Functions

Section 5.3 Functions

Functions are familar mathematical objects from algebra and calculus.

Definition 5.3.1.

A function, \(f:X\rightarrow Y\text{,}\) is a map in which each input is related to one and only one output.

We say \(X\) is the domain and \(Y\) is the codomain of \(f\text{.}\)

Example 5.3.2. Exploring the Definition of a Function.

Let \(f: X\rightarrow Y\) be a map as shown in the figure.

Figure 5.3.3.
Figure 5.3.3 shows a map that is not a function since \(x_2\) maps to two different outputs.
Let \(g: X\rightarrow Y\) be a map as shown in the figure.

Figure 5.3.4.
Figure 5.3.4 shows a map that is not a function since \(x_1\) does not map to any output.
Let \(h: X\rightarrow Y\) be a map as shown in the figure.

Figure 5.3.5.
Figure 5.3.5 shows a map that is a function since each \(x_i\) maps to exactly one \(y_i\text{.}\)

For a given \(x\in X\text{,}\) \(f(x)\) is

the output of \(f\text{,}\)
the value of \(f\) at \(x\text{,}\)
the image of \(x\) under \(f\text{.}\)

Definition 5.3.6.

The image or range of a set \(X\) under \(f\) is the set of outputs of \(f\) corresponding to inputs from \(X\text{:}\)

\begin{equation*} \text{Im}(f)=\{y\in Y : y=f(x) \text{ for some } x\in X\}. \end{equation*}

In Section 4.1 we defined sequences as lists. Now we can define them as functions. A sequence is a function from \(\mathbb{Z}^+\) to \(\mathbb{R}\text{.}\) For example, \(f(n)=\frac{1}{n}\text{.}\)

Example 5.3.7. Addition as a Function.

We may have functions involving Cartesian products of sets. This is common when we want to define algebraic operations. For example, we can define addition on the integers as the function \(f:\mathbb{Z}\times\mathbb{Z}\rightarrow\mathbb{Z}\) given by \(f(a, b)=a+b\text{.}\)

Find \(f(3, 2), f(5, 5), f(2, -4)\text{.}\)
Answer.
\(5, 10, -2\)
What is the image of \(\mathbb{Z}\times\mathbb{Z}\) under \(f\text{?}\)
Answer.
\(\mathbb{Z}\)
Find all ordered pairs that map to 0.
Answer.
\(\{(n, -n) : n\in \mathbb{Z}\}\)

Activity 5.3.1. Multiplication as a Function.

Define \(f:\mathbb{Z}\times\mathbb{Z}\rightarrow\mathbb{Z}\) by \(f(a, b)=ab\text{.}\)

(a)

Find \(f(0, 2), f(1, 1), f(2, 0), f(3, 2), f(n, 0)\text{.}\)

(b)

What is the image of \(\mathbb{Z}\times\mathbb{Z}\) under \(f\text{?}\)

(c)

Find all ordered pairs that map to 0.

Since a function needs to satisfy the property that each \(x\in X\) can only map to one \(y \in Y\text{,}\) we say a function is well-defined if whenever \(a=b\text{,}\) we have \(f(a)=f(b)\text{.}\) Most of the functions you've seen in calculus are clearly well-defined since when \(a=b\text{,}\) \(f(a)=f(b)\text{.}\) This property is really only interesting when elements of \(X\) have multiple representations. In other words, when two equal elements in \(X\) have different forms. The most familiar set where this happens is \(\mathbb{Q}\text{.}\) For example, \(\frac{1}{2}=\frac{2}{4}\text{.}\)

Example 5.3.8. A Map That is Not Well-Defined.

Let \(f:\mathbb{Q}\rightarrow \mathbb{Z}\) be given by \(f(p/q)=p+q\text{.}\)

Then \(1/2=2/4\) in \(\mathbb{Q}\text{,}\) but \(f(1/2)=1+2=3\) and \(f(2/4)=2+4=6\text{.}\) Thus, \(1/2=2/4\) but \(f(1/2)\neq f(2/4)\text{.}\)

Therefore \(f\) is not well-defined, and hence, \(f\) is not a function.

