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Applied Discrete Structures

Al Doerr, Ken Levasseur

Section 7.3 Function Composition

Now that we have a good understanding of what a function is, our next step is to consider an important operation on functions. Our purpose is not to develop the algebra of functions as completely as we did for the algebras of logic, matrices, and sets, but the reader should be aware of the similarities between the algebra of functions and that of matrices. We first define equality of functions.
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Subsection 7.3.1 Function Equality

Definition 7.3.1. Equality of Functions.

Let \(f, g:A \rightarrow B\text{;}\) that is, let \(f\) and \(g\) both be functions from \(A\) into \(B\text{.}\) Then \(f\) is equal to \(g\) (denoted \(f=g\)) if and only if \(f(x) = g(x)\) for all \(x \in A\text{.}\)
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Two functions that have different domains cannot be equal. For example, \(f: \mathbb{Z}\to \mathbb{Z}\) defined by \(f(x)=x^2\) and \(g: \mathbb{R}\to \mathbb{R}\) defined by \(g(x)=x^2\) are not equal even though the formula that defines them is the same.
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On the other hand, it is not uncommon for two functions to be equal even though they are defined differently. For example consider the functions \(h\) and \(k\text{,}\) where \(h: \{-1,0,1,2\}\to \{0,1,2\}\) is defined by \(h(x)=|x|\) and \(k: \{-1,0,1,2\}\to \{0,1,2\}\) is defined by \(k(x) = -\frac{x^3}{3}+x^2+\frac{x}{3}\) appear to be very different functions. However, they are equal because \(h(x)= k(x)\) for \(x = -1, 0, 1, \text{ and } 2\text{.}\)
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Subsection 7.3.2 Function Composition

One of the most important operations on functions is that of composition.
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Definition 7.3.2. Composition of Functions.

Let \(f:A \rightarrow B\) and \(g:B \rightarrow C\text{.}\) Then the composition of \(f\) followed by \(g\text{,}\) written \(g\circ f\text{,}\) is a function from \(A\) into \(C\) defined by \((g \circ f)(x) = g(f(x))\text{,}\) which is read “\(g\) of \(f\) of \(x\text{.}\)”
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The reader should note that it is traditional to write the composition of functions from right to left. Thus, in the above definition, the first function performed in computing \(g \circ f\) is \(f\text{.}\) On the other hand, for relations, the composition \(r s\) is read from left to right, so that the first relation is \(r\text{.}\)
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Example 7.3.3. A basic example.

Let \(f:\{1, 2, 3\}\rightarrow \{a, b\}\) be defined by \(f(1) = a\text{,}\) \(f(2) = a\text{,}\) and \(f(3) = b\text{.}\) Let \(g:\{a, b\} \rightarrow \{5, 6, 7\}\) be defined by \(g(a) = 5\) and \(g(b) = 7\text{.}\) Then \(g\circ f: \{1, 2, 3\}\rightarrow \{5, 6, 7\}\) is defined by \((g\circ f)(1)= 5\text{,}\) \((g\circ f)(2)= 5,\) and \((g\circ f)(3)= 7\text{.}\) For example, \((g\circ f)(1)= g(f(l)) = g(a) = 5\text{.}\) Note that \(f\circ g\) is not defined. Why?
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Let \(f:\mathbb{R} \rightarrow \mathbb{R}\) be defined by \(f(x) = x^3\) and let \(g:\mathbb{R} \rightarrow \mathbb{R}\) be defined by \(g(x) = 3x+1\text{.}\) Then, since
\begin{equation*} (g\circ f)(x) = g(f(x)) = g\left(x^3\right) = 3x^3 + 1 \end{equation*}
we have \(g\circ f: \mathbb{R} \rightarrow \mathbb{R}\) is defined by \((g\circ f)(x)= 3x^3 + 1\text{.}\) Here \(f\circ g\) is also defined and \(f\circ g:\mathbb{R}\rightarrow \mathbb{R}\) is defined by \((f\circ g)(x)=(3x + 1)^3\) . Moreover, since \(3x ^3+ 1 \neq (3x + 1)^3\) for at least one real number, \(g\circ f \neq f\circ g\text{.}\) Therefore, the commutative law is not true for functions under the operation of composition. However, the associative law is true for functions under the operation of composition.
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Proof.

