0
$\begingroup$

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.

[Added]

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.

$\endgroup$
4
$\begingroup$

$\Phi$ 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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.