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0.3: Functions

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Definition:  Cartesian Product

Let A be a set. Then A×A={(a1,a2)|a1,a2A} is the Cartesian product of A.

Example 0.3.1

If A=R then this is the Cartesian plane -  Like {x,y:x,yA}.

Definition: Function

A function f is a subset (symbol is ) of A×B such that for every aA there exists a unique bB such that f(a)=b

Denoted by f:AB or AfB.

A is the domain of the function f.  

B is the co-domain of the function f.

The range of f is f(A)={f(x):xA}B=Im(f), the image of f.

A well-defined function is one that for every value of A, there is one and only one value of B.

Screen Shot 2023-06-27 at 12.56.17 PM.png

 

Definition: 

Let f be a function from AB.

  1. f is called onto (surjective) if f(A)=B.

  2. f is called one to one (injective) if f(x1)=f(x2),x1,x2A then x1=x2.

  3. f is bijective if it is surjective and injective.

  4. A function can be injective, surjective, bijective (both injective & surjective), or neither.

 Note: 

Let f:AB.  If f is 1-1 and onto then f is a bijection from AB. 

If there is a bijection between two finite sets, the number of elements in each set is the same.  Notationally:  |A| = |B|.

Example 0.3.2

Screen Shot 2023-06-27 at 12.58.35 PM.png

Example 0.3.3

Consider the mapping:

Screen Shot 2023-06-27 at 1.00.57 PM.png

Is this mapping 

a) Well defined?

b)  Is it surjective? 

c) Is it injective? 

d) Is it bijective?

Solution

The function is well-defined. Not injective. Not subjective, not bijective.

 

Example 0.3.4

Consider the mapping:

Screen Shot 2023-06-27 at 1.36.38 PM.png

Is this mapping 

a) Well defined?

b)  Is it surjective? 

c) Is it injective? 

d) Is it bijective?

Solution

This is well defined bijective function.

Example 0.3.5

Given f:QZ where Q={ab:a,bZ,b0} and f(ab)=a.  Is this a well-defined function?  


Solution

Let A and B be sets such that A={ab:a,bZ,a0} and B=Z.

Using a counterexample, f(ab) is not a well-defined function since f(12)=1 and f(24)=2, but 12=24.  This is an example of an element in A that maps to more than one element in B.

 

Example 0.3.6

Let

 Q={ab:b>0,and gcd(a,b)=1}

and f:QZ that is defined by

 f(ab)=a.

 a) Is f a well-defined function? b)  Is it surjective?  c) Is it injective?  d) Is it bijective?

Solution

  1. The function is well-defined.

Let A and B be set.

Let aA.

We need to show that if f(a)=b then bB and if f(a)=b1 that b=b1.

We are given f:QZ, which means that this is a function thus f(a) will map to one and only one element of B.  Thus if f(a)=b and if f(a)=b1 then b=b1.

In b) below, we have shown that the function is surjective, thus f(a)=b,

 
  1. The function is surjective

    Proof:

We shall show that f(Q)=Z by showing that  f(Q)Z and  Zf(Q).

Since we are given f:QZf(Q)Z.

Next, we shall show that Zf(Q).

Let nZ, then n1Q and f(n1)=n.

Hence nf(Q).

Hence Zf(Q).

Since Zf(Q) and f(Q)Z,f(Q)=Z.◻

 
  1. The function is not injective.

    Counterexample:

Let A={ab:b>0,and gcd(a,b)=1} and B=Z.

Note 14,13A and that f(14),f(13)B and that f(14)=f(13)=1

Since 1413 , the function is not injective.

 
  1. The function is not bijective since it has been shown not to be both injective and surjective.

 

Operation on functions: Composition

 

Definition:Composition of functions

Let f:AB and g:BC. Then the composition functions of g and f, gf:AC, defined by 

(gf)(a)=g(f(a)),aA.

Screen Shot 2023-06-27 at 1.51.29 PM.png

Definition: Identity function

Identity function idA:AA defined by idA(x)=x,xA.

Definition: Inverse of a function

Let f:AB then f has an inverse g:BA if (fg)=idB and (gf)=idA.  Note the domain of fg is the domain of g.