Note: In order to prove that two functions are equal, we must use the definition of equality of functions. Assuming that the functions have the same domain, they are equal if, for each domain element, the images of that element under the two functions are equal.
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We wish to prove that \((h\circ (g\circ f))(x) = ((h\circ g)\circ f)(x)\) for all \(x \in A\text{,}\) which is the domain of both functions.
\begin{equation*} \begin{split} (h\circ (g\circ f))(x) &= h((g\circ f) (x))\textrm{ by the definition of composition}\\ &=h(g(f(x)))\textrm{ by the definition of composition} \end{split} \end{equation*}
Similarly,
\begin{equation*} \begin{split} ((h\circ g)\circ f)(x) &= (h\circ g)(f(x))\textrm{ by the definition of composition}\\ & =h(g(f(x)))\textrm{ by the definition of composition} \end{split}\text{.} \end{equation*}
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Notice that no matter how the functions in the expression \(h\circ g\circ f\) is grouped, the final image of any element of \(x\in A\) is \(h(g(f(x)))\) and so \(h\circ (g\circ f) = (h\circ g)\circ f\text{.}\)
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If \(f\) is a function on a set \(A\text{,}\) then the compositions \(f\circ f\text{,}\) \(f\circ f\circ f, \dots \) are valid, and we denote them as \(f^2\) , \(f^3, \dots \text{.}\) Repeated compositions of \(f\) with itself can be defined recursively. We will discuss this form of definition in detail in Section 8.1.
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Definition 7.3.5. Powers of Functions.

Let \(f: A\to A\text{.}\)
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  • \(f^1= f\text{;}\) that is, \(f^1(a) = f(a)\text{,}\) for \(a \in A\text{.}\)
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  • For \(n\geq 1\text{,}\) \(f^{n+1}= f\circ f^n\text{;}\) that is, \(f^{n+1}(a)=f\left( f^n(a)\right)\) for \(a \in A\text{.}\)
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Two useful theorems concerning composition are given below. The proofs are left for the exercises.
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We would now like to define the concepts of identity and inverse for functions under composition. The motivation and descriptions of the definitions of these terms come from the definitions of the terms in the set of real numbers and for matrices. For real numbers, the numbers 0 and 1 play the unique role that \(x + 0 = 0 + x = x\) and \(x \cdot 1 = 1 \cdot x = x\) for any real number \(x\text{.}\) 0 and 1 are the identity elements for the reals under the operations of addition and multiplication, respectively. Similarly, the \(n \times n\) zero matrix \(0\) and the \(n \times n\) identity matrix \(I\) are such that for any \(n \times n\) matrix \(A\text{,}\) \(A + 0 = 0 + A=A\) and \(A I = I A = A\text{.}\) Hence, an elegant way of defining the identity function under the operation of composition would be to imitate the above well-known facts.
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Definition 7.3.8. Identity Function.

For any set \(A\text{,}\) the identity function on \(A\) is a function from \(A\) onto \(A\text{,}\) denoted by \(i\) (or, more specifically, \(i_A\)) such that \(i(a) = a\) for all \(a\in A\text{.}\)
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Based on the definition of \(i\text{,}\) we can show that for all functions \(f:A\to A\text{,}\) \(f\circ i=i\circ f = f\text{.}\)
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Example 7.3.9. The identity function on \(\{1,2,3\}\).

If \(A = \{1, 2, 3\}\text{,}\) then the identity function \(i:A \to A\) is defined by \(i(1) = 1\text{,}\) \(i(2) = 2,\) and \(i(3)= 3\text{.}\)
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Example 7.3.10. The identity function on \(\mathbb{R}\).

