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 \(x_1=x_2\text{.}\)
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.
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.
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.
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 \(x=\frac{-1}{3}\notin \mathbb{Z}\text{.}\)
Hence no element in \(\mathbb{Z}\) will map to \(y=1\text{.}\) Therefore, \(f\) is not onto \(\mathbb{Z}\text{.}\)
Example 7.2.11. Prove or Disprove Onto.
Let \(f:\mathbb{R}\rightarrow \mathbb{R}\) be given by \(f(x)=x^2-1\text{.}\) Prove or disprove \(f\) is onto \(\mathbb{R}\text{.}\)
Let \(y=-2\text{.}\) Then if \(f\) is onto \(\mathbb{R}\text{,}\) we could find \(x\) with \(f(x)=-2\text{.}\)
But if \(f(x)=-2\text{,}\) then \(x^2-1=-2\text{,}\) or \(x^2=-1\text{.}\) We know there are no real solutions to this equation. Hence no element in \(\mathbb{R}\) will map to \(y=-2\text{.}\) Therefore, \(f\) is not onto \(\mathbb{R}\text{.}\)
Activity 7.2.1.
\(f:\mathbb{R}\rightarrow\mathbb{R}\)\(f(x)=5x\text{.}\)
(a)
Prove or disprove \(f\) is one-to-one.
(b)
Prove or disprove \(f\) is onto \(\mathbb{R}\text{.}\)
Activity 7.2.2.
\(f:\mathbb{Z}\rightarrow\mathbb{Z}\)\(f(n)=5n\text{.}\)
(a)
Prove or disprove \(f\) is one-to-one.
(b)
Prove or disprove \(f\) is onto \(\mathbb{Z}\text{.}\)
Activity 7.2.3.
\(g:\mathbb{Z}\rightarrow\{0, 1, 2\}\)\(g(n)=r\)\(r\)\(n\)
(a)
Prove or disprove \(g\) is one-to-one.
(b)
Prove or disprove \(g\) is onto \(\{0, 1, 2\}\text{.}\)
Activity 7.2.4.
\(h:\mathbb{Z}\times\mathbb{Z}\rightarrow\mathbb{Z}\)\(h(a, b)=a-b\text{.}\)
(a)
Prove or disprove \(h\) is one-to-one.
(b)
Prove or disprove \(h\) is onto \(\mathbb{Z}\text{.}\)
Activity 7.2.5.
\(d:\mathbb{Z}\rightarrow\mathbb{Z}\times\mathbb{Z}\)\(d(a)=(a, a)\text{.}\)
(a)
Prove or disprove \(d\) is one-to-one.
(b)
Prove or disprove \(d\) is onto \(\mathbb{Z}\times\mathbb{Z}\text{.}\)
Definition 7.2.12.
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.
False.
True or false: \(f:\mathbb{R}\rightarrow \mathbb{R}, f(x)=5x-2\) is one-to-one.
2.
True.
False.
True or false: \(f:\mathbb{R}\rightarrow \mathbb{R}, f(x)=5x-2\) is onto \(\mathbb{R}\text{.}\)
3.
True.
False.
True or false: \(f:\mathbb{Z}\rightarrow \mathbb{Z}, f(x)=5x-2\) is one-to-one.
4.
True.
False.
True or false: \(f:\mathbb{Z}\rightarrow \mathbb{Z}, f(x)=5x-2\) is onto \(\mathbb{Z}\text{.}\)
5.
True.
False.
True or false: \(f:\mathbb{R}\rightarrow \mathbb{R}, f(x)=x^3-3\) is one-to-one.
6.
True.
False.
True or false: \(f:\mathbb{R}\rightarrow \mathbb{R}, f(x)=x^3-3\) is onto \(\mathbb{R}\text{.}\)
7.
True.
False.
True or false: \(f:\mathbb{Z}\rightarrow \mathbb{Z}, f(x)=x^3-3\) is one-to-one.
8.
True.
False.
True or false: \(f:\mathbb{Z}\rightarrow \mathbb{Z}, f(x)=x^3-3\) is onto \(\mathbb{Z}\text{.}\)
9.
True.
False.
True or false: \(f:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}, f(x, y)=x\) is one-to-one.
10.
True.
False.
True or false: \(f:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}, f(x, y)=x\) is onto \(\mathbb{Z}\text{.}\)
11.
True.
False.
True or false: \(f:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}, f(x, y)=(y, x)\) is one-to-one.
12.
True.
False.
True or false: \(f:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}\times\mathbb{Z}, f(x, y)=(y, x)\) is onto \(\mathbb{Z}\times\mathbb{Z}\text{.}\)
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 \(\mathbb{R}\times\mathbb{R}\text{.}\)
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