An Onto Function A Function That Is Not Onto
Advanced Functions
One-To-One Functions | Onto Functions | One-To-One Correspondences | Inverse Functions
One-To-One Functions
Let f: A  B, a function from a set A to a set B. f is called a one-to-one function or injection, if, and only if, for all elements a1 and a2 in A, if f(a1) = f(a2), then a1 = a2 Equivalently, if a1 |
Conversely, a function f: A B is not a one-to-one function 
elements a1 and a2 in A such that f(a1) = f(a2) and a1
a2.
In terms of arrow diagrams, a one-to-one function takes distinct points of the domain to distinct points of the co-domain. A function is not a one-to-one function if at least two points of the domain are taken to the same point of the co-domain. Consider the following diagrams:
 |  |
| A one-to-one function | A function that is not one-to-one |
One-To-One Functions on Infinite Sets
To prove a function is one-to-one, the method of direct proof is generally used. Consider the example:
Example: Define f : R R by the rule
f(x) = 5x - 2 for all x 
R
Prove that f is one-to-one.
Proof: Suppose x1 and x2 are real numbers such that f(x1) = f(x2). (We need to show x1 = x2 .)
5x1 - 2 = 5x2 - 2
Adding 2 to both sides gives
5x1 = 5x2
Dividing by 5 on both sides gives
x1 = x2
We have proven that f is one-to-one.
On the other hand, to prove a function that is not one-to-one, a counter example has to be given.
Example: Define h: R R is defined by the rule h(n) = 2n2. Prove that h is not one-to-one by giving a counter example.
Counter example:
Let n1 = 3 and n2 = -3. Then
h(n1) = h(3) = 2 * 32 = 18 and
h(n2) = h(-3) = 2 * (-3)2 = 18
Hence h(n1) = h(n2) but n1 n2, and therefore h is not one-to-one.
 Practice Exercises |  |
Onto Functions
| Let f: AB be a function from a set A to a set B. f is called onto or surjective if, and only if, all elements in B can find some elements in A with the property that y = f(x), where y B and x A. f is onto |
Conversely, a function f: A B is not onto y in B such that 
x A, f(x) y.
In arrow diagram representations, a function is onto if each element of the co-domain has an arrow pointing to it from some element of the domain. A function is not onto if some element of the co-domain has no arrow pointing to it. Consider the following diagrams:
 |  |
| An onto function | A function that is not onto |
Proving or Disproving That Functions Are Onto
Example: Define f : R R by the rule f(x) = 5x - 2 for all xR. Prove that f is onto.
Proof: Let y R. (We need to show that x in R such that f(x) = y.)
If such a real number x exists, then 5x -2 = y and x = (y + 2)/5. x is a real number since sums and quotients (except for division by 0) of real numbers are real numbers. It follows that
f(x) = 5((y + 2)/5) -2 by the substitution and the definition of f
= y + 2 -2
= y by basic algebra
Hence, f is onto.
Example: Define g: Z Z by the rule g(n) = 2n - 1 for all n Z. Prove that g is not onto by giving a counter example.
Counter example:
The co-domain of g is Z by the definition of g and 0 Z. However, g(n) 0 for any integer n.
If g(n) = 0, then
2n -1 = 0
2n = 1 by adding 1 on both sides
n = 1/2 by dividing 2 on both sides
But 1/2 is not an integer. Hence there is no integer n for g(n) = 0 and so g is not onto.
| Practice Exercises | |
b in B, there is an element a in A such that f(a) = b as f is onto and there is only one such b as f is one-to-one. In this case, the function f sets up a pairing between elements of A and elements of B that pairs each element of A with exactly one element of B and each element of B with exactly one element of A. This pairing is called one-to-one correspondence or bijection. When depicted by arrow diagrams, it is illustrated as below:
A function which is a one-to-one correspondence
Inverse Functions
If there is a function f which has a onIMG SRC="images//I> correspondence from a set A to a set B, then there is a function from B to A that "undoes" the action of f. This function is called the inverse function for f.
| Suppose f: A B is a one-to-one correspondence (f is one-to-one and onto). Then there is a function f -1: B -1. Solution: By the definition of f -1, f -1(y) = x such that f(x) = y But ; f(x) = y
Hence f -1(y) = (y + 2)/5.
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