The identity function on \(\mathbb{R}\) is \(i : \mathbb{R} \rightarrow \mathbb{R}\) defined by \(i(x) = x\text{.}\)
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Subsection 7.3.3 Inverse Functions

We will introduce the inverse of a function with a special case: the inverse of a function on a set. After you’ve taken the time to understand this concept, you can read about the inverse of a function from one set into another. The reader is encouraged to reread the definition of the inverse of a matrix in Section 5.2 (Definition 5.2.5) to see that the following definition of the inverse function is a direct analogue of that definition.
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Definition 7.3.11. Inverse of a Function on a Set.

Let \(f: A\rightarrow A\text{.}\) If there exists a function \(g : A \rightarrow A\) such that \(g\circ f = f\circ g = i\text{,}\) then \(g\) is called the inverse of \(f\) and is denoted by \(f^{-1}\) , read “\(f\) inverse.”
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Notice that in the definition we refer to “the inverse” as opposed to “an inverse.” It can be proven that a function can never have more than one inverse (see exercises).
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An alternate description of the inverse of a function, which can be proven from the definition, is as follows: Let \(f: A \rightarrow A\) be such that \(f(a) = b\text{.}\) Then when it exists, \(f^{-1}\) is a function from \(A\) to \(A\) such that \(f^{-1}(b)=a\text{.}\) Note that \(f^{-1}\) “undoes” what \(f\) does.
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Example 7.3.12. The inverse of a function on \(\{1,2,3\}\).

Let \(A = \{1, 2, 3\}\) and let \(f\) be the function defined on \(A\) such that \(f(1) = 2\text{,}\) \(f(2) = 3\text{,}\) and \(f(3) = 1\text{.}\) Then \(f^{-1} : A \rightarrow A\) is defined by \(f^{-1}(1) = 3\text{,}\) \(f^{-1}(2) = 1\text{,}\) and \(f^{-1}(3) = 2\text{.}\)
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Example 7.3.13. Inverse of a real function.

If \(g : \mathbb{R} \rightarrow \mathbb{R}\) is defined by \(g(x) =x^3\) , then \(g^{-1}\) is the function that undoes what \(g\) does. Since \(g\) cubes real numbers, \(g^{-1}\) must be the “reverse” process, namely, takes cube roots. Therefore, \(g^{-1} : \mathbb{R} \rightarrow \mathbb{R}\) is defined by \(g^{-1}(x)=\sqrt[3]{x}\text{.}\) We should show that \(g^{-1}\circ g = i\) and \(g\circ g^{-1}=i\text{.}\) We will do the first, and the reader is encouraged to do the second.
\begin{equation*} \begin{split} \left(g^{-1}\circ g\right)(x) &= g^{-1}(g(x)) \quad \textrm{ Definition of composition}\\ &= g^{-1}\left(x^3\right)\quad \text{Definition of } g\\ &=\sqrt[3]{x^3}\quad\textrm{Definition of } g^{-1}\\ & = x\quad\text{Definition of cube root}\\ &= i(x)\quad\text{Definition of the identity function} \end{split} \end{equation*}
Therefore, \(g^{-1}\circ g = i\text{.}\) Why?
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The definition of the inverse of a function alludes to the fact that not all functions have inverses. How do we determine when the inverse of a function exists?
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Proof.

(\(\Rightarrow\)) In this half of the proof, assume that \(f^{-1}\) exists and we must prove that \(f\) is one-to-one and onto. To do so, it is convenient for us to use the relation notation, where \(f(s)=t\) is equivalent to \((s,t)\in f\text{.}\) To prove that \(f\) is one-to-one, assume that \(f(a)=f(b) = c\text{.}\) Alternatively, that means \((a, c)\) and \((b, c)\) are elements of \(f\) . We must show that \(a =b\text{.}\) Since \((a, b), (c, b) \in \text{ }f\text{,}\) \((c, a)\) and \((c,b)\) are in \(f^{-1}\text{.}\) By the fact that \(f^{-1}\) is a function and \(c\) cannot have two images, \(a\) and \(b\) must be equal, so \(f\) is one-to-one.
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Next, to prove that \(f\) is onto, observe that for \(f^{-1}\) to be a function, it must use all of its domain, namely A. Let \(b\) be any element of \(A\text{.}\) Then \(b\) has an image under \(f^{-1}\) , \(f^{-1}(b)\text{.}\) Another way of writing this is \(\left(b,f^{-1}(b)\right)\in f^{-1}\text{,}\) By the definition of the inverse, this is equivalent to \(\left(f^{-1}(b), b\right) \in f\text{.}\) Hence, \(b\) is in the range of \(f\text{.}\) Since \(b\) was chosen arbitrarily, this shows that the range of \(f \) must be all of \(A\text{.}\)
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(\(\Leftarrow\) ) Assume \(f\) is one-to-one and onto and we are to prove \(f^{-1}\) exists. We leave this half of the proof to the reader. \(\square\)
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Definition 7.3.15. Permutation.

