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# Function types

Module by: Sunil Kumar Singh. E-mail the author

The relation between two sets under a rule has two perspectives. We can look at the relation in the direction from domain set “A” to co-domain set “B”. This is the function view. But, we can also look this relation in opposite direction from “B” to “A”.

When we see function relation from domain “A” to co-domain “B”, then we find following possibilities :

• Every element of domain is related to different element of co-domain (one to one function or injection)
• More than one elements of domain is related to an element of co-domain (many to one function)

When we see relation from co-domain “B” to domain “A”, then we find following possibilities :

• There are elements in co-domain, which are not related to any of the elements in domain (into function).
• There are no elements in co-domain, which are not related to elements in domain (onto function or surjection).
• There are elements in co-domain, which are related to exactly one element in domain. This statement is an equivalent statement (one to one function).
• There are elements in co-domain, which are related to more than one element in domain. This statement is an equivalent statement (many to one function).

Thus, we see that there are many ways in which a function - as a relation - can be unique and hence different. This gives rise to function types, which – as we shall see – are reflection of different possibilities that we have enumerated above.

## One - one function (Injection)

As is evident, this function describes relation in which something can be related distinctly to something. The countries have unique and distinct capital. It is evident that a function, based on this relation, would be an injection.

Definition 1: One - one function (Injection)
A function f : A B f : A B is an injection, if different elements of domain set “A” have different images in co-domain set “B”.

In plain words, every “x” in “A” associates with a distinct “y” in “B”. We can yet put it like this : Every argument (x) is related to distinct value (y).

In order to represent the condition of injectivity symbolically, we can think of two different elements “x” and “y” in set “A”. Then, two images f(x) and f(y) in “B”, corresponding to these elements in “A” are not equal. We capture this intent in constructing condition for an injection as :

f : A B is an injection x y , f x f y for all x , y A f : A B is an injection x y , f x f y for all x , y A

We can also interpret injection by asserting that if two images are equal, then it means that they are images of the same pre-image. The map diagram, corresponding to an injection, is shown in the figure. Note that elements in “A” are mapped to different elements in “B”.

### Example

Problem 1: Consider a function defined as :

f : Z Z b y f x = x 2 + 1 for all x Z f : Z Z b y f x = x 2 + 1 for all x Z

Determine whether the function is an injection?

Solution : We consider two arbitrary elements of the domain set such that :

f x = f y f x = f y

We have deliberately considered a contradictory supposition with respect to the requirement of injectivity. If this supposition yields x = y, then the given function is an injection; otherwise not. Here,

x 2 + 1 = y 2 + 1 x 2 + 1 = y 2 + 1

x 2 = y 2 x 2 = y 2

x = ± y x = ± y

This is not an unique solution. Here, “x” is not uniquely equal to “y”. We conclude that given function is not an injection. As a matter of fact, we can infer a check on our conclusion as,

f 1 = f - 1 = 2 f 1 = f - 1 = 2

Thus, we see that two pre-images relate to one image, which is contradictory to the requirement of an injection.

### Increasing and decreasing functions

The fact that function value is different for different arguments has an important bearing on the nature of injection plot. Consider two plots shown in the figure. In the plot shown on the left, a straight line parallel to x-axis intersects the plot only once. In the second plot, a line parallel to x-axis intersects the plot at two points for x = x 1 x = x 1 and x = x 2 x = x 2 . The function represented by second plot is not an injection as two values of arguments map to a single value of function – not two different values as required for an injection function.

It means that intersection plot intersects a line parallel to x-axis only once. This is possible only if the function is either (i) continuously increasing or (ii) continuously decreasing. Note the use of word “continuously”. An increasing plot, for example, if drops, then we can always find a line parallel ot x-axis, which intersects it at two points.

Hence, an injection graph is either an increasing or decreasing type. We can associate these characteristics with differential calculus. We can say that :

Either

d y d x > 0 for all x d y d x > 0 for all x

or,

d y d x < 0 for all x d y d x < 0 for all x

As a matter of fact the derivative can be equal to zero for certain values of "x" - not for an interval of "x". Thus, we can write the condition of increasing function : iif function is continuous and

d y d x 0 for all x ; equality holding for certain values of x d y d x 0 for all x ; equality holding for certain values of x

Similarly, we can write the condition of decreasing function : iif function is continuous and

d y d x 0 for all x ; equality holding for certain values of x d y d x 0 for all x ; equality holding for certain values of x

## Many – one function

More than one pre-images of a function are related to same image.

Definition 2: Many - one function
A function f : A B f : A B is an many – one function, if two or more elements of domain set “A” have the same images in co-domain set “B”.

The test of condition for many-one function is easy : if a function is not one-one, then it is many-one. Alternatively, we can check literally going by the definition – whether there exist such many-one relation. A map diagram showing the relation will reveal such instances of many-one relation.

