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Quotient map closed?

  1. Aug 11, 2007 #1
    1. The problem statement, all variables and given/known data
    THe quotient map f is open but is it also closed?




    3. The attempt at a solution
    I think it is. Consider f: X->Y

    FOr every open set V in Y there exists by definition an open set f^-1(V) in X. There is a one to one correspondence between open sets in X and open sets in Y by definition.

    So for every closed set V complement in Y there exists a closed set f^-1(V) complement in X. So f is both closed and open.
     
    Last edited: Aug 11, 2007
  2. jcsd
  3. Aug 13, 2007 #2

    morphism

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    Are you asking whether or not an open quotient map is closed? If so, then what you did does not prove that. There isn't a 1-1 correspondence between open sets in X and open sets in Y. Look carefully at the definition. If f:X->Y is a quotient map, then U is open in Y iff f^-1(U) is open in X. This does not exhaust all the open sets in X. So you haven't proved that f takes closed sets to closed sets.

    Anyway, this is false. For a counterexample, let [itex]\pi_1 : \mathbb{R}^2 \to \mathbb{R}[/itex] be the projection map onto the first coordinate. Then [itex]\pi_1[/itex] is an open surjection, so it's a quotient map. However it's not closed. (I'll let you find a closed set that doesn't get mapped to a closed set.)
     
    Last edited: Aug 13, 2007
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