A bijection of a set \(A\) into itself is called a permutation of \(A\text{.}\)
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Next, we will consider the functions for which the domain and codomain are not necessarily equal. How do we define the inverse in this case?
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Definition 7.3.16. Inverse of a Function (General Case).

Let \(f:A \rightarrow B\text{,}\) If there exists a function \(g:B \rightarrow A\) such that \(g \circ f = i_A\) and \(f\circ g = i_B\) , then \(g\) is called the inverse of \(f\) and is denoted by \(f^{-1}\) , read “\(f\) inverse.”
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Note the slightly more complicated condition for the inverse in this case because the domains of \(f\circ g\) and \(g \circ f\) are different if \(A\) and \(B\) are different. The proof of the following theorem isn’t really very different from the special case where \(A=B\text{.}\)
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Example 7.3.18. Another inverse.

Let \(A =\{1,2, 3\}\) and \(B = \{a, b, c\}\text{.}\) Define \(f:A \rightarrow B\) by \(f(1) = a\text{,}\) \(f(2) = b\text{,}\) and \(f(3) = c\text{.}\) Then \(g: B \rightarrow A\) defined by \(g(a) = 1\text{,}\) \(g(b) = 2\text{,}\) and \(g(c) = 3\) is the inverse of \(f\text{.}\)
\begin{equation*} \left. \begin{array}{c} (g\circ f)(1)= 1 \\ (g\circ f)(2)=2 \\ (g\circ f)(3)=3 \\ \end{array} \right\}\Rightarrow \text{ }g\circ f = i_A \textrm{ and } \left. \begin{array}{c} (f\circ g)(a)=a \\ (f\circ g)(b)=b \\ (f\circ g)(c)=c \\ \end{array} \right\}\Rightarrow \text{ }f\circ g = i_B \end{equation*}
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Exercises 7.3.4 Exercises

1.

Let \(A = \{1,2, 3, 4, 5\}\text{,}\) \(B = \{a, b, c, d, e,f\}\text{,}\) and \(C = \{+, -\}\text{.}\) Define \(f: A \to B\) by \(f(k)\) equal to the \(k^{th}\) letter in the alphabet, and define \(g : B \rightarrow C\) by \(g(\alpha ) = +\) if \(\alpha\) is a vowel and \(g(\alpha ) = -\) if \(\alpha\) is a consonant.
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  1. Find \(g\circ f\text{.}\)
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  2. Does it make sense to discuss \(f\circ g\text{?}\) If not, why not?
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  3. Does \(f^{-1}\) exist? Why?
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  4. Does \(g^{-1}\) exist? Why?
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Answer.
  1. \(g\circ f:A\to C\) is defined by \((g\circ f)(k)=\begin{cases} + & \textrm{ if } k=1 \textrm{ or } k=5 \\ - & \textrm{ otherwise} \end{cases}\)
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  2. No, since the domain of \(f\) is not equal to the codomain of \(g\text{.}\)
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  3. No, since \(f\) is not surjective.
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  4. No, since \(g\) is not injective.
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2.

Let \(A = \{1, 2, 3\}\text{.}\) Define\(f:A\rightarrow A\) by \(f(1) = 2\text{,}\) \(f(2) = 1\text{,}\) and \(f(3) = 3\text{.}\) Find \(f^2\text{,}\) \(f^3\text{,}\) \(f^4\) and \(f^{-1}\text{.}\)
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3.

