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Math Help - Quotient maps

  1. #1
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    Quotient maps

    Hello math experts...
    Let X,Y,Z be topological spaces.
    Let \pi : X \rightarrow Y be quotient map.
    Let f : X \rightarrow Z be continues.
    Find a necessary and sufficient condition on f for the existence of a continues function g: Y \rightarrow Z such that g \circ \pi = f

    thanks!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by aharonidan View Post
    Hello math experts...
    Let X,Y,Z be topological spaces.
    Let \pi : X \rightarrow Y be quotient map.
    Let f : X \rightarrow Z be continues.
    Find a necessary and sufficient condition on f for the existence of a continues function g: Y \rightarrow Z such that g \circ \pi = f

    thanks!
    It just needs to be constant on the fibers \pi^{-1}(\{y\}), right? Suppose that f is constant on the fibers and define g:Y\to Z by taking g(y)=f(c(\pi^{-1}(\{y\}))) where c is some choice function (in simpler terms, if f is constant on the fibers of \pi then define a map from Y\to Z by taking each element y\in Y to the unique value of f taken on \pi^{-1}(\{y\}). This evidently satisfies g\circ\pi=f and is thus continuous by the universal property of quotient maps. The converse is equally as easy since evidently if f=g\circ \pi ( g must be continuous by the universal property again, there is no need to assume it is) then for any x_1,x_2\in\pi^{-1}(\{y\}) one has that f(x_1)=g(\pi(x_1))=g(\pi(x_2))=f(x_2) so that f is constant on the fibers.
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