Functions and Relations (CSCI 2824, Spring 2015)

This page covers the following concepts.

Determining whether a function is injective
Determining whether a function is surjective
Proving a function is injective/surjective

These topics can be found in Section 4.3 of the textbook.

Injective and surjective functions

Recall the definition of injective and surjective functions.

Injective, surjective and bijective functions

A function F:Ato B is said to be

injective (or one-one) if for any distinct , ;
surjective (or onto) if for any , such that .
bijective if is both injective and surjective.

More descriptively,

is injective if maps every element of to a unique element in . In other words no element of are mapped to by two or more elements of .
- . (This is the contrapositive of the implication we have in the definition.)
is surjective if every element of is mapped to by some element of . In other words, nothing is left out.

Example 1

Classify the following functions $f_j: mathbb{N} rightarrow mathbb{N}$ between natural numbers as one-to-one and onto.

	One-to-One?	Onto?
	Yes	No
	Yes	No
$f_3(n) = lfloor sqrt{n} rfloor$	No	Yes
$f_4(n) = left{begin{array}{l} n-1, mbox{odd} n n+1, mbox{even} n end{array} right.$ .	Yes	Yes

It helps to visualize the mapping for each function to understand the answers.

Reasons

is not onto because it does not have any element such that , for instance.
is not onto because no element such that , for instance.
is not one-to-one since .

Guessing whether a function is injective/surjective

If a given function has both the domain and codomain being the set of real numbers mathbb{R} , then one visual way of guessing whether that function is injective/surjective can be done by using the graph. We imagine swiping a horizontal rule up or down and the number of intersection this horizontal rule makes with the graph.

A function is injective if and only if the horizontal rule intersects the graph at most once at any fixed -value.
A function is surjective if and only if the horizontal rule intersects the graph at least once at any fixed -value.

(As an aside, the vertical rule can be used to determine whether a relation is well-defined: at any fixed -value, the vertical rule should intersect the graph of a function with domain mathbb{R} exactly once.)

Proving that a function is injective/surjective

Let f: Ato B be a function.

To prove that is injective, we proceed like this…
- Let be such that .
- Infer from the equality and use the definition of to show that .
To prove that is surjective, we proceed like this…
- Fix any .
- Consider the equality and express in terms of .
- Double check: plug in your expression for into , and evaluate .

Example 2

Prove that the function $f:mathbb{R}_+tomathbb{R}: nmapsto n^2$ is one-to-one.

Proof

Fix any with .
Then implies that .
Then .
Splitting cases on , we have
- For , $-n notin mathbb{R}_+$ , therefore for this case.
- For , we have .
Therefore, it follows that for both cases.
This proves that is one-to-one. QED.

Example 3

Prove that the function $f:mathbb{R}_+tomathbb{R}_+: tmapsto frac{1}{1+t}$ is injective.

Proof

Fix any $s,tinmathbb{R}_+$ with .
Then implies that $frac{1}{1+s} = frac{1}{1+t}$ .
Inverting the fractions on both sides of the equality, we get .
Canceling 1 on both sides, we get .
This proves that is injective. QED.

Example 4

Prove that the function f:mathbb{R}tomathbb{R}: xmapsto -2x + 3 is surjective.

Proof

Fix any .
Consider the equality . Expressing in terms of , we get $x=frac{1}{2}(3-y)$ .
Now note that $x=frac{1}{2}(3-y)inmathbb{R}$ and $f(x)= -2left(frac{1}{2}(3-y)right) +3 = y$ .
This shows that is surjective. QED.

Example 5

Prove that the function f:mathbb{N}tomathbb{N}: nmapsto lfloor sqrt{n} rfloor is onto.

Proof

Fix any .
Then $m^2inmathbb{N}$
and $f(m^2)=lfloor sqrt{m^2} rfloor = lfloor m rfloor = m$ .
Therefore, all are mapped onto.
Hence is onto. QED.