Timeline for Factoring a topological universal cover
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Oct 31, 2019 at 16:18 | comment | added | user43326 | Or, you can simply use the classification of coverings. | |
| Oct 31, 2019 at 15:35 | comment | added | user43326 | Since there is a continuous map from one to the other... | |
| Oct 31, 2019 at 15:17 | comment | added | Jackson Morrow | @user43326 I believe this this should work, but I am a bit confused about how you show that the topology of $\tilde{X}/\mathbb{Z}$ is finer than that of $Y$? Could you expand on this point? | |
| Oct 31, 2019 at 10:27 | comment | added | user43326 | The topology of $Y$ is finer than that of $X$. But the topology of $\tilde{X}/{\mathbb Z}$ is finer than that of $Y$. However, $\tilde{X}/{\mathbb Z}$ is nothing but $X$. So $X$ and $Y$ should be homeomorphic, or am I missing something? | |
| Oct 31, 2019 at 0:30 | review | Close votes | |||
| Nov 4, 2019 at 3:05 | |||||
| Oct 30, 2019 at 22:45 | comment | added | YCor | @user44191 what you prove is what I just said, namely that $a$ is bijective. A little more step is needed to obtain that $a$ is a homeomorphism. | |
| Oct 30, 2019 at 22:37 | comment | added | Jackson Morrow | @YCor Sorry about that! I meant a morphism of topological spaces | |
| Oct 30, 2019 at 22:36 | history | edited | Jackson Morrow | CC BY-SA 4.0 |
added details to question
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| Oct 30, 2019 at 22:09 | comment | added | YCor | It easily follows from the assumptions that $a$ is bijective. | |
| Oct 30, 2019 at 21:53 | comment | added | YCor | @KevinCasto $X$ is a group by assumption (I agree the notation is misleading). | |
| Oct 30, 2019 at 21:50 | comment | added | YCor | What do you mean by "morphism $Y\to X$"? morphism of what? | |
| Oct 30, 2019 at 21:49 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Oct 30, 2019 at 20:27 | history | asked | Jackson Morrow | CC BY-SA 4.0 |