Activity 5.3.2. Showing a Map is Not Well-Defined.

Let \(f:\mathbb{Q}\rightarrow\mathbb{Z}\) be given by \(f(m/n)=m\text{.}\) Show \(f\) is NOT well-defined by finding two equivalent fractions in \(\mathbb{Q}\) that map to two different integers.

Now we will look at specific properties of functions. We will learn how to prove a function is one-to-one and/or onto its codomain. These properies are important as they are the exact properties we need in order for a function to have an inverse function.

Definition 5.3.9.

A function, \(f:X\rightarrow Y\text{,}\) is one-to-one or injective if for all \(x_1, x_2\in X\text{,}\) if \(f(x_1)=f(x_2)\) then \(x_1=x_2\text{.}\)

Equivalently, \(f\) is one-to-one if \(x_1\neq x_2\) implies \(f(x_1)\neq f(x_2)\text{.}\) We note, this is just the contrapositive of the definition.

Although it is easier to prove a function is one-to-one using the definition, the contrapositive can be helpful for deciding if a function is one-to-one.

Proving and Disproving One-to-One.

To prove \(f:X\rightarrow Y\) is one-to-one:

Assume \(f(x_1)=f(x_2).\)
Show \(x_1=x_2.\)

To prove \(f:X\rightarrow Y\) is NOT one-to-one:

Find a counterexample.
In particular, find \(x_1, x_2\in X\) with \(x_1\neq x_2\) and \(f(x_1)=f(x_2)\text{.}\)

Note that a one-to-one proof is really a proof of uniqueness. We are proving there is a unique input \(x\) for each output.

Example 5.3.10. Not One-to-One.

Figure 5.3.11. A function which is not one-to-one.

The function given in Figure 5.3.11 in not one-to-one since \(c\) and \(d\) both map to the same value in \(Y\text{.}\)

Example 5.3.12. Proving One-to-One.

Let \(f:\mathbb{R}\rightarrow \mathbb{R}\) be given by \(f(x)=3x+2\text{.}\) Prove \(f\) is one-to-one.

Proof.

Assume \(f(x_1)=f(x_2)\text{.}\) Then

\begin{align*} 3x_1+2&=3x_2+2\\ 3x_1&=3x_2\\ x_1&=x_2 \end{align*}

which is what we wanted to show.

Example 5.3.13. Disproving One-to-One.

Let \(f:\mathbb{Z}\rightarrow \mathbb{Z}\) be given by \(f(x)=x^2-1\text{.}\) Disprove \(f\) is one-to-one.

We need a counterexample with \(x_1\neq x_2\) and \(f(x_1)=f(x_2)\text{.}\) Let \(x_1=2, x_2=-2\text{.}\)

Then \(f(2)=4-1=3\) and \(f(-2)=4-1=3\text{.}\) So \(f(x_1)=f(x_2)\text{,}\) but \(2\neq-2\text{.}\)

Definition 5.3.14.

A function, \(f:X\rightarrow Y\text{,}\) is onto \(Y\) or surjective if for all \(y\in Y\) there exists \(x\in X\) such that \(f(x)=y\text{.}\)

Although we need the definition for onto to be able to write a proof, the concept of onto is easier to understand without the definition. Basically, we need every \(y\in Y\) to get mapped to by some \(x\in X\text{.}\) We can also think about onto in terms of sets. A function is onto \(Y\) if \(Y\) is the range or image of \(f\text{.}\)

Proving and Disproving Onto.

To prove \(f:X\rightarrow Y\) is onto \(Y\text{:}\)

Let \(y\) be a general element of \(Y\text{.}\) You should not be using any information about the function at this point.
Find \(x\in X\) such that \(f(x)=y\text{.}\) Finding \(x\) may involve scratchwork.
In your proof, state \(x\text{,}\) show \(x\in X\text{,}\) and show \(f(x)=y\text{.}\)

To prove \(f:X\rightarrow Y\) is NOT onto \(Y\text{:}\)

Find a counterexample.
In particular, find \(y\in Y\) such that no \(x\in X\) will map to \(y\text{.}\)

An onto proof is really an existence proof: state your candidate \(x\text{,}\) show \(x\) satifies the required properties. In particular we show \(x\in X\) and \(f(x)=y\text{.}\)

Example 5.3.15. Not Onto.