Let f:AB be a bijection.  Then the inverse of f, denoted by f1:BA defined by f1(b)=a if f(a)=b if aA and bB. In this case, ff1=idA, and f1f=idB.

 

Screen Shot 2023-06-27 at 2.25.13 PM.png 

Example 0.3.7

Let f:AB and g:BC. Then gf:AC. 

  1. If f and g are surjective, then gf is surjective.

Proof:

We will show that gf is surjective.

Let cC . Since g is surjective, there exists bB  such that g(b)=c.

Since f is surjective, there exists aA  such that f(b)=a.

Hence there exists aA such that gf(f(a))=c.

Therefore, if f and g are surjective, then gf  is surjective.◻

(better proof) If f is onto and g is onto then gf is onto.

Proof:

We shall show that (gf)(A)=C.

Since f and g are onto, by definition of a function, (gf)(A)C.

Now we need to show that C(gf)(A).

Let cC.

Since g is onto bB s.t. c=g(b).

Since f is onto, aA s.t. b=f(a).

Consider (gf)(a)=g(f(a)) (by definition)

        =g(b)

        =c.

Therefore c(gf)(A).

Hence C(gf)(A).

Since (gf)(A)C and C(gf)(A), then gf is onto.◻

 
Example 0.3.8
  1. If f and g are injective, then gf is injective. That is,

If f is 1-1 and g is 1-1 then gf is 1-1.

Proof:

Let   a1,a2A  such that gf(a1)=gf(a2).  Then g(f(a1))=g(f(a2)).

Since  g is 1-1, f(a1)=f(a2).

Since,  f is 1-1, a1=a2.

 

Hence gf  is injective.

 

 

Example 0.3.9

Given A is a matrix and TA:R2R2 defined by TA[xy]=[1101][xy].  Note this is a matrix transformation which is linear.

  1. Is TA 1-1?

Yes, TA is 1-1.

Proof:

Let [x1y1]and [x2y2]R2 s.t. TA[x1y1]=TA[x2y2].

We will show that [x1y1]=[x2y2].

TA[xy]=[x+yy].

Thus  [x1+y1y1]=[x2+y2y2].

Thus x1+y1=x2+y2 and y1=y2.

Thus x1=x2 and y1=y2.

Hence [x1y1]=[x2y2].

Thus TA is 1-1.◻

 
  1. Is TA[xy] onto?

Yes, TA[xy] is onto.

Proof:

We know by definition that TA[xy]R2.

We need to show that R2TA[xy].

Let [ab]R2.

Find [xy]R2 s.t. [1101][xy]=[ab].

[x+yy]=[ab].

Thus y=b and x=ab.

Thus TA[abb]=[ab].

Thus TA is onto.◻

 
  1. Does TA have an inverse?  If so, what is it?

a) Yes TA has an inverse since it is a bijection (1-1 and onto) ∴ T1A exists.

Proof:  By definition?  Obvious?

b) T1A:R2R2.

T1A[xy]=[1101][xy]

    =[xyy].

Check: T1ATA=[1101][1101]=[1001].

Example 0.3.10

Let f:AB and g:BC. Then gf:AC  . Given gf is injective and f is onto, prove that g is injective.

Proof:

Assume gf is injective.  (by definition)

Let x1,x2B s.t. g(x1)=g(x2).

We shall show that x1=x2.

Let a1,a2A.

Since f is surjective (by definition), we know that f(a1)=x1 and f(a2)=x2.

Since x1B and f is surjective, a1A s.t. f(a1)=x1.

Since x2B and f is surjective, a2A s.t. f(a2)=x2.

Since g(x1)=g(x2), then g(f(a1))=g(f(a2)) and gf)(a1)=(gf)(a2).

Since gf) is injective, a1=a2.

Thus f(a1)=f(a2) since f is a function.

Therefore, x1=x2.

Hence g is injective.

Example 0.3.11

Consider the mapping α:

 Screen Shot 2023-06-27 at 1.36.38 PM.png

Find α2 , α3 and α1.


This page titled 0.3: Functions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pamini Thangarajah.

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