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# Suppose the relation R  A  B is a function - PowerPoint PPT Presentation

Inverse function . . Suppose the relation R  A  B is a function. It means that for any a  A there exists a unique b  B, ( a , b )  R , or we have f ( a )= b. . Then the inverse relation R -1 = { ( b , a ) | b  B , a  A and f ( a )= b }  B  A .

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Suppose the relation R  A B is a function

It means that for anya A there exists a uniqueb B,

(a, b) R , or we have f (a)=b.

Then the inverse relation

R-1 = { (b, a) | bB, aA and f (a)=b}  BA

also defines a function only if f : A B is a bijection.

R-1

R is a function

R-1is not a function. Why?

R-1

R

R is a function

R-1is not a function. Why?

Theorem. Let f : A  B be a bijection. Then the inverse relation

of f , defined from B to A as g = {(b, a)| bB, aA and f (a)=b}, is

a function from B to A such that gf (a)=a for all a A and

fg (b)=b for all b B.

The function g is called the inverse function of f and is denoted f –1.

f g (b)

b

B

A

f

a

g

gf (a)

The converse of the previous Theorem is also true.

Theorem . Let f : A  Band g : BAbe two functions.

If gf (a)= a for all aA , and fg (b)=b for all bB, then

both f and g are bijections, and they are inverse functions of

each other, that is f = g-1 and g = f –1 .

We can first prove that if gf (a)= a for all aA,

and fg (b)=b for all bB, then f is a bijection and g = f –1 .

a1= a2

g(f (a1)= a1

a1

f (a1) = f (a2)

g(f (a2)= a2

a2

Sketch of the proof that f is a bijection, given

f : A  Band g : BA

a A, g  f (a)= a

b B, f  g (b)= b

f is bijection = injection + surjection

f is injection: f (a1) = f (a2)  a1= a2

Since g is a function,f (a1) = f (a2)  g(f (a1)=g(f (a2),

i. e. a1= a2 .

Proof that if gf (a)= a for all aA , and fg (b)=b for all bB, then f is an injection.

Suppose f (a1)=f (a2) for a1, a2A. We want to show that a1 = a2.

We have gf (a1) = gf (a2), since g is a function, g : BA.

But gf (a1) = a1 and gf (a2)= a2 (gf (a)= a for all aA is given).

It implies thata1 = a2 .

Sketch of the proof that f is a surjection, given

f : A  Band g : BA

a A, g  f (a)= a

b B, f  g (b)= b

f is surjective: for any bB there exists aA such that f (a)=b

a ?

fg (b)=b

g

bB

Since g is a function, a =g(b) is defined for bB

f (a) = f (g (b))= fg (b)=b

Proof that if gf (a)= a for all aA , and fg (b)=b for all bB, then f is surjection.

To prove that f is a surjection, take any bB to show that

there always exists aA such that f (a)=b. Let a=g(b), then

f (a)=f (g (b))=b, according to fg (b)=b for all bB.

So, f is a bijection and f –1 defined as f –1 (b)=a  f (a) = b,

is a function.

g = f -1

definition of inverse

f -1 (b)=a  f (a)=b

?

g(b)=a  f (a)=b

?

?

bB, g(b)=a  f (a)=b and a A, f (a)=b  g(b)=a

Given:bB, fg (b)=b, so a=g(b), f (a)=b

Given: aA, gf (a)=a, so b=f(a), g (b)=a

Proof that if gf (a)= a for all aA , and fg (b)=b for all bB, then g = f –1 .

To prove that g = f –1 , we need to prove that g(b)=a  f (a) = b.

But fg (b)=b for any b B, that is g (b)=a  f (a)=b.

In the same way,we havegf (a)= a for all aA, i. e.

f (a)=b  g (b)=a .

QED.

g  f (a2)

a1

f (a1)= f (a2)

g  f (a1)

a2

b1

g

f

a

b2

Suppose f : A  Band g : BA and only one condition

holds: a A, g  f (a)= a .

Is it sufficient to imply any properties of g or f ?

Can f be not injective?

No. f must be injective.

Can f be not surjective?

Yes.

b1

f

g

a1

g

•b2

a1

g

b

f

a2•

Suppose f : A  Band g : BA and only one condition

holds: a A, g  f (a)= a .

Is it sufficient to imply any properties of g or f ?

Can g be not injective?

Yes.

No, g must be surjective.

Can g be not surjective?

of a composition.

Theorem . If f : A  Band g : BC are two bijections, then

(gf )-1 = f -1 g -1.

Two functions f : A  Band g : BA are inverse

of each other iff we have the relationship

gf (a)=a for allaA and f g(b)=bfor allbB.

It is equivalent to say, that gf =IA and f g(b)= IB, where

IAdenotes identity function on A ( IA(a) = a for all aA )

and similarly, IB is identity function on B.

B

C

g

f

g -1

f -1

gf

Theorem . If f : A  Band g : BC are two bijections, then

(gf )-1 = f -1 g -1

Proof. To prove that (gf )-1 = f -1 g -1, it suffices to prove that

(f -1 g -1)(gf )=IA and (gf )(f -1 g -1) =IC

(gf )(f -1 g -1) = ((gf )f -1 )g -1

by associative property of relation composition

= (g(f f -1))g -1 ,by associative property

= (gIB)g -1, since f f -1 =IB

= gg -1, since gIB= g

= IC

Similarly we can prove that (f -1 g -1)(gf )=IA

f

f (C)

A

CA

Proofs involving functions.

Let f: AB be a function. Then for any subset of A, CA, we can

define the set of images for elements of C that we denote as f (C):

f (C) = {f (x) | xC }.

f

DB

f-1(D)

A

For any subset of B, DB, we can define the set of pre-images

of elements from D, which we denote as f-1(D):

f-1(D) = {x | xA and f (x)D }.

Pay attention that f-1(D) is just a notation for the set of pre-images,

which does not assumes the existence of inverse function for f.

A

a

x

b

y

c

z

Example: A = {a, b, c}, B ={x, y, z}, and f (a)=x, f (b)=y, f (c)=y.

In this example, function f is neither injective nor surjective.

So, inverse function f -1 does not exist (in other words, inverse

relation of f is not a function). But we can find a set of pre-images

for any subset of B. For example:

f -1({z})=; f -1({x, z})={a}; f -1({x, y}) = {a, b, c}.

• Is it always true, that f (A) = B?

Ans. It is true only if f is surjective. Otherwise f (A)B.

• Is it always true that f -1(B)=A?

• Ans. Yes, since f is a function, any element of A has an image

• that belongs B. As a result any element of A appears as a pre-image

• of some element of B.

• Is it true that if f(C)=f(D), where C, DA, then C=D?

Ans. This is true only if f is injective.

• Is it true that if f -1(C)=f -1(D), where C, D  B, then C=D?

Ans. This is true only if f is surjective

• Example. Prove that if f is injective, then XY= implies

• f(X)f(Y)=.

f

B

A

X

f (X)

Y

f (Y)

Proof by contradiction. Assume that XY= and

f(X)f(Y), (1).

(1) implies that there exists some common element,

zf(X) and z f(Y), (2).

From the definition of the set f(X) and f (Y) (2) implies that

there exists xX, such that f(x)=z (3)

and there exists yY, such that f(y)=z (4).

So, we have f(x)= f(y)=z, (5)

and since f is injective (5) implies that

x=y, (6).

Since xX and yY , (6) implies that XY in contradiction

with assumption.

The contradiction proves the initial statement.