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In Chapter 13 of Thurston's book on manifolds, he alludes to a particular construction of the universal covering of a space $X$, with the property that for all $x \in X$, there is a neighborhood $U$ so that for any cover $p : X' \to X$, $p$ trivializes over $U$. (He brings this up because he immediately starts to work with this idea in the context of orbifolds. He doesn't provide any further explanation.)

First, he remarks that if $(X',p')$ and $(X'',p'')$ are two pointed coverings of $(X,p)$ then the component of $(p,p'')$ in their fiber product is a covering of $X'$ and $X''$. This makes sense tome.

Next, he suggests that one can take a suitable inverse limit over some set of pointed, connecting coverings spaces of $X$, which represent all isomorphism classes, and that this inverse limit is a universal cover for $X$.

Question: Can someone explain how this works, or provide a reference? The main point (see my thoughts below) is that this inverse limit is still a cover of $X$. (Can one abstractly prove the existence of a universal covering this way, without going through all the business with spaces of paths.)


My thoughts: In different language, I expect he means -- let $C$ be the category of connected, pointed coverings of $X$, with morphisms pointed covering maps, including the identity covering. This category $C$ is a subcategory of the category of topological spaces, $TOP$. So, he suggests taking an limit over the diagram $C$ inside of $TOP$, and then presumably taking the connected component of the base point. (For size issues, we can replace $C$ with a skeleton subcategory that has a set of objects. Since the fundamental group of $X$ is a set, and there is one isomorphism type of pointed cover per subgroup...)

The resulting space exists, call it $\tilde{X}$, because we can take limits of small diagrams in $TOP$. (As a subspace of the product, of points satisfying all the compatibility conditions.) $\tilde{X}$ will have a (unique) projection to each covering, and because of the compatibility criterion, this will be compatible with the covering of the identity covering.

So, all that needs to be shown is that the map from $\tilde{X}$ to $X$ is a covering map. I notice that we haven't used the observation that covers are preserved by connected component of the fiber products, except in that this guarantees that $C$ is a directed set, so the limit over the diagram $C$ is an inverse limit.

Therefore I suspect that what one needs to show is that an inverse limit of connected covering maps is still a cover.

I know that each cover is trivial over $U$, so each is described by a discrete set $F$, and is of the form $U \times F$, with projection, up to isomorphism. But one can take inverse limits of discrete sets and get more complicated spaces, such as with the p-adic integers. Or more simply, the product of an infinite collection of discrete sets is not discrete in the product topology. I suspect that a reason why we insist on connected coverings is to avoid this issue, in some kind of way. Alternatively, if we have the existence of a universal cover, then these difficulties disappear.

A natural strategy for proof would be : Suppose that $(X_i, p_i, \phi_i)$ , $i \in I$, are pointed covers of $(X,p)$ , $\phi_i$ are the covering map. Let $\tilde{X}$ denote the fiber product of this diagram, over $X$. For an $x \in X$, suppose that $U$ is a neighborhood of $x$so that all covers above $U$ are trivial. We pull back the fiber diagram along $U$. Then the preimage in $\tilde{X}$ is the fiber product of sets of the form $U \times F_i$ over $U$,for $F_i$ discrete. This is naturally $U \times \Pi F_i$. However, $\Pi F_i$ is not discrete when $I$ is infinite.


So, maybe the box topology is meant to be used? This has too many open sets to be the limit in $TOP$, but I guess that is not a problem in this context, since we don't seem to care very much for the universal property of the fiber product.

In addition to Thurstons book, I found the same sketch of covering space theory via fiber products on page 8 here, https://arxiv.org/pdf/math/0107172.pdf

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  • $\begingroup$ I think you are way more mathematical sophisticated than I am, so i would just want to point things that may be trivial for you but you may have overlooked. Let $f, g : Y \to Z$ two maps in $Cov(X)$ and let $y_0 \in Y$ . Assume that $Y$ is connected and $f(y_0) = g(y_0)$ then $f = g$ by this lemma we can make a natural transformation. Let $x_0 \in X$ be a point and let $f : Y \to X$ be a universal cover and $y_0$ a lift of $x_0$ we have a natural transformation: $Hom(Y,-):Z \mapsto Hom(Y,Z) $ and $Fib_{x_0}:Z \mapsto f^{-1}_Z (x_0)$ this is natrual transformation is bijective $\endgroup$ – Elad Jun 29 '17 at 8:03
  • $\begingroup$ @Elad If you have an extended comment it is perfectly fine to post it as an answer - it may be easier to format. I think the only thing about my account that might project sophistication is the occasional use of "fancy" terminology - this is not a reliable certificate of my mathematical sophistication. (In fact if you poke around in my history you will find some truly embarrassing stuff.) I appreciate any help or insights you have to offer, whether or not you think they are trivial. Thanks for reading my question. :) $\endgroup$ – Lorenzo Jun 29 '17 at 8:08
  • $\begingroup$ the thing is that i just started learning the field, but i tried to google my intuition that the universal cover is a representative of a functor and got to this thing that might help: qchu.wordpress.com/2013/06/23/… $\endgroup$ – Elad Jun 29 '17 at 8:37
  • $\begingroup$ You might look at page 85 here: fenix.tecnico.ulisboa.pt/downloadFile/395139415315/… I think this might work. Tell me if its any help because i'm curious to know if i even understood your question. $\endgroup$ – Elad Jun 29 '17 at 8:48
  • $\begingroup$ @Elad As far as I understand, the Grothendieck story is about the functor that takes a covering of $X$ and pulls it back to a covering over the universal cover of $X$ (where it becomes trivial, hence is basically determined by its components) (one can also pick a point in $X$ and look at the fiber) ... maybe one can show that pulling back covers to the limit I describe behaves like pulling it back to the universal cover, so one can try to conclude by a Yoneda lemma argument... that seems like a reasonable idea, thanks $\endgroup$ – Lorenzo Jun 29 '17 at 9:30

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