Timeline for Does anyone know a basepoint-free construction of universal covers?
Current License: CC BY-SA 4.0
35 events
| when toggle format | what | by | license | comment | |
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| Nov 13, 2019 at 12:54 | answer | added | John Klein | timeline score: 5 | |
| Nov 10, 2019 at 13:27 | comment | added | David Roberts♦ | @Tom ah, that seems likely, now you mention it. I'll raise this at the nForum. I think I recall who wrote that (a few years back now!) | |
| Nov 10, 2019 at 12:31 | comment | added | Tom Goodwillie | @David Roberts I think that that nlab discussion has an error (in the part that begins by saying that the dependence on the basepoint is spurious). That equalizer is taking the covering spaces $B_b$ associated to all possible basepoints $b\in B$ and gluing them together using all possible paths between one basepoint and another. This is equivalent (in the connected case) to taking a universal cover and dividing by the group of deck transformations. | |
| Nov 10, 2019 at 2:26 | vote | accept | Kim | ||
| Nov 10, 2019 at 2:26 | vote | accept | Kim | ||
| Nov 10, 2019 at 2:26 | |||||
| Nov 9, 2019 at 6:25 | comment | added | David Roberts♦ | @Tom I believe you, but the OP asked for a construction of the universal covering space without choice of basepoint, and the page I linked does that. It doesn't claim that the projection map is natural or anything stronger than merely constructing $\widetilde{X}$. | |
| Nov 8, 2019 at 22:02 | comment | added | HJRW | @DavidESpeyer — yes, I forgot to mention that you should pass to a connected component. I don’t know if Zorn is needed — your Zornless argument seems OK to me. | |
| Nov 8, 2019 at 20:11 | answer | added | Kevin Walker | timeline score: 7 | |
| Nov 8, 2019 at 16:03 | answer | added | Tom Goodwillie | timeline score: 27 | |
| Nov 8, 2019 at 14:32 | comment | added | David E Speyer | Now I do need to single out a point $x \in X$ for a moment: We know that $F$ is nonempty because it contains the point which is $(x,0)$ in every $Y_i$. But now we can throw that point away again and choose a connected component of $F$ (fortunately, since $X$ is connected, I only need to choose once). It shouldn't be too hard to show that this is a cover and every cover factors through it; of course, without a basepoint, there is no uniqueness to the factorization. | |
| Nov 8, 2019 at 14:27 | comment | added | David E Speyer | Here is a version which I think does not use Zorn. Let $X$ be a nonempty topological space. Let $\mathcal{I}$ be the set of all topologies on $X \times \mathbb{Z}$ for which the projection map $\pi: X \times \mathbb{Z} \to X$ is a cover. Since a topology is a subset of the power set of $X$, the collection $\mathcal{I}$ really is a set. For $i \in I$, I'll write $Y_i$ for the corresponding topological space and $\pi_i : Y_i \to X$ for the covering map. Let $F = \{ (y_i) \in \prod_{i \in \mathcal{I}} Y_i : \pi_i(y_i) = \pi_j(y_j) \forall i,j \in \mathcal{I} \}$. (continued) | |
| Nov 8, 2019 at 14:19 | comment | added | David E Speyer | @HJRW A correction and a comment on HJRW's answer: A fiber product of connected covers need not be connected, so you need to take a connected component of the fiber product. Also, I at first was skeptical that Zorn was needed, but then I realized I wasn't sure how to prove the product over all covers was nonempty without Zorn, so maybe you do? | |
| Nov 8, 2019 at 13:23 | comment | added | HJRW | Probably this is implicit in some of the things already written, but perhaps it's worth saying explicitly. The answer depends on what you think a universal cover is. If it's a cover that covers all covers, then you can just note that the fibre product of a family of covers is a common cover, and then apply Zorn's lemma. But it's not clear that the resulting space is simply connected, and I suspect you need base points for that. | |
| Nov 8, 2019 at 12:57 | comment | added | Tom Goodwillie | @David Roberts: It is explained at nlab that covering spaces are equivalent to functors from the fundamental groupoid to Set. But this means that making a universal covering space (of connected $X$) without using a basepoint corresponds to taking any connected groupoid and canonically making a functor to Set such that for each object in the groupoid the associated action of the corresponding group is free and transitive. If you could do that in a way that was functorial, even with respect to equivalences of groupoids, then I believe you would have a contradiction. | |
