Quotient maps

**aharonidan** · May 29th 2011, 12:25 AM

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!

**Drexel28** · May 29th 2011, 01:05 AM

Originally Posted by aharonidan

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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