# When Is A Function Onto? (2 Key Ideas)

We know that functions are useful in algebra and calculus, but there are many different distinctions we can make.  One of these is the idea of an onto function (also called a surjection).

So, when is a function onto?  A function f:A→B is onto if f(A) = B (the image of f is equal to the codomain of f). Each possible output y (in the set B) has an input x (in the set A) so that f(x) = y. In other words, we can always find an input value from the set A so that f can “reach” a particular output value in the set B.

Of course, we will need to examine a function, its domain, its image, and its codomain to determine whether it is onto or not.

In this article, we’ll talk about when a function is onto and how to prove it.  We’ll also look at some examples of onto functions (and also functions that are not onto).

Let’s get going.

## When Is A Function Onto?

A function f:A→B is onto if f(A) = B.  This notation means that the image of f is equal to the codomain of f). An onto function (surjection) f from X to Y is one where each value of y of the codomain (outputs, right set) has a corresponding value x in the domain (inputs, left set) such that f(x) = y.

Intuitively, this means that for any element in the set B, we can find a corresponding element in the set A.  So, for each value of y in B, we can find a value of x in A so that f(x) = y.

The domain and codomain make a big difference when we are trying to determine if a function is onto or not.

### Example 1: f(x) = 2x Is Onto for f:R→R

The function f(x) = 2x is onto when we consider its domain (all real numbers) and codomain (all real numbers).

This is easy to see: for any real number y, we simply divide by 2 to get x: x = y/2.  This value of x will map to the desired value of y.

So, if we want to map to y = 2, we choose x = 1: f(1) = 2(1) = 2.

If we want to map to y = 7, we choose x = 3.5: f(3.5) = 2(3.5) = 7.

If we want to map to y = 12, we choose x = 6: f(6) = 2(6) = 12.

### Example 1: f(x) = 2x Is Not Onto for f:Z→Z

The function f(x) = 2x is not onto when we consider its domain (all integers) and codomain (all integers).

Remember that an integer has no fraction or decimal part: …,-3, -2, -1, 0, 1, 2, 3, …

So, the trick we used in the last example will not work:

If we want to map to y = 7, there is no way to do it.  When we divide 7 by 2, we get 3.5, which is not an integer.

So, there is no number x in the set of integers that will map to y = 7 for the function f(x) = 2x.

Thus, there are numbers in the codomain (integers) that we cannot “reach” with the function f(x) = 2x and its domain (integers).  Specifically, we cannot map to (reach) any odd numbers in the codomain.

## How To Prove A Function Is Onto Or Not

To prove a function f:A→B is onto, we must prove that each output y in the set B has an input in the set A so that f(x) = y.

### Example 1: Prove f(x) = x + 1 is onto for f:[0,1]→[1, 2]

First, consider an element y in the codomain of f: so, y is in the set [1, 2].

So, we need to find an element x in the domain of f so that f(x) = y.

Using the definition of the function f(x), we get:

• f(x) = y  [need an element x that maps to y]
• x + 1 = y [definition of the function f(x)]
• x = y – 1  [solved for x]

So, given an element y, all we need to do is subtract 1 to find the value of x that maps to y.

Since y is in the set [1, 2], we can write the following inequalities:

• 1 <= y <= 2  [since y is in the set [1, 2]]
• 1 – 1 <= y – 1 <= 2 – 1  [subtract 1 from all three parts of this inequality chain]
• 0 <= y – 1 <= 1  [simplify]
• 0 <= x <= 1  [since x = y – 1, which is what we solved for earlier]

This means that x is between 0 and 1, so x is in the set [0, 1], which is the domain of the function f.

Thus, f is onto for this domain and codomain.

### Example 2: Prove f(x) = x2 is not onto for f:R→R

To prove a function is not onto, all we need to do is find one element y in the set B (the codomain of f) that does not have an element x in A (the domain of f) with f(x) = y.

In this example, we can graph the function to see that it is never negative.

[insert graph of f(x) = x2]

Since xis real, then x2 >= 0.  Thus, the function f(x) = x2 always has an output greater than or equal to 0.

So, f(x) cannot map to any negative value y in the codomain (the real numbers).

We can easily see this by solving a value of x that maps to y = -1:

• f(x) = y  [need an element x that maps to y]
• x2 = y [definition of the function f(x)]
• x2 = -1  [we want to find an x that maps to y = -1]
• x = i  [take the square root of both sides of the equation]

This implies x = i (the square root of -1) is a solution.  However, i is an imaginary number – it is not contained in the domain of f (the set of real numbers).

Thus, there is no value x in the domain of f that maps to y = -1.  Thus, f(x) is not onto for this domain (real numbers) and codomain (real numbers).

