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Let $X$ be a real manifold (for simplicity). The standard construction of the universal cover $\varphi: \widetilde{X} \longrightarrow X$ involves fixing a basepoint $p \in X$ and considering homotopy classes of paths from $p$ to $x \in X$.

Is there an alternative construction of $\varphi$ that avoids choosing a basepoint?

Kim
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    Must $X$ be connected? – Noam D. Elkies Nov 08 '19 at 00:55
  • @NoamD.Elkies I'm happy to assume connected, if it helps. – Kim Nov 08 '19 at 01:01
  • 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 $\xi_q$. – LSpice Nov 08 '19 at 01:53
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    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 $\pi_1(X)$. – R. van Dobben de Bruyn Nov 08 '19 at 01:58
  • 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. – B K Nov 08 '19 at 02:56
  • @BK What do you mean? Doesn't a universal cover come with the data of a map $X \longrightarrow Y$? – Kim Nov 08 '19 at 03:13
  • @ 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 ;-) – B K Nov 08 '19 at 03:22
  • 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. – B K Nov 08 '19 at 03:30
  • @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)? – Kim Nov 08 '19 at 03:37
  • 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". – B K Nov 08 '19 at 03:40
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    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 $S^1$. – Tom Goodwillie Nov 08 '19 at 05:13
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    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]. – R. van Dobben de Bruyn Nov 08 '19 at 05:23
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    @TomGoodwillie How do we prove this? – Kim Nov 08 '19 at 07:11
  • @Kim: perform the natural map associated to the universal cover for every translation of the circle. – Ryan Budney Nov 08 '19 at 07:35
  • @RyanBudney So it amounts to a failure to embed $S^1$ into $\mathbf{R}$? – Kim Nov 08 '19 at 07:41
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    Sure there is: https://ncatlab.org/nlab/show/reconstruction+of+covering+spaces+from+monodromy#InTermsOfACoend – David Roberts Nov 08 '19 at 11:58
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    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. – Tom Goodwillie Nov 08 '19 at 12:49
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    @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. – Tom Goodwillie Nov 08 '19 at 12:57
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    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. – HJRW Nov 08 '19 at 13:23
  • @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? – David E Speyer Nov 08 '19 at 14:19
  • 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) – David E Speyer Nov 08 '19 at 14:27
  • 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. – David E Speyer Nov 08 '19 at 14:32
  • @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. – HJRW Nov 08 '19 at 22:02
  • @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 $\widetilde{X}$. – David Roberts Nov 09 '19 at 06:25
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    @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. – Tom Goodwillie Nov 10 '19 at 12:31
  • @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!) – David Roberts Nov 10 '19 at 13:27

5 Answers5

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I think that homotopy-theorists often fall into the habit of working mainly with based spaces, even when they don't need to. It can be instructive to notice when the use of a basepoint is unnecessary, even artificial. But it's also important to notice the parts of the subject where the use of a basepoint is necessary. This (the topic of universal covering spaces) is one of those parts.

By "universal covering space" of a connected manifold $M$ I assume we mean a simply connected manifold $\tilde M$ with covering map $p:\tilde M\to M$. (By "simply connected" I mean connected and having trivial $\pi_1$ for one, hence any, basepoint. The empty space is not connected.)

There is always a universal covering space, and to explain how to make one we usually start by picking a point $x\in M$. Any two universal covering spaces, no matter how they are constructed, are related by an isomorphism, by which I mean a diffeomorphism that respects the projection to $M$. But this isomorphism is not unique, because for any such $(\tilde M,p)$ there is a group of isomorphisms $\tilde M\to \tilde M$ (i.e. deck transformations), a nontrivial group except in the case when $M$ itself is simply connected.

Suppose that there were a way of making a universal covering space $\tilde M$ that did not depend on a choice of basepoint (or any other arbitrary choice), and suppose that for $x\in M$ there was a canonical isomorphism between this $\tilde M$ and the one determined by $x$.

But this would imply that when we use two points $x\in M$ to make two universal covering spaces of $M$ then there is a canonical isomorphism between these.

Every homotopy class of paths from $x$ to $y$ in $M$ (homotopy with endpoints fixed) determines an isomorphism between the two covering spaces, and every isomorphism arises from exactly one such homotopy class. So if we had a canonical isomorphism we would have a canonical homotopy class of paths from $x$ to $y$. And surely we don't.