Modulus function is one such many-one function. The function yields same value for positive and negative arguments of same magnitude.

f x = | x | f x = | x |

f - 1 = | - 1 | = 1 f - 1 = | - 1 | = 1

f 1 = | 1 | = 1 f 1 = | 1 | = 1

We should understand that a reverse function of the type “one to many” is barred from the very definition of function. The element of domain can be related to exactly one element in co-domain.

## Onto function (surjection)

The definition of function puts the restriction on domain that every element in it is related. If we extend this restriction to co-domain also, then we get a function called “onto” or “surjection”.

Definition 3: Onto function (surjection)
A function f : A B f : A B is an onto function or surjection, if every element of co-domain set is the image of some element in the domain set “A”.

One of the implications of surjection is that all elements of co-domain is related. It reduces the co-domain to range set. In other words, co-domain is also the range of the function.

Co-domain of "f" = Range of "f" Co-domain of "f" = Range of "f"

This equality of sets form one of the condition for testing a function to be surjection. Alternatively, we can check surjectivity by evaluating the rule of the function for the argument “x”. If the expression of “x” is valid for elements in co-domain, then the function is a surjection.

### Example

Problem 2 : Consider a function defined as :

f : R R b y f x = x 3 + 1 for all x R f : R R b y f x = x 3 + 1 for all x R

Determine whether the function is a surjection?

Solution : We solve the rule for argument “x” as :

y = x 3 + 1 y = x 3 + 1

x = y 1 1 / 3 x = y 1 1 / 3

We see that expression on the right hand side is a valid real expression for all values of “y” in "R" i.e co-domain. Hence, given function is an onto function or surjection.

## Into function

We have discussed in the beigining of this module that there is possibility that some of the elements of co-domains are not related. In that case, function is said to be into function.

Definition 4: Onto function (surjection)
A function f : A B f : A B is an into function, if there exists element in co-domain set, which has no pre-image in the domain set “A”.

One of the implications is that all elements of co-domain are not related to elements in domain set. In other words, range of the function is subset of the co-domain :

Range of "f" Co-domain of "f" Range of "f" Co-domain of "f"

We can check whether a given function is an into function or not by checking whether the function is an onto set or not. If the function is not an onto function, then it an into function. Alternatively, we can check the equality of co-domain and range set. If they are not equal, then the function is into function.

### Into function

Problem 3 : Consider a function defined as :

f : R R b y f x = x 2 + 1 f o r a l l x R f : R R b y f x = x 2 + 1 f o r a l l x R

Determine whether the function is an into function?

Solution : The rule of the function is :

y = x 2 + 1 y = x 2 + 1

The square of a real number is a positive number for all real number. Hence,

y = x 2 + 1 > 0 y = x 2 + 1 > 0

It means that images are only the right half of the real number i.e. from zero to infinity. But, the co-domain of the function is “R”. It means that left half of the co-domain i.e. from negative infinity to less than zero has no image in “A”. Therefore, the given function is an into function.

## One – one onto function (Bijection)

The bijection presents the most stringent condition for a function. Every element of both domain and co-domain is related to distinct element. This requirement is fulfilled when a function is both an injection and surjection.

The injection means that every element of domain is related to a distinct element in co-domain. On the other hand, surjection means that every element of co-domain is related, both distinctly and commonly. When conditions of injection and surjection are taken together, then it is also ensured that elements of co-domains are also related to distinct elements only.

### One – one onto function (Bijection)

Problem 4 : Consider a function defined as :

f : A B b y f x = x - 2 x - 3 f : A B b y f x = x - 2 x - 3

Determine domain (A) and co-domain(B) of the function so that it is a bijection.

Solution : For determining domain of the function, we inspect the given rule,

f x = x 2 x 3 f x = x 2 x 3

We observe that the given rational function is defined for all values of “x” except for x = 3. Hence, its domain is :

Domain = R { 3 } Domain = R { 3 }

In order that the given function is a bijection, it should be both an injection and a surjection. For infectivity, we put :

f x = f y f x = f y

x - 2 x - 3 = y - 2 y - 3 x - 2 x - 3 = y - 2 y - 3

x 2 y 3 = x 3 y 2 x 2 y 3 = x 3 y 2

x y 3 x 2 y + 6 = x y 2 x 3 y + 6 x y 3 x 2 y + 6 = x y 2 x 3 y + 6

3 x 2 y = 2 x 3 y 3 x 2 y = 2 x 3 y

x = y x = y

Hence, function is an injection for the domain as determined above. Now, for surjection we solve the rule of the function for the argument (x),

y = x 2 x 3 y = x 2 x 3

x y 3 y = x 2 x y 3 y = x 2

x y 1 = 3 y 2 x y 1 = 3 y 2

x = 3 y 2 y - 1 x = 3 y 2 y - 1

This equation is valid for all real values of “y” except “1”. Hence, function is surjection for all real values of “y” except for “1”. Hence, co-domain for the function to be a surjection is :

Co-domain = R { 1 } Co-domain = R { 1 }

Thus, we conclude that the given function is bijection for the domain and co-domain as determined above.

Domain = R { 3 } Domain = R { 3 }

Co-domain = R { 1 } Co-domain = R { 1 }

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