## Can A Constant Function Be Onto?

A constant function can be onto if the codomain contains only the value of the constant function.  Note that this will be a codomain with only one element.

### Example: A Constant Function That Is Onto

Consider the function f(x) = 1 for f:R→{1}

In this case, the function is onto, since the image of f is equal to the codomain of f:

• The image of f is the single value {1}, since any value of x in the domain (real numbers) maps to y = 1.
• The codomain of f is given as {1}.

## Is A Linear Function Onto?

A linear function can be onto in certain cases.  It will depend on the given codomain of the function.

### Example 1: A Linear Function That Is Onto

Consider the linear function f(x) = 3x + 1 for f:RR.

In this case, f(x) is onto.  To prove this, consider any element y in the domain (real numbers).

Using the definition of the function f(x), we get:

• f(x) = y  [need an element x that maps to y]
• 3x + 1 = y [definition of the function f(x)]
• 3x = y – 1
• x = (y – 1) / 3  [solved for x]

So, given an element y, all we need to do is subtract 1 and divide the result by 3 to find the value of x that maps to y.

Since y is a real number, then y – 1 is real, and so (y – 1) / 3 is also real.  So, x = (y – 1) / 3 is a real number, so there is a value of x that the function f will map to the value of y.

Thus, f(x) is onto.

### Example 2: A Linear Function That Is Not Onto

Consider the linear function f(x) = 2 for f:RR.

In this case, f(x) is not onto.  To prove this, all we need to do is find an output that f cannot map to.

Let’s choose y = 3, which is in the codomain (real numbers).  The function f(x) = 2 can only map to 2, so it can never have an output of 3.

Thus, f(x) = 2 is not onto for this domain (real numbers) and codomain (real numbers).

## Is A Quadratic Function Onto?

A quadratic function can be onto in certain cases.  It will depend on the given codomain of the function.

### Example 1: A Quadratic Function That Is Onto

Consider the quadratic function f(x) = x2 + 1 for f:[1, 2]→[2, 5]

In this case, f(x) is onto.  To prove this, consider any element y in the domain [2, 5].

Using the definition of the function f(x), we get:

• f(x) = y  [need an element x that maps to y]
• x2 + 1 = y [definition of the function f(x)]
• x2 = y – 1
• x = √(y – 1)  [solved for x and took positive or principal square root]

So, given an element y, all we need to do is subtract 1 and take the square root of the result to find the value of x that maps to y.

Since y is in the set [2, 5], then we can write the following inequalities:

• 2 <= y <= 5  [since y is in the set [2, 5]]
• 2 – 1 <= y – 1 <= 5 – 1  [subtract 1 from each part of the inequality chain]
• 1 <= y – 1 <= 4
• √1 <= √(y – 1) <= √4  [take the principal square root of each part of the inequality chain]
• 1 <= √(y – 1) <= 2
• 1<= x <= 2 [since x = √(y – 1)]

So, the value of x that we solve for is in the domain [1, 2].  Thus, f(x) is onto.

### Example 2: A Quadratic Function That Is Not Onto

Consider the quadratic function f(x) = x2 for f:RR.  In this case, the function is not onto.

We can see this by considering the value y = -1 in the codomain (real numbers)

There is no value of x in the domain (real numbers) that will map to y = -1.

The value x = i will work, but i is an imaginary number, not a real number, so it is not in the domain of f.

Thus, f(x) is not onto in this case.

## Can A Function Be Onto & Not One To One?

A function can be onto and not one-to-one.  Remember that a one-to-one function has only one x value that maps to a given y value.

### Example: A Function That Is Onto & Not One-To-One

Consider the function f(x) = x2 for f:[-1, 1][0,1].  This function is onto, but not one-to-one.

The function is onto because for any y value in the codomain [0, 1], we can simply take the square root to find a value of x that will map to y.  Since the square root of a number in [0, 1] will also be in [0, 1], we know that x is in the domain of f, which is [-1, 1].

However, f is not one-to-one, since there are two input values that map to the same output.  For example, consider x = 1 and x = -1:

• f(1) = 12 = 1
• f(-1) = (-1)2 = 1

Both inputs x = 1 and x = -1 map to the same output y = 1.  So, f is not a one-to-one function.

## When Is A Function Not Onto?

A function f:A→B is not onto when there is at least one element y in the set B which has no element x in A that maps to y. A function that is not onto has at least one element in the codomain (here, C in the right set) that has no element in the domain (here, the left set) that maps to it.

Of course, a function that is not onto can have more than one element in the codomain that cannot be mapped to from the domain.

## Conclusion

Now you know when a function is onto and when it is not.  You also know how to prove a function is onto or how to prove it is not with a counterexample.