One-to-One, Onto, Inverse Functions

Section 7.2 One-to-One, Onto, Inverse Functions

In this section 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 7.2.1.

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 a Function is 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 \(f(x_1)=f(x_2)\text{.}\)

Example 7.2.2. Arrow Diagram: Not One-to-One.

Figure 7.2.3. A function which is not one-to-one

The function given in Figure 7.2.3 in not one-to-one since \(c\) and \(d\) both map to the same value in \(Y\text{.}\)

Example 7.2.4. 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 7.2.5. 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 7.2.6.

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 of \(f\text{.}\)

Proving a Function is 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 \(y\text{.}\)

Example 7.2.7. Arrow Diagram: Not Onto.

Figure 7.2.8. A function which is not onto \(Y\)

The function given in Figure 7.2.8 in not onto \(Y\) since 2 is not mapped to by any value in \(X\text{.}\)

Example 7.2.9. 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 7.2.10. 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 one-to-one correspondence.

Theorem 7.2.13.

If \(f:X\rightarrow Y\) is a one-to-one correspondence, 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 7.2.14. 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 7.2.15.

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 7.2.6.

Which of the functions from the activities, Activity 7.2.1, Activity 7.2.2, Activity 7.2.3, Activity 7.2.4 and Activity 7.2.5 are bijections? If a function is a bijection, find its inverse function. Is the inverse one-to-one and onto?

Reading Questions Check Your Understanding

1.

True or false: \(f:\mathbb{R}\rightarrow \mathbb{R}, f(x)=5x-2\) is one-to-one.

True.
False.

2.

True or false: \(f:\mathbb{R}\rightarrow \mathbb{R}, f(x)=5x-2\) is onto \(\mathbb{R}\text{.}\)

True.
False.

3.

True or false: \(f:\mathbb{Z}\rightarrow \mathbb{Z}, f(x)=5x-2\) is one-to-one.

True.
False.

4.

True or false: \(f:\mathbb{Z}\rightarrow \mathbb{Z}, f(x)=5x-2\) is onto \(\mathbb{Z}\text{.}\)

True.
False.

5.

True or false: \(f:\mathbb{R}\rightarrow \mathbb{R}, f(x)=x^3-3\) is one-to-one.

True.
False.

6.

True or false: \(f:\mathbb{R}\rightarrow \mathbb{R}, f(x)=x^3-3\) is onto \(\mathbb{R}\text{.}\)

True.
False.

7.

True or false: \(f:\mathbb{Z}\rightarrow \mathbb{Z}, f(x)=x^3-3\) is one-to-one.

True.
False.

8.

True or false: \(f:\mathbb{Z}\rightarrow \mathbb{Z}, f(x)=x^3-3\) is onto \(\mathbb{Z}\text{.}\)

True.
False.

9.

True or false: \(f:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}, f(x, y)=x\) is one-to-one.

True.
False.

10.

True or false: \(f:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}, f(x, y)=x\) is onto \(\mathbb{Z}\text{.}\)

True.
False.

11.

True or false: \(f:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}, f(x, y)=(y, x)\) is one-to-one.

True.
False.

12.

True or false: \(f:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}, f(x, y)=(y, x)\) is onto \(\mathbb{Z}\times\mathbb{Z}\text{.}\)

True.
False.

Exercises Exercises

1.

All but two of the following are correct ways to express the fact that a function \(f\) is onto. Find the two that are incorrect.

\(f\) is onto if and only if every element in its codomain is the image of some element in its domain.
\(f\) is onto if and only if every element in its domain has a corresponding image in its codomain.
\(f\) is onto if and only if \(\forall y\in Y, \exists x\in X\) such that \(f(x)=y\text{.}\)
\(f\) is onto if and only if \(\forall x\in X, \exists y\in Y\) such that \(f(x)=y\text{.}\)
\(f\) is onto if and only if the range of \(f\) is the same as the codomain of \(f\text{.}\)

2.

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.

3.

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.

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 \(g(x)=4x-5\text{.}\)

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

6.

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 \(\mathbb{Z}\text{.}\)

7.

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 \(\mathbb{R}\times\mathbb{R}\text{.}\)

8.

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 \(\mathbb{R}\times\mathbb{R}\text{.}\)

You have attempted of activities on this page.

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