[Math] Injective function: example of injective function that is not surjective.

algebra-precalculuselementary-set-theoryexamples-counterexamplesfunctions

Does there exist an injective function that is not surjective? Could I have an example, please?

Best Answer

There are many examples. It just all depends on how your define the range and domain.

For example $\operatorname{f} : \mathbb{R} \to \mathbb{R}$ given by $\operatorname{f}(x)=x^3$ is both injective and surjective. But then I can change the image and say that $\operatorname{f} : \mathbb{R} \to \mathbb{C}$ is given by $\operatorname{f}(x) = x^3$. Now it is still injective but fails to be surjective.

Similarily, the function $\operatorname{g} : \mathbb{R} \to \mathbb{R}$ given by $\operatorname{g}(x)=x^2$ is neither surjective nor injective. But if I change the range and domain to $\operatorname{g}: \mathbb{R}^+ \to \mathbb{R}^+$ then it is both injective and surjective.

Edit: As requested by the OP:

An example of an injective function $\mathbb{R}\to\mathbb{R}$ that is not surjective is $\operatorname{h}(x)=\operatorname{e}^x$. This "hits" all of the positive reals, but misses zero and all of the negative reals. But the key point is the the definitions of injective and surjective depend almost completely on the choice of range and domain.

Motivation

When I go hiking, I want to be able to retrace my steps and get back to my starting point.

When I go to a store, I want to be able to return home.

When I swim out from the shore, I want to be able to get back to the shore.

Going somewhere and then coming back to where one started is important in life.

And it is important in mathematics.

And it is important in functional programming.

Domain, Codomain, and Inverse

If a function maps a set of elements (the domain) to a set of values (the codomain) then it is often useful to have another function that can take the elements in the codomain and send them back to their original domain values. The latter is called the inverse function.

In order for a function to have an inverse function, it must possess two important properties, which I explain now.

The Injective Property

Let the domain be the set of days of the week. In Haskell one can create the set using a data type definition such as this:

data Day = Monday | Tuesday | Wednesday | Thursday | Friday | Saturday | Sunday

Let the codomain be the set of breakfasts. One can create this set using a data type definition such as this:

data Breakfast = Eggs | Cereal | Toast | Oatmeal | Pastry | Ham | Grits | Sausage

Now I create a function that maps each element of the domain to a value in the codomain.

Here is one such function. The first line is the function signature and the following lines is the function definition:

f  ::  Day  ->  Breakfast
f  Monday       =  Eggs
f  Tuesday      =  Cereal
f  Wednesday    =  Toast
f  Thursday     =  Oatmeal
f  Friday       =  Pastry
f  Saturday     =  Ham
f  Sunday       =  Grits

An important thing to observe about the function is that no two elements in the domain map to the same codomain value. This function is called an injective function.

[Definition] An injective function is one such that no two elements in the domain map to the same value in the codomain.

Contrast with the following function, where two elements from the domain -- Monday and Tuesday -- both map to the same codomain value -- Eggs.

g  ::  Day  ->  Breakfast
g  Monday       =  Eggs
g  Tuesday      =  Eggs
g  Wednesday    =  Toast
g  Thursday     =  Oatmeal
g  Friday       =  Pastry
g  Saturday     =  Ham
g Sunday        =  Grits

The function is not injective.

Can you see a problem with creating an inverse function for g :: Day -> Breakfast?

Specifically, what would an inverse function do with Eggs? Map it to Monday? Or map it to Tuesday? That is a problem.

[Important] If a function does not have the injective property then it cannot have an inverse function.

In other words, I can't find my way back home.

The Surjective Property

There is a second property that a function must possess in order for it to have an inverse function. I explain that next.

Did you notice in the codomain that there are 8 values:

data Breakfast = Eggs | Cereal | Toast | Oatmeal | Pastry | Ham | Grits | Sausage

So there are more values in the codomain than in the domain.

In function f :: Day -> Breakfast there is no domain element that mapped to the codomain value Sausage.

So what would an inverse function do with Sausage? Map it to Monday? Tuesday? What?

The function is not surjective.

[Definition] A surjective function is one such that for each element in the codomain there is at least one element in the domain that maps to it.

[Important] If a function does not have the surjective property, then it does not have an inverse function.

[Important] In order for a function to have an inverse function, it must be both injective and surjective.

Injective + Surjective = Bijective

One final piece of terminology: a function that is both injective and surjective is said to be bijective. So, in order for a function to have an inverse function, it must be bijective.

Recap

If you want to be able to come back home after your function has taken you somewhere, then design your function to possess the properties of injectivity and surjectivity.

[Math] Functions $\mathbb{N} \to \mathbb{N}$ that are injective but not surjective, and vice versa

It may be that the downvotes are because your work does not even exhibit that you understand the concepts of injective and surjective, the definitions of which you may well be expected to use in your answers for b) and c).

Both of your answers are dead-wrong: the function listed in b) is NOT from $\Bbb N \to \Bbb N$ (it has the wrong co-domain). For example, $f(1) = \frac{1}{2}$ is NOT a natural number.

The function you give in c) IS surjective, but it also is injective, To see this, suppose:

$f(x) = f(y) \implies x - 1 = y - 1 \implies (x - 1) + 1 = (y - 1) + 1 \implies x = y$.

(EDIT: as pointed out in the comments, $f$ is not even a function from $\Bbb N \to \Bbb N$, as one can see by noting $f(0) = -1 \not\in \Bbb N$).

I suggest you try some function $f$ for b) that "skips" values in $\Bbb N$, you want "gaps" in the co-domain. For c), you might try using the floor function, somehow.

a) is the most important question, here though. Finiteness is key, that's what b) and c) are supposed to convince you of. Start by assuming $f$ is surjective. What happens if you assume (by way of contradiction), that $f$ is not injective? (hint: compare the cardinalities of the range, and the domain).