Let $X$ be a CW complex and $\Phi : D \rightarrow \bar e$ be the characteristic map for an open cell $e$.

I wonder whether $\Phi$ is a quotient map. I konw it is surjective. But I cannot prove that $\bar e$ has the quotient topology induced by $\Phi$. I can not find a counter example for this.

If it is true, I want a proof. If not, a counter example.


I saw two different definitions of Weak topology. One is that $A \subset X$ is open if and only if $A \cap \bar e$ is open for each $e$[Lee's Topological manifolds]. The other is that $A \subset X$ is open if and only if $\Phi ^{-1}(A)$ is open for each $\Phi$[Brendon's Toplogy and Geometry]. If these are equivalent, it implies,I think, that $\Phi$ is a quotient map.


$\Phi$ is surjective and closed, hence is a quotient map.

Note that the closed sets in its domain are compact so are sent by $\Phi$ to compact sets wich on their turn are closed set in its codomain wich is Hausdorff.

  • $\begingroup$ Thanks!! It is just a closed map lemma! $\endgroup$ – Jeong Jul 27 '15 at 10:42
  • $\begingroup$ I looked at your profile and noticed that you never accepted answers to your questions (15). Why not? I am not urging you to accept my answer, but you must be attended on this fact. $\endgroup$ – drhab Jul 27 '15 at 10:52

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