Results 1 to 8 of 8

Math Help - Proving bijective functions

  1. #1
    Newbie
    Joined
    Nov 2007
    Posts
    7

    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).


    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by IIuvsnshine View Post
    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).


    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.
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2007
    Posts
    7
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,908
    Thanks
    1759
    Awards
    1
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2007
    Posts
    7
    I understand the definitions...I'm struggling with wording for the proof.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Aug 2007
    Posts
    239
    Its an  \Leftrightarrow statement so you have 2 directions to go in.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,908
    Thanks
    1759
    Awards
    1
    Quote Originally Posted by IIuvsnshine View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Nov 2007
    Posts
    7
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Bijective functions
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: April 1st 2011, 02:14 PM
  2. Range and inverse of bijective functions
    Posted in the Algebra Forum
    Replies: 22
    Last Post: November 16th 2009, 01:25 PM
  3. Replies: 1
    Last Post: November 1st 2009, 03:35 PM
  4. Replies: 1
    Last Post: September 21st 2009, 09:01 PM
  5. Proving : existance of the inverse/ bijective
    Posted in the Calculus Forum
    Replies: 8
    Last Post: May 13th 2009, 08:28 PM

Search Tags


/mathhelpforum @mathhelpforum