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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?

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    $\begingroup$ 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 $\pi_1(X)$. $\endgroup$ – R. van Dobben de Bruyn Nov 8 at 1:58
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    $\begingroup$ 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 $S^1$. $\endgroup$ – Tom Goodwillie Nov 8 at 5:13
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    $\begingroup$ 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. $\endgroup$ – Tom Goodwillie Nov 8 at 12:49
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    $\begingroup$ @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. $\endgroup$ – Tom Goodwillie Nov 8 at 12:57
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    $\begingroup$ 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. $\endgroup$ – HJRW Nov 8 at 13:23
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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 $y$.)

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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.

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    $\begingroup$ I had something simpler in mind. See my later comment. $\endgroup$ – Tom Goodwillie Nov 8 at 13:22
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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.

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    $\begingroup$ 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. $\endgroup$ – Tom Goodwillie Nov 8 at 13:21
  • $\begingroup$ 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!) $\endgroup$ – Ronnie Brown Nov 8 at 14:46
  • $\begingroup$ 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? $\endgroup$ – Tom Goodwillie Nov 8 at 15:32
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    $\begingroup$ 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 $X\times X$. $\endgroup$ – Tom Goodwillie Nov 8 at 16:32
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    $\begingroup$ 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. $\endgroup$ – Ronnie Brown Nov 8 at 17:20
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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 $X$.

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    $\begingroup$ Here $p: C(X) \to X$ is what I have called $\sigma: \pi_1(X) \to X$. $\endgroup$ – Ronnie Brown Nov 8 at 20:51
  • $\begingroup$ Thanks. I had not noticed that you were describing essentially the same thing. $\endgroup$ – Kevin Walker Nov 9 at 14:06
  • $\begingroup$ I think it is also useful to investigate the target category - covering morphisms of groupoids. $\endgroup$ – Ronnie Brown Nov 12 at 9:54
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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.

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  • $\begingroup$ 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. $\endgroup$ – Julian Rosen Nov 13 at 15:15
  • $\begingroup$ That is of course correct (both assertions). $\endgroup$ – John Klein Nov 13 at 20:13

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