(That's not rigorous, because what does "canonical" mean? But surely if one had an actual recipe for making an $\tilde M$ for $M$ without first making some arbitrary choice then for any diffeomorphism $h:M_1\cong M_2$ the choice of canonical path classes in $M_1$ would be related by $h$ to the corresponding choice in $M_2$. In particular this would be the case for a reflection $S^1\to S^1$ that fixes two points $x$ and $y$ but of course does not fix any class of paths from $x$ to $y$.)

Tom Goodwillie
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Part of 10.5.8 of Topology and Groupoids is, in a more usual notation, essentially the following, in which $\sigma, \tau$ are the source and target maps, $St_G x$ is $\sigma ^{-1} x$, by $N$ is totally disconnected is meant that $N(x,y)$ is empty for $x \ne y$, and normality of $N$ in $G$ also means that $N,G$ have the same set of objects:

Let $X$ be a space which admits a universal cover, and let $N$ be a totally disconnected normal subgroupoid of the fundamental groupoid $\pi_1( X) $, Then the set of elements of the quotient groupoid $\pi_1(X)/N$ may be given a topology such that the projection $$q = (\sigma, \tau) : \pi_1(X)/N \to X \times X$$ is a covering map and for $x \in X$ the target map $\tau :St_{\pi_1( X)/N} \to X$ is the covering map determined by the normal subgroup $N(x)$ of $\pi_1(X, x)$.

So this uses all the points of $X$ and puts all these covers into a covering space, which means you don't make a choice of base point; instead you use all the choices. Further, $\pi_1(X)/N$ with this topology is actually a topological groupoid.

This may be the optimal way of answering the question.

I believe that you can do a similar trick with getting a bundle of $n$-th homotopy groups over $X$ if $X$ admits a universal cover, and that this was to be in the Dyer and Eilenberg book on algebraic topology.

Ronnie Brown
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    I don't understand. It appears that the points mapping to $x\in X$ can be made to correspond in a definite way with the elements of $\pi_1(X,x)$. But what is this canonical action of the group $\pi_1(X,x)$ on the set $\pi_1(X,x)$? I suspect that it is conjugation, rather than being free and transitive. – Tom Goodwillie Nov 08 '19 at 13:21
  • Any group $G$ acts on its underlying set $U(G)$ by left multiplication. Won;t that do? In a groupoid $G$, and $x \in Ob(G)$ the group $G(x)$ acts on the set $St_G (x)$. in the same way.(One needs to be careful on conventions for writing multiplication!) – Ronnie Brown Nov 08 '19 at 14:46
  • What is your functor from $G$ to Set such that for every object x this functor restricted to $G(x)$ gives you a group acting on itself by left multiplication? – Tom Goodwillie Nov 08 '19 at 15:32
  • How can a morphism $x\to y$ in $G$ possibly give you an isomorphism of sets $U(G(x))\to U(G(y))$ in a canonical way, except by conjugation? – Tom Goodwillie Nov 08 '19 at 15:35
  • @Tom Hi! Maybe introducing $U$ for groups is misleading.for the groupoid case.In this latter case , if $g: x \to y$ and $a: x \to x$ then $ga: x to y$, and that gives the action of the group $G(x)$ on the set $St_G (x)$ (according to the conventions of T&G), Of course $G(x)$ is bijective with $(G(x,y)$ but not with $St_Gx$. – Ronnie Brown Nov 08 '19 at 16:22
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    Okay, I misunderstood you. But you are not asserting that there is a way of constructing a universal (connected) covering space of a connected space $X$ without choosing a base point. You are basically saying that if you make such a covering space once for every base point in $X$ and form their disjoint union then you can topologize that in such a way that it is a covering space of $X\times X$. – Tom Goodwillie Nov 08 '19 at 16:32
  • Exactly! One hopes that such specific description will be of use. – Ronnie Brown Nov 08 '19 at 17:01
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    I add that there was no connectivity assumption (except locally). The intention of that Chapter was also to advertise the notion of covering morphisms of groupoids as algebraic models of covering morphisms of spaces. – Ronnie Brown Nov 08 '19 at 17:20
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[UPDATE: As Tom Goodwillie points out, this is much more complicated than necessary and misunderstands the line of argument that he had in mind. Still, it has some interesting features so I will leave it here.]

