label{ex:propfcn-18}\)

Prove that \(C \subseteq f^{-1}(f(C))\) for any function \(f :{A}\to{B}\), and \(C\subseteq A\).