Let \(A = \{1, 2, 3\}\text{.}\)
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  1. List all permutations of \(A\text{.}\)
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  2. Find the inverse and square of each of the permutations of part a, where the square of a permutation, \(f\text{,}\) is the composition \(f \circ f\text{.}\)
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  3. Show that the composition of any two permutations of \(A\) is a permutation of \(A\text{.}\)
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  4. Prove that if \(A\) is any set where \(\lvert A\rvert= n\text{,}\) then the number of permutations of \(A\) is \(n!\text{.}\)
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Answer.
  1. The permutations of \(A\) are \(i,r_1,r_2,f_1,f_2,\) and \(f_3\text{,}\) defined in Table 15.3.1.
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  2. \begin{equation*} \begin{array}{ccc} g & g^{-1} & g^2 \\ i & i & i \\ r_1 & r_2 & r_2 \\ r_2 & r_1 & r_1 \\ f_1 & f_1 & i \\ f_2 & f_2 & i \\ f_3 & f_3 & i \\ \end{array} \end{equation*}
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  3. If \(f\) and \(g\) are permutations of \(A\text{,}\) then they are both injections and their composition, \(f\circ g\text{,}\) is a injection, by Theorem 7.3.6. By Theorem 7.3.7, \(f\circ g\) is also a surjection; therefore, \(f\circ g\) is a bijection on \(A\text{,}\) a permutation.
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  4. Proof by induction: Basis: \((n=1)\text{.}\) The number of permutations of \(A\) is one, the identity function, and 1! \(=1\text{.}\)
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    Induction: Assume that the number of permutations on a set with \(n\) elements, \(n\geq 1\text{,}\) is \(n\text{!.}\) Furthermore, assume that \(|A|=\)\(\text{ }n+1\) and that \(A\) contains an element called \(\sigma\text{.}\) Let \(A'=A-\{\sigma\}\text{.}\) We can reduce the definition of a permutation, \(f\text{,}\) on \(A\) to two steps. First, we select any one of the \(n\text{!}\) permutations on \(A'\text{.}\) (Note the use of the induction hypothesis.) Call it \(g\text{.}\) This permutation almost completely defines a permutation on \(A\) that we will call \(f\text{.}\) For all \(a\) in \(A'\text{,}\) we start by defining \(f(a)\) to be \(g(a)\text{.}\) We may be making some adjustments, but define it that way for now. Next, we select the image of \(\sigma\text{,}\) which can be done \(n+1\) different ways, allowing for any value in \(A\text{.}\) To keep our function bijective, we must adjust \(f\) as follows: If we select \(f(\sigma)=y \neq \sigma\text{,}\) then we must find the element, \(z\text{,}\) of \(A\) such that \(g(z)=y\text{,}\) and redefine the image of \(z\) to \(f(z)=\sigma\text{.}\) If we had selected \(f(\sigma)=\sigma\text{,}\) then there is no adjustment needed. By the rule of products, the number of ways that we can define \(f\) is \(n!(n+1)=(n+1)!\) \(\square\)
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4.

Define \(s\text{,}\) \(u\text{,}\) and \(d\text{,}\) all functions on the integers, by \(s(n) = n^2\) , \(u(n) = n + 1\text{,}\) and \(d(n) = n-1\text{.}\) Determine:
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  1. \(\displaystyle u \circ s \circ d\)
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  2. \(\displaystyle s \circ u\circ d\)
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  3. \(\displaystyle d \circ s \circ u\)
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5.