Figure 5.3.16. A function which is not onto \(Y\)

The function given in Figure 5.3.16 in not onto \(Y\) since 2 is not mapped to by any value in \(X\text{.}\)

Example 5.3.17. Proving Onto.

Let \(f:\mathbb{R}\rightarrow \mathbb{R}\) be given by \(f(x)=3x+2\text{.}\) Prove \(f\) is onto \(\mathbb{R}\text{.}\)

Proof.

Let \(y\in\mathbb{R}\text{.}\)

[Scratchwork: we want to find \(x\) so that \(f(x)=y\text{.}\) So we want \(3x+2=y\text{,}\) or \(x=\frac{y-2}{3}\text{.}\)]

Let \(x=\frac{y-2}{3}\text{.}\) Then since \(y\in \mathbb{R}, x\in\mathbb{R}\text{.}\) Furthermore,

\begin{equation*} f(x)=f(\frac{y-2}{3})=3(\frac{y-2}{3})+2=y-2+2=y, \end{equation*}

which is what we wanted to show.

Example 5.3.18. Disproving Onto.

Let \(f:\mathbb{Z}\rightarrow \mathbb{Z}\) be given by \(f(x)=3x+2\text{.}\) Prove \(f\) is NOT onto \(\mathbb{Z}\text{.}\)

Let \(y\in\mathbb{Z}\text{.}\)

We saw in the previous example \(x=\frac{y-2}{3}\text{.}\) But \(x\) is not necessarily in \(\mathbb{Z}\text{.}\) So for our counterexample, let \(y=1\text{.}\) Then we would need \(x=\frac{-1}{3}\notin \mathbb{Z}\text{.}\)

A function \(f:X\rightarrow Y\) is a one-to-one correspondence or bijection if \(f\) is one-to-one and onto \(Y\text{.}\)

We showed in the above examples that \(f:\mathbb{R}\rightarrow\mathbb{R}\) given by \(f(x)=3x+2\) is one-to-one and onto \(\mathbb{R}\text{.}\) Thus, it is an example of a bijection.

Theorem 5.3.21.

If \(f:X\rightarrow Y\) is a bijection, then there exists a function \(f^{-1}:Y\rightarrow X\) such that

\begin{equation*} f^{-1}(y)=x\Leftrightarrow f(x)=y. \end{equation*}

If it exists, we say \(f^{-1}\) is the inverse of \(f\text{.}\)

Example 5.3.22. Inverse Function.

Since \(f:\mathbb{R}\rightarrow\mathbb{R}\) given by \(f(x)=3x+2\) is one-to-one and onto, it has an inverse. We can find the inverse as we did in calculus.

Let \(y=3x+2\text{,}\) solve for \(x\text{.}\)

We get \(x=\frac{y-2}{3}\text{.}\) Thus \(f^{-1}(x)=\frac{x-2}{3}\text{.}\)

Theorem 5.3.23.

If \(f:X\rightarrow Y\) is one-to-one and onto \(Y\text{,}\) then \(f^{-1}:Y\rightarrow X\) is one-to-one and onto \(X\text{.}\)

Proof.

Show \(f^{-1}\) is one-to-one: assume \(f^{-1}(y_1)=f^{-1}(y_2)\text{.}\) Then \(f^{-1}(y_1)=f^{-1}(y_2)=x\) for some \(x\in X\text{.}\) Thus, \(f(x)=y_1\) and \(f(x)=y_2\text{.}\) Since \(f\) is a function, \(y_1=y_2\text{.}\)

Show \(f^{-1}\) is onto \(X\text{.}\) Let \(x\in X\text{.}\) Then there exists \(y\in Y\) such that \(f(x)=y\) since \(f\) is a function from \(X\text{.}\) Now, \(f^{-1}(y)=x\text{.}\) Therefore, there exists \(y\in Y\) such that \(f^{-1}(y)=x\text{.}\)

Activity 5.3.7. Bijections.

