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Discrete Mathematics: An Active Approach to Mathematical Reasoning

Jennifer Firkins Nordstrom

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.
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Definition 7.2.1.

A function, f:X→Y, is one-to-one or injective if for all x1,x2∈X, if f(x1)=f(x2) then x1=x2.
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Equivalently, f is one-to-one if x1≠x2 implies f(x1)≠f(x2). We note, this is just the contrapositive of the definition.
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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.
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Proving a Function is One-to-One.

To prove f:X→Y is one-to-one:
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  • Assume f(x1)=f(x2).
  • Show x1=x2.
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To prove f:X→Y is not one-to-one:
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  • Find a counterexample.
  • In particular, find x1,x2∈X with x1≠x2 and f(x1)=f(x2).
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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.

Example 7.2.4. Proving One-to-One.

Let f:R→R be given by f(x)=3x+2. Prove f is one-to-one.

Proof.

Assume f(x1)=f(x2). Then
3x1+2=3x2+23x1=3x2x1=x2
which is what we wanted to show.

Example 7.2.5. Disproving One-to-One.

Let f:Z→Z be given by f(x)=x2−1. Disprove f is one-to-one.
We need a counterexample with x1≠x2 and f(x1)=f(x2). Let x1=2,x2=−2.
Then f(2)=4−1=3 and f(−2)=4−1=3. So f(x1)=f(x2), but 2≠−2.

Definition 7.2.6.

A function, f:X→Y, is onto Y or surjective if for all y∈Y there exists x∈X such that f(x)=y.
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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∈Y to get mapped to by some x∈X. We can also think about onto in terms of sets. A function is onto Y if Y is the range of f.
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Proving a Function is Onto.

To prove f:X→Y is onto Y:
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  • Let y be a general element of Y. You should not be using any information about the function at this point.
  • Find x∈X such that f(x)=y. Finding x may involve scratchwork.
  • In your proof, state x, show x∈X, and show f(x)=y.
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To prove f:X→Y is not onto Y:
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  • Find a counterexample.
  • In particular, find y∈Y such that no x∈X will map to y.
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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.

Example 7.2.9. Proving Onto.

Let f:R→R be given by f(x)=3x+2. Prove f is onto R.

Proof.

Let y∈R.
[Scratchwork: we want to find x so that f(x)=y. So we want 3x+2=y, or x=y−23.]
Let x=y−23. Then since y∈R,x∈R. Furthermore,
f(x)=f(y−23)=3(y−23)+2=y−2+2=y,
which is what we wanted to show.

Example 7.2.10. Disproving Onto.

Let f:Z→Z be given by f(x)=3x+2. Prove f is not onto Z.
Let y∈Z.
We saw in the previous example x=y−23. But x is not necessarily in Z. So for our counterexample, let y=1. Then we would need x=−13∉Z.
Hence no element in Z will map to y=1. Therefore, f is not onto Z.

Example 7.2.11. Prove or Disprove Onto.

Let f:R→R be given by f(x)=x2−1. Prove or disprove f is onto R.
Let y=−2. Then if f is onto R, we could find x with f(x)=−2.
But if f(x)=−2, then x2−1=−2, or x2=−1. We know there are no real solutions to this equation. Hence no element in R will map to y=−2. Therefore, f is not onto R.

Activity 7.2.1.

f:R→Rf(x)=5x.

(a)

Prove or disprove f is one-to-one.
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(b)

Prove or disprove f is onto R.
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Activity 7.2.2.

f:Z→Zf(n)=5n.

(a)

Prove or disprove f is one-to-one.
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(b)

Prove or disprove f is onto Z.
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Activity 7.2.3.

g:Z→{0,1,2}g(n)=rrn

(a)

Prove or disprove g is one-to-one.
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(b)

Prove or disprove g is onto {0,1,2}.
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Activity 7.2.4.

h:Z×Z→Zh(a,b)=a−b.

(a)

Prove or disprove h is one-to-one.
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(b)

Prove or disprove h is onto Z.
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Activity 7.2.5.

d:Z→Z×Zd(a)=(a,a).

