These notes will assume you are a little comfortable with the idea of a function1 Also called a map or a mapping.
The key fact about a function is that every permitted input has exactly one output.
Understanding this key fact is vital for moving onto the difficult properties known as injectivity and surjectivity.
Here’s a sample diagram showing a function which takes \(a\), \(b\), \(c\) or \(d\) as inputs and gives out integer values between \(1\) and \(5\). Notice every point on the left (from the domain) has exactly one line coming out of it, going into exactly one point on the right.
If we call this function \(f\) then
\[f(a)=5,\; f(b)=4,\; f(c)=1,\; f(d)=3\]
Firstly we must discuss the ridiculous number of names for the same adjectives describing a function.
The following three statements all mean exactly the same thing
The term one-to-one is particularly misleading, as you are about to discover.
Definition
A function is injective (or one-to-one) if
Whenever you pick one value achieved by the function on right-hand side, investigation shows that it’s achieved from only one input.
This is very difficult to grasp when you first learn it because it sounds very similar to our Key Fact above about functions5 which is the reason many people dislike the term one-to-one and prefer injective. Note the difference in these two statements:
Every individual input to a function results in a single output value.
Every output value from a function came from only one input value.
The first box is a statement of fact about every function6 it’s what it means to be a function!.
The second box is not always a fact. If it happens to be true for a function then we say that function is one-to-one or injective.
Which of the following four diagrams, each representing a different function, has the injective property7 property is used here like the word characteristic, we say metals have certain properties to describe what they are like. The same is true here, but with a function.?
For a discussion of the answers, see the bottom of these notes.
Click this line to jump straight to the injectivity answers
This property is actually easier to understand as it is less easily confused with the definition of what it means to be a function.8 it’s still a little weird!
This property only has two standard names. The following two statements both mean exactly the same thing
The slightly weird thing about being surjective is that whether a function is surjective or not depends upon the set of values you think it can take!11 You might be screaming now that this sounds subjective! In the diagrams above the set drawn on the right-hand side is the set we care about, this is often called the co-domain12 with the left hand set of inputs called the domain. However, some authors call it the range which can be confusing13 let’s not open that can of worms here!.
Definition
A function is surjective (or onto) if
Every value in your identified set of possible output values is achieved by at least one of your inputs.
In terms of the picture diagrams above, this one is very easy… does every point in the set on the right-hand side have at least one arrow coming into it? Are they any points in the right hand set with no arrows coming in?
If all right-hand side points have an arrow coming in… then it’s surjective.
The quite strange thing here is that suppose you have a function which is not surjective because there are some points with no arrows coming in. You might complain and ask… ‘Then why bother to draw them in the first place?’ or ‘Why not pretend this points aren’t possible outputs?’
These are both great questions, and the answer is that if you are allowed to go back and change you co-domain after seeing all the function output values, so that there are no lonely points then you have turned your function into a surjective one!14 You’re just not normally allowed to do this, especially when comparing multiple functions with similar output sets
Now for some examples for you to try…
Which of the following four diagrams, each representing a different function, has the surjective property?
Bonus question: do any of the eight function diagrams above show a function which is both injective and surjective?
Definition
A function which is both injective and surjective is called bijective15 notice ‘bi’ meaning two, because it’s both the ‘jectives’?!. Such functions are generally very nice.
The property bijective functions have is that you can move left to right then back to left again; or right to left and back to right again on the mapping diagram and always get back to where you started. This means that the procedure of un-doing the function and re-doing it, or doing the function and un-doing it always end up back where you started. This is the idea of invertibility discussed in many areas of maths. It’s vital for being able to reverse engineer the answer to a question, for example in decoding.
Our plan is to identify injectivity when it occurs. This means making sure no elements in the right-hand set of all the diagrams have more than one arrow coming into them. It’s that easy, once you know how!
Function 1: Clearly not injective as both \(1\) and \(4\) have more than one input that result in them. e.g. \(f(b)=f(d)=1\) an example of two inputs both giving the same input. Not injective.
Function 2: There’s only one problem value here but it’s still not injective. You need every output to have only one source input. Here the output of \(3\) comes from both \(b\) and \(d\). Not injective.
Function 3: This is definitely not injective! There are four different inputs all leading the same output of \(4\)!
Function 4: This is injective. Every input is mapped to a different output, so every output only has one arrow coming in. Injective.
In the Function 4 discussion there is another sentence which can be useful to remember:
Every input is mapped to a different output
This is another way to check if a function is injective (but veers closer to the earlier confusion, so I’ve not mentioned it until now). It’s not a shortcut to showing a function is injective, because you don’t just need to find two inputs that go to different places. You need to show every pair of inputs go to different outputs.16 It does provide an easy way to show a function is not injective. Just find two input values that go to the same place. e.g. the \(f(x)=x^2\) function for all integers. We know that \(f(-3)=f(3)=9\) so it is not injective.
Our plan is to identify surjectivity when it occurs. It’s as easy as looking at the sets on the right-hand side and seeking out any examples of elements which have zero incoming arrows.
Function 5: Not surjective. The value \(1\) is not the output of any of the inputs.
Function 6: (Very!) Not surjective. The value \(1\) is not the output of any of the inputs17 Values \(2\) and \(4\) are also not outputs, but any one of them having this property was enough to be not surjective.
Function 7: This is surjective. You’ll notice that \(1\), \(2\), \(3\) and \(4\) all have exactly one arrow coming in. For every possible output you can find (at least one18 more than one is okay, we’re talking surjectivity not injectivity here!) input that reaches it.
Function 8: This example is also surjective. Notice that I removed \(1\) from the output set, every element in the output set19 called the co-domain that we’re considering, is reached by at least one of the inputs20 \(2\) is reached from two different inputs, but we don’t care.
Were there any functions that were both surjective and injective. You’ll need to go back and answer the questions for the other sets of functions to find the answer. None of the first four functions was surjective, but Function 7 was surjective and a quick inspection shows every output reached from only one input, so it’s also injective. So Function 7 is bijective!
As a test, pick any point on the left of Function 7, follow any arrow21 there’ll only be 1 because it’s a function! to the right and pick any arrows back22 there’ll only be one because it’s injective and you’ll be back where you started.
Now also try the same with the right-hand side, pick any starting element from the right, follow any arrow23 exactly 1, because the function is surjective and injective back to the input set, and then following the only function line out of that input gets you back to where you began. Well done!