Properties of Composite Functions

[sara_87]

Question:

Let f: A -> B be a bijective function. Prove that

f o f^(-1) = IB

(IB = I lowercase B )

Answer:

Let x be an arbitrary element in the set A and f(x)=y . Therefore x is in A because f: A -> B.

f o f^(-1)
=f(f^(-1) (x))

and that's as far as i got

any help would be appreciated

thank you

 

[cristo]

Well, try taking a in A and b in B s.t f(a)=b. Use the fact that f is bijective, and try applying f^-1 to b, and then f to the result of this.

edit: you could, instead, carry on from here:

Let x be an arbitrary element in the set A and f(x)=y .

f o f^(-1)
=f(f^(-1) (x))

However, apply the compostie function to y. What does the bijectivity of f say about f^-1(y)?

 

[sara_87]

'Well, try taking a in A and b in B s.t f(a)=b. Use the fact that f is bijective, and try applying f-1 to b, and then f to the result of this.'


i said (what you said) let a belong to A and b belong to B such that f(a) = b

f is bijective => it is surjective so every element in A has an image in B

s.t f(a) = b

f^(-1) b = a

and (f o f^(-1)) = f(a)
f(a) = b

so f o f^(-1) = IB

it works!

thanx

 

[cristo]

'Well, try taking a in A and b in B s.t f(a)=b. Use the fact that f is bijective, and try applying f-1 to b, and then f to the result of this.'


i said (what you said) let a belong to A and b belong to B such that f(a) = b

f is bijective => it is surjective so every element in A has an image in B

s.t f(a) = b

f^(-1) b = a

and (f o f^(-1)) b= f(a)
f(a) = b

so f o f^(-1) = IB

it works!

thanx

You're welcome. Note that you missed out the element the function is acting on in the above proof (the bit in bold). Apart from that, it's correct.

 

[HallsofIvy]

By the way, you didn't mean "I lowercase B". That would be Ib and it wouldn't make sense to write it as IB to begin with. What you meant was "I subscript B". And it would have been even better to just write "the identity function from B to itself". Actually, the fact that f(f^-1)= I_b or that f(f^-1(x)= x is part of the definition of f^-1. What you are really proving is that f^-1 exists.

 

[sara_87]

By the way, you didn't mean "I lowercase B". That would be Ib and it wouldn't make sense to write it as IB to begin with. What you meant was "I subscript B". And it would have been even better to just write "the identity function from B to itself".

oh...next time i'll right it properly

Actually, the fact that f(f^-1)= I_b or that f(f^-1(x)= x is part of the definition of f^-1. What you are really proving is that f^-1 exists.

i don't know what that means but because you tried to help thanx!

 

[HallsofIvy]

What is the definition of f^-1?

 

[sara_87]

you know the proof that i wrote does it not prove what the question asks us to?
actually i think i know what you're saying now...i didn't show that f o f^-1 of an element goes to the identity function B

 

[cristo]

Yes, it does prove what the questions asks you. You have shown that f(f^-1(x))=x for all x in A. Hence f o f^-1=I_B.

I think what HOI meant was that the definition of f^-1 states that when composed with f it is the identity function. (This is the point of the inverse function.) However, this is only if the inverse exists. The question really asks you to show that it exists.

 

[HallsofIvy]

Definition of inverse functions:f and g are "inverse" functions if and only if f(g(x))= x for all x in the domain of g and g(f(x))= x for all x in the domain of y. That was my point. The fact that f(f^-1(x))= x is part of the definition. What you are really proving is that a function satisfying that exists.

 

[sara_87]

You know this could of stopped at post #4, but i know HOI you wanted me to open my eyes and think about what i did and the question...so whilst answering the question i proved:

f^(-1) when composed with f it is the identity function?

i understood this from the beginning...this is so confusing!

anway i did it right and if something similar comes in the exam i think i will know what to do

(no one knows this but: I'm actually only studying maths because it is convenient; i really find some parts of it interesting but i only did it because i got good grades (i know my posts don't show this but it's true) not because i enjoy it, otherwise i would have studied art)