# Proving bijective functions

• Nov 29th 2007, 11:11 AM
IIuvsnshine
Proving bijective functions
I have 2 I am struggling with:

1. Let f:A to B. Prove that f is a bijection iff for all b that is an element of B, there exists exactly one a that is an element of A f(a)=f(b).:confused:

2. Prove: If f:A to B is a bijection and g:B to C is a bijection, then the composition of g of f is a bijection.:confused:
• Nov 29th 2007, 11:17 AM
topsquark
Quote:

Originally Posted by IIuvsnshine
I have 2 I am struggling with:

1. Let f:A to B. Prove that f is a bijection iff for all b that is an element of B, there exists exactly one a that is an element of A f(a)=f(b).:confused:

2. Prove: If f:A to B is a bijection and g:B to C is a bijection, then the composition of g of f is a bijection.:confused:

You appear to have a typo. Shouldn't it read:
"Let f:A to B. Prove that f is a bijection iff for all b that is an element of B, there exists exactly one a that is an element of A f(a)=b."?

I'm not sure how much of a help this will be as I don't know what problems you are having with these, but in general to show a function is bijective you need to show that it is both injective and surjective. My advice is to simply apply the definition of each.

For example, the function in problem 1 is clearly surjective, since for every element b of B there exists (at least) one element of A such that f(a) = b.

-Dan
• Nov 29th 2007, 11:48 AM
IIuvsnshine
Yes, I did have a typo.

My concern is the proof (lol).

Do I need to approach this by cases?

For the 1st part I have:

Suppose f is a bijection. Then f is injective and surjective.
Case 1: Suppose f is injective. Then for all a,b that is an element of A, f(a)=f(b).
Case 2: Suppose f is surjective. Then for all of b that is an element of B, there exists an a such that f(b) = a.

I'm not sure I'm headed down the correct path.
• Nov 29th 2007, 11:54 AM
Plato
A bijection is surjective injection.

A function is a surjection if and only if each element of the final set, in this case B, is the image of some element in the initial set A.

A function is an injection if and only if no two of its pairs have the same second term.

From the given does the function f have those two properties?
• Nov 29th 2007, 12:05 PM
IIuvsnshine
I understand the definitions...I'm struggling with wording for the proof.
• Nov 29th 2007, 12:08 PM
shilz222
Its an $\Leftrightarrow$ statement so you have 2 directions to go in.
• Nov 29th 2007, 12:11 PM
Plato
Quote:

Originally Posted by IIuvsnshine
I understand the definitions

Well that is not at all clear to me that you do understand how to use them.

You are given that each b is the image of some a therefore f is onto.

You are given that each b is the image of exactly one a therefore f is one-to-one.
• Nov 29th 2007, 12:31 PM
IIuvsnshine
It is sometimes difficult to state exactly what the meaning of words typed into a thread.
I do understand the definitions. I apologize if that is not clear to you. I also understand that i'm given a b that is an image of a such that f is onto, and that i'm given a b that is the image of exactly one a such that f is one-to-one.