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F maps A onto A=>f is one to one

  1. Sep 25, 2009 #1
    Say f:A->A where A is a metric space and f is onto. I think it should be true that this implies that f is also one to one. Is there a way to formally prove this?
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  3. Sep 25, 2009 #2


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    There is no way to prove it since it is false, and it is impossible to prove untrue things.
    Try to construct a counter example. Hint it is easy.
  4. Sep 25, 2009 #3


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    Let's try starting with something simple!

    The contrapositive is often something useful to consider:
    f is not one-to-one ==> A is not a metric space​

    What is the simplest example you can think of of an onto function that is not one-to-one? Can you prove the domain of that function cannot have a metric space structure on it?

    (P.S. what is f? Any function at all?)
  5. Sep 25, 2009 #4
    what does metric space even have to do with it?

    i was about to construct a counter example but maybe there's something to this.

    if A={a,b,c} and the range of f is all of A i don't see how one can define f in such a way that it's not injective because f(a) = a and f(a) = (b) isn't allowed right?
  6. Sep 25, 2009 #5


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    If A is finite and f is onto, then f must be injective
  7. Sep 25, 2009 #6
    yea i figured that out shortly after posting
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