Consider the following functions on the set of bit strings of length 4. In these definitions, addition is done modulo 2, so that \(1+1=0\text{.}\) Which of these functions has an inverse? For those that have an inverse, what is it? For those that don’t explain why.
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  1. \(\displaystyle f_1(b_1 b_2 b_3 b_4) = b_2 b_3 b_4 b_1\)
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  2. \(\displaystyle f_2(b_1 b_2 b_3 b_4) = b_4 b_3 b_2 b_1\)
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  3. \(\displaystyle f_3(b_1 b_2 b_3 b_4) = (b_1+b_2)(b_2+b_3)(b_3+b_4)(b_4+b_1)\)
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  4. \(\displaystyle f_4(b_1 b_2 b_3 b_4) = b_1(b_1+b_2)(b_1+b_2+b_3)(b_1+b_2+b_3+b_4)\)
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Answer.
  1. \(f_1\) has an inverse. \(f_1^{-1}= f_1^3\text{.}\)
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  2. \(f_2\) has an inverse. \(f_2^{-1}= f_2\text{.}\)
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  3. \(f_3\) does not have an inverse. One way to verify this is to note that \(f_3\) is not one-to-one because \(f_3(0000) = 0000 = f_3(1111)\text{.}\)
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  4. \(f_4\) has an inverse. \(f_4^{-1}=f_4^3.\)
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6.

Inverse images. If \(f\) is any function from \(A\) into \(B\text{,}\) we can describe the inverse image as a function from \(B\) into \(\mathcal{P}(A)\text{,}\) which is also commonly denoted \(f^{-1}\text{.}\) If \(b \in B\text{,}\) \(f^{-1}(b) = \{a \in A\mid f(a) = b\}\text{.}\) If \(f\) does have an inverse, the inverse image of \(b\) is \(\left\{f^{-1}(b)\right\}\text{.}\)
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  1. Let \(g : \mathbb{R} \to \mathbb{R}\) be defined by \(g(x) = x^2\text{.}\) What are \(g^{-1}(4)\text{,}\) \(g^{-1}(0)\) and \(g^{-1}(-1)\text{?}\)
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  2. If \(r: \mathbb{R}\to \mathbb{Z}\text{,}\) where \(r(x) = \lceil x\rceil\text{,}\) what is \(r^{-1}(1)\text{?}\)
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7.

Let \(f,\) \(g\text{,}\) and \(h\) all be functions from \(\mathbb{Z}\) into \(\mathbb{Z}\) defined by \(f(n) = n + 5\text{,}\) \(g(n) = n - 2,\) and \(h(n)=n^2\text{.}\) Define:
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  1. \(\displaystyle f\circ g\)
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  2. \(\displaystyle f^3\)
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  3. \(\displaystyle f\circ h\)
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Answer.
  1. \(\displaystyle f\circ g(n)=n+3\)
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  2. \(\displaystyle f^3(n)=n+15\)
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  3. \(\displaystyle f\circ h(n)=n^2+5\)
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8.

Define the following functions on the integers by \(f(k) = k + 1\text{,}\) \(g(k) = 2k\text{,}\) and \(h(k)=\lceil k/2\rceil\)
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  1. Which of these functions are one-to-one?
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  2. Which of these functions are onto?
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  3. Express in simplest terms the compositions \(f\circ g\text{,}\) \(g \circ f\text{,}\) \(g \circ h\text{,}\) \(h \circ g\text{,}\) and \(h^2\text{.}\)
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9.

Let \(A\) be a nonempty set. Prove that if \(f \) is a bijection on \(A\) and \(f \circ f=f\text{,}\) then \(f\) is the identity function, \(i\)
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Hint.
You have seen a similar proof in matrix algebra.
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10.

For the real matrix \(A=\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)\text{,}\) \(\det(A)= a d-b c\text{.}\)
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Recall that a bijection from a set to itself is also referred to as a permutation of the set. Let \(\pi\) be a permutation of \(\{a,b,c,d\}\) such that \(a\) becomes \(\pi (a)\text{,}\) \(b\) becomes \(\pi (b)\text{,}\) etc.
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Let \(B=\left( \begin{array}{cc} \pi(a)& \pi(b)\\ \pi(c)& \pi(d)\\ \end{array} \right)\text{.}\) How many permutations of \(\pi\) leave the determinant of \(A\) invariant, that is, \(\det A = \det B\text{?}\)
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11.

