Cover $~E \to B~$ gives a homeomorphism $~E/Aut(E) \to B~$

Question

Let $p : E \rightarrow B$ be a cover s.t. Aut(E) acts transitively on $p^{-1}(b)$ for some fix $b \in B$. Then $E/Aut(E)$ is homeomorphic to $B$ where $Aut(E) \subset Cov(E,E)$. The problem is that I don't even understand how the map $~E/Aut(E) \to B~$ is defined, would someone please give me this definition?

Apparently it is "clear" but I fail to see this.

$\widetilde{x} \mapsto p(x)$? (Where $\widetilde{x}$ is the class of $x$ in $E/!\mathrm{Aut}(E)$.) — Agustí Roig, Sep 17 '10 at 13:20

score 5 · Accepted Answer · answered Sep 22 '10 at 10:44

5

Let's remove this question from the list of unanswered questions by up-voting this community wiki answer stating that Agusti Roig's comment above contains the right answer.

answered Sep 22 '10 at 10:44

Rasmus

18,404

score -4 · Answer 2 · edited Apr 13 '17 at 12:21

-4

Let's honor Rasmus' creative handling of the software restrictions by downvoting this self-referential community wiki answer.

edited Apr 13 '17 at 12:21

Community

1

answered Sep 22 '10 at 15:07

Who

270

Funny, for a moment I thought you got this idea from http://mathoverflow.net/questions/39626/is-there-a-general-setting-for-self-reference but apparently that post turned up later. – Rasmus Sep 22 '10 at 17:31

Cover $~E \to B~$ gives a homeomorphism $~E/Aut(E) \to B~$

2 Answers2