Let $\mathcal{M}$ be the category of connected smooth manifolds and smooth maps, let $\mathcal{M}_1$ be the subcategory with the same objects whose morphisms are the diffeomorphisms, and let $J\colon\mathcal{M}_1\to\mathcal{M}$ be the inclusion. Suppose we have a functor $U\colon\mathcal{M}_1\to\mathcal{M}$ and a natural map $p\colon UM\to JM$ that is a universal cover for all $M$. Consider $S^1$ as the usual subspace of $\mathbb{C}$, and choose a point $a\in p^{-1}\{1\}\subset U(S^1)$. For $z\in S^1$ we can define $\mu_z\in\mathcal{M}_1(S^1,S^1)$ by $\mu_z(u)=zu$, and then define $s(z)=U(\mu_z)(a)\in U(S^1)$. This defines a section $s$ of the map $p\colon U(S^1)\to S^1$. If we make enough additional assumptions to ensure that $s$ is continuous, then we arrive at a contradiction.

I think that in fact no additional assumptions are needed, but that needs a slightly different approach. We can identify $S^1$ with $\mathbb{R}P^1$, and then we have an action of the group $G=PSL_2(\mathbb{R})$. Let $H$ be the upper triangular subgroup, which is the stabiliser of the basepoint $1\in S^1$. For $h\in H$ there is a unique $h'\colon U(S^1)\to U(S^1)$ with $ph'=hp$ and $h'(a)=a$. The map $Fh$ need not obviously fix $a$ so it need not coincide with $h'$, but it must have $Fh=\phi(h)\circ h'$ for some deck transformation $\phi(h)$. The group of deck transformations can be identified with $\pi_1(S^1,1)=\mathbb{Z}$, and $H$ acts on this in a natural way (independent of the supposed existence of $U$). Using the connectivity of $H$ we see that this action is trivial. I think it follows that $\phi\colon H\to\mathbb{Z}$ is a homomorphism, but any element $h\in H$ has $n$'th roots for all $n>0$, and this forces $\phi$ to be trivial, so $Fh=h'$ for all $h$. This proves that $Fh$ depends continuously on $h$ for $h\in H$. Moreover, one can find $h_z,k_z\in H$ such that the entries are rational functions of $z$ and $\mu_z=h_z\mu_{-1}k_z$. It follows that $F(\mu_z)$ depends continuously on $z$ except possibly at finitely many values of $z$. These possible exceptions can then be removed by an auxiliary argument with the group structure.

Neil Strickland
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If you want something functorial and base-point-independent, one option is the following $\widetilde X$ bundle over $X$. It combines all the base-point-dependent covering spaces into one gadget.

Let $C(X)$ be the space of all maps $I \to X$, modulo homotopy-rel-end-points. Let $p:C(X) \to X$ be the evaluation at the initial endpoint of $I$. It's easy to see that $p^{-1}(x)$ is the usual universal cover $\widetilde X_x$ contructed using the base point $x\in X$. So $p : C(X) \to X$ is an $\widetilde X$ bundle over $X$.

The assignment $$ X \; \mapsto \; (p : C(X) \to X) $$ is functorial in $X$.

Kevin Walker
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Here is another attempt at pinning down the meaning of "canonical" in reference to Tom's answer.

  1. Let $X$ be a nice space (connected, locally path-connected and semi-locally simply connected).

  2. Let $\pi_X$ be the fundamental groupoid of $X$: this is a category whose objects are points $x\in X$, where a morphism $x\to y$ is a homotopy class of path fixing the endpoints.

  3. Let $U_X$ be the groupoid of universal covers: an object is a universal cover $X_1 \to X$ and a morphism $X_1 \to X_2$ is an isomorphism of covers over the identity map of $X$.

There is a functor $$f:\pi_X\to U_X$$ (i.e., a homomorphism of groupoids) given by the usual construction of a universal cover. Then $f$ is an equivalence of categories (covering space theory).

Let $$g: U_X \to \pi_X$$ be its adjoint (which is defined up to unique isomorphism).

This means for any $\tilde X\in U_X$, with $g(\tilde X) = x\in X$ we have a preferred isomorphism $$ f(x) \cong \tilde X\, . $$

In other words, a universal cover determines a basepoint and a basepoint determines a universal cover.

John Klein
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  • The correspondence {universal covers} $\leftrightarrow$ {basepoints} depends on the choice of $g$, right? I agree $g$ is unique up to unique isomorphism, but it is not literally unique on the nose, and since there is only one isomorphism class in $\pi_X$, knowing $g(\tilde{X})$ only up to isomorphism is no information. I believe it is possible to choose $g$ to be constant on the level of objects. – Julian Rosen Nov 13 '19 at 15:15
  • That is of course correct (both assertions). – John Klein Nov 13 '19 at 20:13