| Nov 8, 2019 at 12:49 | comment | added | Tom Goodwillie | To justify my assertion: The group of diffeomorphisms $S^1\to S^1$ has no compatible action on $\mathbb R$ (i.e. none satisfying the naturality that I require, i.e. no action such that the projection $\mathbb R\to S^1$ intertwines it with the action on $S^1$). Even the subgroup generated by a rotation $R:S^1\to S^1$ of order $2$ has no such action. | |
| Nov 8, 2019 at 12:28 | history | became hot network question | |||
| Nov 8, 2019 at 12:16 | answer | added | Ronnie Brown | timeline score: 11 | |
| Nov 8, 2019 at 11:58 | comment | added | David Roberts♦ | Sure there is: ncatlab.org/nlab/show/… | |
| Nov 8, 2019 at 11:37 | answer | added | Neil Strickland | timeline score: 9 | |
| Nov 8, 2019 at 7:41 | comment | added | Kim | @RyanBudney So it amounts to a failure to embed $S^1$ into $\mathbf{R}$? | |
| Nov 8, 2019 at 7:35 | comment | added | Ryan Budney | @Kim: perform the natural map associated to the universal cover for every translation of the circle. | |
| Nov 8, 2019 at 7:11 | comment | added | Kim | @TomGoodwillie How do we prove this? | |
| Nov 8, 2019 at 5:23 | comment | added | R. van Dobben de Bruyn | I realised that I'm a little unhappy with my earlier comment, because the fundamental group $\pi_1(X)$ itself relies on a choice, so is only well-defined up to inner automorphism. I believe that this choice and the choice of the universal cover, although related, should not cancel out (as should become visible in the abelian case). Maybe someone who speaks $\infty$-groupoids should write a coherent answer [pun intended]. | |
| Nov 8, 2019 at 5:13 | comment | added | Tom Goodwillie | If you stipulate that the construction should be functorial with respect to diffeomorphisms and that the map $\tilde X\to X$ should be natural, then it's impossible even for $S^1$. | |
| Nov 8, 2019 at 3:40 | comment | added | B K | It does, but we can ignore this just as we can forget that $\mathbb R$ has a canonical basepoint. Then $S^1$ is just a circle in the "plane without a coordinate system". | |
| Nov 8, 2019 at 3:37 | comment | added | Kim | @BK If you take that as your definition of $S^1$, doesn't it come with a canonical basepoint (i.e. the identity element 1 in the group)? | |
| Nov 8, 2019 at 3:30 | comment | added | B K | Addition to my previous comment: Here I had the definition $S^1:=\{x\in \mathbb C: |x|=1\}$ in mind. If we define $S^1:=\mathbb R / \mathbb Z$ then of course the covering map is canonical and needs no basepoint as $\mathbb Z$ acts on $\mathbb R$ even if we consider the latter merely as an affine line. | |
| Nov 8, 2019 at 3:22 | comment | added | B K | @ Kim: Of course you're right. I should have more precisely said that I doubt one can write down a covering map $\mathbb R \to S^1$ without choosing a basepoint. I hereby challenge anyone reading this comment to do so ;-) | |
| Nov 8, 2019 at 3:13 | comment | added | Kim | @BK What do you mean? Doesn't a universal cover come with the data of a map $X \longrightarrow Y$? | |
| Nov 8, 2019 at 2:56 | comment | added | B K | Even if there was such a basepoint-free construction, I doubt that one could verify for two given spaces $X$ and $Y$ that $X$ is the universal cover of $Y$ without choosing a basepoint in $Y$. Already for $X=\mathbb R$ (considered merely as an affine line with the canonical basepoint $0$ ``forgotten'') and $Y=S^1$ this seems difficult to me. | |
| Nov 8, 2019 at 1:58 | comment | added | R. van Dobben de Bruyn | Just like there is no the algebraic closure of a field, I think there should not be a the universal cover of a space. Any two constructions are isomorphic, but the set of isomorphisms (over $X$) is a torsor under $\pi_1(X)$. | |
| Nov 8, 2019 at 1:53 | comment | added | LSpice | If $X$ is path connected, and $\tilde X_p$ is the universal cover constructed from the chosen point $p$, then it seems like you could probably give a description of $\bigsqcup_p \tilde X_p$ that doesn't involve choosing $p$ (all paths in $X$, up to endpoints-fixed homotopy, maybe?), and then impose an equivalence relation that puts $\xi_p \in \tilde X_p$ equivalent to $\xi_q \in \tilde X_q$ if there is some path $\xi_{pq}$ from $p$ to $q$ such that $\xi_p$ is the concatenation of $\xi_{pq}$ and $\xi_q$. | |
| Nov 8, 2019 at 1:01 | comment | added | Kim | @NoamD.Elkies I'm happy to assume connected, if it helps. | |
| Nov 8, 2019 at 0:55 | comment | added | Noam D. Elkies | Must $X$ be connected? | |
| Nov 8, 2019 at 0:52 | history | asked | Kim | CC BY-SA 4.0 |