(a)

Prove or disprove d is one-to-one.
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(b)

Prove or disprove d is onto Z×Z.
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Definition 7.2.12.

A function f:X→Y is a one-to-one correspondence or bijection if f is one-to-one and onto Y.
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We showed in the above examples that f:R→R given by f(x)=3x+2 is one-to-one and onto R. Thus, it is an example of a one-to-one correspondence.
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If it exists, we say f−1 is the inverse of f.
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Example 7.2.14. Inverse Function.

Since f:R→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, solve for x.
We get x=y−23. Thus f−1(x)=x−23.

Proof.

Show f−1 is one-to-one: assume f−1(y1)=f−1(y2). Then f−1(y1)=f−1(y2)=x for some x∈X. Thus, f(x)=y1 and f(x)=y2. Since f is a function, y1=y2.
Show f−1 is onto X. Let x∈X. Then there exists y∈Y such that f(x)=y since f is a function from X. Now, f−1(y)=x. Therefore, there exists y∈Y such that f−1(y)=x.

Reading Questions Check Your Understanding

1.

    True or false: f:R→R,f(x)=5x−2 is one-to-one.
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  • True.

  • False.

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2.

    True or false: f:R→R,f(x)=5x−2 is onto R.
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  • True.

  • False.

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3.

    True or false: f:Z→Z,f(x)=5x−2 is one-to-one.
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  • True.

  • False.

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4.

    True or false: f:Z→Z,f(x)=5x−2 is onto Z.
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  • True.

  • False.

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5.

    True or false: f:R→R,f(x)=x3−3 is one-to-one.
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  • True.

  • False.

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6.

    True or false: f:R→R,f(x)=x3−3 is onto R.
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  • True.

  • False.

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7.

    True or false: f:Z→Z,f(x)=x3−3 is one-to-one.
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  • True.

  • False.

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8.

    True or false: f:Z→Z,f(x)=x3−3 is onto Z.
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  • True.

  • False.

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9.

    True or false: f:Z×Z→Z,f(x,y)=x is one-to-one.
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  • True.

  • False.

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10.

    True or false: f:Z×Z→Z,f(x,y)=x is onto Z.
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  • True.

  • False.

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11.

    True or false: f:Z×Z→Z×Z,f(x,y)=(y,x) is one-to-one.
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  • True.

  • False.

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12.

    True or false: f:Z×Z→Z×Z,f(x,y)=(y,x) is onto Z×Z.
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  • True.

  • False.

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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.
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  1. f is onto if and only if every element in its codomain is the image of some element in its domain.
  2. f is onto if and only if every element in its domain has a corresponding image in its codomain.
  3. f is onto if and only if ∀y∈Y,∃x∈X such that f(x)=y.
  4. f is onto if and only if ∀x∈X,∃y∈Y such that f(x)=y.
  5. f is onto if and only if the range of f is the same as the codomain of f.
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2.

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

Let X={1,2,3}, Y={1,2,3,4} and Z={1,2}.
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  1. Define a function f:X→Y that is one-to-one but not onto.
  2. Define a function g:X→Z that is onto but not one-to-one.
  3. Define a function h:X→X that is neither one-to-one nor onto.
  4. Define a function k:X→X that is one-to-one and onto, but is not the identity function.
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4.

Define f:Z→Z by f(n)=4n−5.
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  1. Prove or disprove f is one-to-one.
  2. Prove or disprove f is onto Z.
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5.

Define g:R→R by g(x)=4x−5.
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  1. Prove or disprove g is one-to-one.
  2. Prove or disprove g is onto R.
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6.

Define F:P({a,b,c})→Z by
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F(A)= the number of elements in A.
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  1. Prove or disprove F is one-to-one.
  2. Prove or disprove F is onto Z.
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7.

Define G:R×R→R×R by
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G(x,y)=(2y,−x).
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  1. Prove or disprove G is one-to-one.
  2. Prove or disprove G is onto R×R.
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8.

Define H:R×R→R×R by
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H(x,y)=(x+1,2−y).
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  1. Prove or disprove H is one-to-one.
  2. Prove or disprove H is onto R×R.
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