# Quotient maps

• May 29th 2011, 12:25 AM
aharonidan
Quotient maps
Hello math experts...
Let $\displaystyle X,Y,Z$ be topological spaces.
Let $\displaystyle \pi : X \rightarrow Y$ be quotient map.
Let $\displaystyle f : X \rightarrow Z$ be continues.
Find a necessary and sufficient condition on $\displaystyle f$ for the existence of a continues function $\displaystyle g: Y \rightarrow Z$ such that $\displaystyle g \circ \pi = f$

thanks!
• May 29th 2011, 01:05 AM
Drexel28
Quote:

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

thanks!

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