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Bijection proof (intro analysis)

  1. Nov 24, 2013 #1
    I'm wondering whether my solution to this problem is correct, since the answer in the answer sheet says that this is provable, and I think I found a counterexample... Any help is appreciated :)

    1. The problem statement, all variables and given/known data


    Prove or disprove: ##f:D->K##, where ##D,K != empty##, is a bijection iff there exists a unique function ##g:K->D## such that:
    ##f\circ{g}\circ{f}=f##



    2. Relevant equations



    3. The attempt at a solution



    Lets look at the following function: ##f:\{1\}-->\{1,2\}##, defined by ##f(1)=1##. Clearly the function##g:\{1,2\}-->\{1\}##, defined by ##g(1)=1## and ##g(2)=1## satisfies ##f\circ{g}\circ{f}=f##, since ##f(g(f(1)))=f(g(1))=f(1)##, and 1 is the only element in the domain of f. Lets assume that ##g':\{1,2\}-->\{1\}## is a function that satisfies ##f\circ{g'}\circ{f}=f##. The only way that g' is even a function is if we define it as ##g'(1)=1##and ##g'(2)=1##, so ##g'=g##, so ##g## is a unique function ##K-->D## which satisfies the condition.
    But f is not a surjection, so we have found a function ##f:D-->K## which is not a bijection, but for which there exists a unique function ##g:K-->D## such that ##f\circ{g}\circ{f}=f##, so the proposition does not stand.
     
    Last edited: Nov 24, 2013
  2. jcsd
  3. Nov 24, 2013 #2

    Dick

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    Science Advisor
    Homework Helper

    Sounds correct to me. There is only one function g:{1,2}->{1}, but f is not a surjection.
     
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