Timeline for Constructions unique up to non-unique isomorphism
Current License: CC BY-SA 2.5
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 17, 2011 at 4:10 | comment | added | Tom Goodwillie | There cannot be any such construction. The group of homeomorphisms from the circle to itself has no compatible action on the (or should I say "a") universal covering space. | |
| May 17, 2011 at 1:38 | answer | added | algori | timeline score: 4 | |
| Mar 9, 2011 at 12:40 | answer | added | Emil Jeřábek | timeline score: 6 | |
| Jan 31, 2011 at 18:57 | comment | added | Chris Schommer-Pries | @ David - this coequalizer still just gives B back. Try an explicit example like B = circle. The point is that there are many paths and so the different universal covers are identified with eachother in more than one way. This forces you to take a quotient of the universal covers which is too small (namely B itself) and no longer a universal cover. You can see that this has to be the case because the paths from b to b (up to homotopy) are just $pi_1$ and so this colimit factors through the quotient by the action of $pi_1$. Agreed? Where in the n-lab is this written? | |
| Jan 31, 2011 at 9:02 | answer | added | Fernando Muro | timeline score: 12 | |
| Jan 31, 2011 at 5:31 | answer | added | ARupinski | timeline score: 6 | |
| Jan 30, 2011 at 20:50 | answer | added | David Roberts♦ | timeline score: 4 | |
| Jan 30, 2011 at 20:43 | comment | added | David Roberts♦ | @Chris - hmm, perhaps I worded it wrongly. You take the disjoint union of all the covering spaces of the based spaces, and glue them together by the isomorphisms induced by paths. Here's the relevant bit from the nLab ($B$ is the base space in question) the coequalizer of the pair of arrows $\sum_{[\phi]: b \to c} \sum_c B^{(1)}_c \overset{\to}{\to} \sum_c B^{(1)}_c$ in $Top/B$, where one arrow is projection and the other is given by the action of pulling back along classes of paths | |
| Jan 30, 2011 at 18:36 | comment | added | Nick S | Doesn't this always happen when the construction has not-trivial automorphisms? | |
| Jan 30, 2011 at 16:46 | answer | added | Andrés E. Caicedo | timeline score: 7 | |
| Jan 30, 2011 at 16:14 | answer | added | John Palmieri | timeline score: 8 | |
| Jan 30, 2011 at 15:07 | answer | added | Steven Landsburg | timeline score: 12 | |
| Jan 30, 2011 at 14:46 | comment | added | Chris Schommer-Pries | @ David: "the colimit of this diagram is the universal covering space of X", are you sure this is correct? What you described is essentially the groupoid of universal covering spaces and covering space maps inside Top and you are taking the colimit inside Top. This is the same as taking a single covering space and taking the quotient by all covering automorphisms, so you just get X back. | |
| Jan 30, 2011 at 14:41 | answer | added | Theo Buehler | timeline score: 4 | |
| Jan 30, 2011 at 14:07 | answer | added | Qiaochu Yuan | timeline score: 14 | |
| Jan 30, 2011 at 10:17 | answer | added | Chris Heunen | timeline score: 3 | |
| Jan 30, 2011 at 9:59 | comment | added | David Roberts♦ | It is possible to functorially define the universal covering space of a non-pointed space X: take the diagram $\Pi_1(X) \to Top$ sending a point $x\in X$ to the (pretty well unique) covering space one gets from the pointed space $(X,x)$, and a homotopy class of paths to the map between covering spaces this induces. The colimit of this diagram is the universal covering space of $X$, and this time this is functorial wrt maps $X\to Y$. I learned this from Todd Trimble, I think. | |
| Jan 30, 2011 at 9:46 | answer | added | Martin Brandenburg | timeline score: 13 | |
| Jan 30, 2011 at 8:15 | history | edited | David Feldman | CC BY-SA 2.5 |
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| Jan 30, 2011 at 7:53 | history | asked | David Feldman | CC BY-SA 2.5 |