State and prove a theorem on inverse functions analogous to the one that says that if a matrix has an inverse, that inverse is unique.
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Answer.
If \(f:A\to B\) and \(f\) has an inverse, then that inverse is unique.
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Proof: Suppose that \(g\) and \(h\) are both inverses of \(f\text{,}\) both having domain \(B\) and codomain \(A\text{.}\)
\begin{equation*} \begin{split}g &= g\circ i_B \\ & =g\circ (f\circ h)\\ & =(g\circ f)\circ h\\ & =i_A\circ h\\ & =h\quad \Rightarrow g=h \quad \square \end{split} \end{equation*}
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12.

Let \(f\) and \(g\) be functions whose inverses exist. Prove that \((f\circ g)^{-1}= g^{-1}\circ f^{-1}\text{.}\)
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Hint.
See Exercise 3 of Section 5.4.
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13.

Answer.
Let \(x,x'\) be elements of \(A\) such that \(g\circ f(x)=g\circ f(x')\text{;}\) that is, \(g(f(x))=g(f(x'))\text{.}\) Since \(g\) is injective, \(f(x)=f(x')\) and since \(f\) is injective, \(x=x'\text{.}\) \(\square\)
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Let \(x\) be an element of \(C\text{.}\) We must show that there exists an element of \(A\) whose image under \(g\circ f\) is \(x\text{.}\) Since \(g\) is surjective, there exists an element of \(B\text{,}\) \(y\text{,}\) such that \(g(y)=x\text{.}\) Also, since \(f\) is a surjection, there exists an element of \(A\text{,}\) \(z\text{,}\) such that \(f(z)=y\text{,}\) \(g\circ f(z)=g(f(z))=g(y)=x\text{.}\)\(\square\)
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15.

Prove by induction that if \(n\geq 2\) and \(f_1\text{,}\) \(f_2, \dots, f_n\) are invertible functions on some nonempty set \(A\text{,}\) then \(\left( f_1\circ f_2\circ \cdots \circ f_n \right){}^{-1}= f_n^{-1}\circ \cdots \circ f_2^{-1}\circ f_1^{-1}\text{.}\) The basis has been taken care of in Exercise 7.3.4.12.
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Answer.
Basis: \((n=2)\text{:}\) \(\left(f_1\circ f_2\right){}^{-1}=f_2{}^{-1}\circ f_1{}^{-2}\) by Exercise 7.3.4.12.
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Induction: Assume \(n\geq 2\) and
\begin{equation*} \left(f_1\circ f_2\circ \cdots \circ f_n\right){}^{-1}= f_n{}^{-1}\circ \cdots \circ f_2{}^{-1}\circ f_1{}^{-1} \end{equation*}
and consider \(\left(f_1\circ f_2\circ \cdots \circ f_{n+1}\right)^{-1}\text{.}\)
\begin{equation*} \begin{split} \left(f_1\circ f_2\circ \cdots \circ f_{n+1}\right){}^{-1} &=\left(\left(f_1\circ f_2\circ \cdots \circ f_n\right)\circ f_{n+1}\right){}^{-1}\\ & =f_{n+1}{}^{-1}\circ \left(f_1\circ f_2\circ \cdots \circ f_n\right){}^{-1}\\ & \quad \quad \quad \textrm{ by the basis}\\ &=f_{n+1}{}^{-1}\circ \left(f_n{}^{-1}\circ \cdots \circ f_2{}^{-1}\circ f_1{}^{-1}\right)\\ & \quad \quad \quad \textrm{ by the induction hypothesis}\\ &=f_{n+1}{}^{-1}\circ \cdots \circ f_2{}^{-1}\circ f_1{}^{-1} \quad. \square \end{split} \end{equation*}
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16.

  1. Our definition of cardinality states that two sets, \(A\) and \(B\text{,}\) have the same cardinality if there exists a bijection between the two sets. Why does it not matter whether the bijection is from \(A\) into \(B\) or \(B\) into \(A\text{?}\)
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  2. Prove that “has the same cardinality as” is an equivalence relation on sets.
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17.

Construct a table listing as many “Laws of Function Composition” as you can identify. Use previous lists of laws as a guide.
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18.

Based on the definition of the identity function, show that for all functions \(f:A\to A\text{,}\) \(f\circ i=i\circ f = f\text{.}\)
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