Which of the functions from the activities, Activity 5.3.3, Activity 5.3.4, Activity 5.3.5 and Activity 5.3.6 are bijections? If a function is a bijection, find its inverse function. Is the inverse one-to-one and onto?

Exercises Exercises

1.

Let \(X=\{1, 5, 9\}\) and \(Y=\{3, 4, 7\}\text{.}\)

Define \(f:X\rightarrow Y\) by specifying that \(f(1)=4, f(5)=7, f(9)=4\text{.}\)

Is \(f\) one-to-one? Is \(f\) onto? Explain your answers.
Define \(g:X\rightarrow Y\) by specifying that \(g(1)=7, g(5)=3, g(9)=4\text{.}\)

Is \(g\) one-to-one? Is \(g\) onto? Explain your answers.

2.

Let \(X=\{1, 2, 3\}\text{,}\) \(Y=\{1, 2, 3, 4\}\) and \(Z=\{1, 2\}\text{.}\)

Define a function \(f:X\rightarrow Y\) that is one-to-one but not onto.
Define a function \(g:X\rightarrow Z\) that is onto but not one-to-one.
Define a function \(h:X\rightarrow X\) that is neither one-to-one nor onto.
Define a function \(k:X\rightarrow X\) that is one-to-one and onto, but is not the identity function (does not send every element to itself).

3.

Let \(f\) be a function from the integers to the even integers given by \(f(a)=2a\text{.}\)

Prove \(f\) is one-to-one.
Prove \(f\) is onto.

4.

Define \(f:\mathbb{Z}\rightarrow\mathbb{Z}\) by \(f(n)=4n-5\text{.}\)

Prove or disprove \(f\) is one-to-one.
Prove or disprove \(f\) is onto \(\mathbb{Z}\text{.}\)

5.

Define \(g:\mathbb{R}\rightarrow\mathbb{R}\) by \(f(x)=4x-5\text{.}\)

Prove or disprove \(g\) is one-to-one.
Prove or disprove \(g\) is onto \(\mathbb{R}\text{.}\)

6.

Let \(A={\mathbb R}-\{-1\}\text{.}\) Define \(f:A\rightarrow {\mathbb R}\) by

\begin{equation*} f(a)={2a\over a+1}. \end{equation*}

Prove \(f\) is one-to-one but not onto.

7.

Define \(F:{\cal P}(\{a, b, c\})\rightarrow\mathbb{Z}\) by

\(F(A)=\) the number of elements in \(A\text{.}\)

Prove or disprove \(F\) is one-to-one.
Prove or disprove \(F\) is onto \(\mathbb{Z}\text{.}\)

8.

Let \(f:{\mathbb Z}\times{\mathbb Z}\rightarrow{\mathbb Z}\) be given by \(f(a, b)=a\text{.}\)

Prove or disprove \(f\) is one-to-one.
Prove or disprove \(f\) is onto.

Note: this map is called a projection.

9.

Let \(f:{\mathbb Z}\rightarrow{\mathbb Z}\times{\mathbb Z}\) be given by \(f(a) =(a, a)\text{.}\)

Prove or disprove \(f\) is one-to-one.
Prove or disprove \(f\) is onto.

Note: this map is called an embedding.

10.

Define \(G:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}\times\mathbb{R}\) by

\(G(x, y)=(2y, -x)\text{.}\)

Prove or disprove \(G\) is one-to-one.
Prove or disprove \(G\) is onto \(\mathbb{R}\times\mathbb{R}\text{.}\)

11.

Define \(H:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}\times\mathbb{R}\) by

\(H(x, y)=(x+1, 2-y)\text{.}\)

Prove or disprove \(H\) is one-to-one.
Prove or disprove \(H\) is onto \(\mathbb{R}\times\mathbb{R}\text{.}\)

12.

Let \(f:{\mathbb Q}\times{\mathbb Q}\rightarrow{\mathbb Q}\times{\mathbb Q}\) be given by \(f(a, b) =(2a, a+b)\text{.}\)

Prove or disprove \(f\) is one-to-one.
Prove or disprove \(f\) is onto.