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Homework Help: Continuous topology problem

  1. Jun 2, 2010 #1
    Let [itex](X; T ) [/itex] be a topological space. Given the set Y and the function [itex]f : X \rightarrow Y [/itex], define

    [itex]U := {H\inY \mid f^{-1}(H)\in T}[/itex]

    Show that U is the finest topology on Y with respect to which f is continuous.

    2. Relevant equations



    3. The attempt at a solution

    I was wondering is this implying that [itex]U[/itex] is the Quotient topology?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 2, 2010 #2

    Dick

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    Re: Topology

    No, it's not a quotient topology. You aren't identifying any points. Suppose you give Y a finer topology than U? Use the topological definition of continuity.
     
  4. Jun 2, 2010 #3
    Re: Topology

    Thanks
     
  5. Jun 2, 2010 #4

    Hurkyl

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    Re: Topology

    Yes he is, via the equivalence relation P~Q iff f(P)=f(Q).
     
  6. Jun 2, 2010 #5
    Re: Topology

    So if it is a quotient topology will I need to prove the iff statement
     
  7. Jun 2, 2010 #6
    Re: Topology

    I mean do I have to show that if there is an equivalence relation I need to show that there is a function which is a homeomorhism.
     
  8. Jun 2, 2010 #7

    Dick

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    Re: Topology

    What is ~ supposed to mean?! f:R->R with the usual topology. f(x)=x^2. You are defining a topology on Y. Of course, f(1)=f(-1). How is 1~(-1)?
     
  9. Jun 3, 2010 #8

    Hurkyl

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    Re: Topology

    ~ is a symbol denoting an equivalence relation, which I subsequently defined.

    For any function f : X -> Y, Ker(f) is the set of pairs (a,b) for which f(a)=f(b).

    Theorem (First isomorphism theorem): Ker(f) is an equivalence relation on X. The corresponding set of equivalence classes is canonically bijective with the image of f


    For any surjective continuous function f:X -> Y, it's fair to ask if the canonical bijection X/Ker(f) -> Y is a homeomorphism, in which case it would be fair to say that Y has the quotient topology.
     
  10. Jun 3, 2010 #9

    Dick

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    Re: Topology

    Ok, so it's a quotient in some sense on XxY if I get your drift. I still think this is a bit of a distraction from the gloriously simple approach of just picking a topology strictly finer than U and then showing f is not continuous in that topology.
     
  11. Jun 4, 2010 #10

    Hurkyl

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    Re: Topology

    I wasn't trying to suggest an approach to the problem; I just didn't want him to be misinformed about quotient topologies. (Incidentally, for the OP, the definition Wikipedia gives of "quotient topology" is exactly the topology you wrote... with the extra condition that f is supposed to be surjective)
     
  12. Jun 4, 2010 #11
    Re: Topology

    Thanks for your help guys. In my textbook is says that the quotient topology is the finest. However in my tutorials when ever they talked about quotient topologies they always mentioned something about an equivalence relation. I think this question was meant to make me think about it which I did. Once again thanks
     
  13. Jun 4, 2010 #12

    Dick

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    Re: Topology

    I see what you mean now. Thanks for clarifying.
     
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