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Math Help - one to one function proof?

  1. #1
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    one to one function proof?

    Let f: A-->B, g:B--->C be functions. Prove that if g o f is one to one then f is one to one.

    Proof by contrapositive

    Assume that f is not one to one then f(a1)≠f(a2) where a1,a2 belong to set A. We also assume a1≠a2. Let f(a1)=r and f(a2)=s where r and s belong to set B. Then that means r≠s. Hence (g o f)(a1)≠(g o f)(a2)=g(f(a1)≠g(f(a2)=
    g(r)≠g(s). Thus g o f is not one to one.

    Did I do this right? When a function is not one to one do i get to assume that f(x)
    ≠f(y) where x,y belong to domain and furthermore x≠y?

    ----------------------------------
    Remark
    I did this wrong. I forgot that when you negate the one to one definition it becomes A function f: A--->B is not one one when f(x)=f(y) where x,y are elements of A and x
    ≠y

    Edit to proof
    Assume that f is not one to one then f(a1)=f((a2) where a1,a2 belong to set A. We also assume a1≠a2. Let f(a1)=r and f(a2)=s where r and s belong to set B. Then that means f(a1)=f(a2). Hence (g o f)(a1)=(g o f)(a2)<=>g(f(a1)=g(f(a2)<=>
    g(r)=g(s) but r≠s. Thus g o f is not one to one.


    is it right now?
    Last edited by bonfire09; December 11th 2012 at 06:38 PM.
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  2. #2
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    Re: one to one function proof?

    Quote Originally Posted by bonfire09 View Post
    Let f: A-->B, g:B--->C be functions. Prove that if g o f is one to one then f is one to one.
    Why not use the simple straightforward proof,
    \begin{gathered}  f(a) = f(b) \hfill \\  g \circ f(a) = g \circ f(b) \hfill \\  a = b \hfill \\ \end{gathered}
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  3. #3
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    Re: one to one function proof?

    Well the problem wants me to do it as many different ways I can. I already did a direct proof.
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  4. #4
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    Re: one to one function proof?

    Quote Originally Posted by bonfire09 View Post
    Well the problem wants me to do it as many different ways I can. I already did a direct proof.
    Then that your problem not ours!
    You should have made that clear in the OP.
    Good luck!
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  5. #5
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    Re: one to one function proof?

    wait but i originally wrote to check if my proof was right in my remark. So please check it and give me any feedback.
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