It is often said that a sheaf on a topological space $X$ is a "continuously-varying set" over $X$, but the usual definition does not reflect this because a sheaf is not a continuous map from $X$ to some "space of sets". (Such a space must have a proper class of points!) However, I recently had the epiphany that this can be made to work, if one is willing to give up some generality and focus on locally constant sheaves, a.k.a. covering maps.

Let $X$ be a connected CW complex. If I understand correctly, an $n$-fold covering map of $X$ is the same thing as a $S_n$-structured fibre bundle with typical fibre a discrete set of $n$ points, and so their isomorphism classes naturally correspond to isomorphism classes of principal $S_n$-bundles on $X$, which are in turn classified by an Eilenberg–MacLane space $\mathrm{B} S_n = K(S_n, 1)$.

Question 1. Is there a universal $n$-fold covering map of $\mathrm{B} S_n$, i.e. a $n$-fold covering map $T_n \to \mathrm{B} S_n$ such that every $n$-fold covering map of $X$ is obtained (up to isomorphism) as a pullback of $T_n \to \mathrm{B} S_n$ along the classifying map?

It seems to me that once this is done, we can improve the situation slightly and get a classifying space for all finite covering maps by considering $\coprod_{n \in \mathbb{N}} \mathrm{B} S_n$.

Question 2. Does the obvious generalisation work, i.e. does $\mathrm{B} S_{\kappa}$ classify $\kappa$-fold covering maps for each cardinal $\kappa$?

  • $\begingroup$ (Technical nitpick: I don't think there is anything wrong with having (or quantifying over) maps with proper classes as codomains. They just can't be surjective then.) $\endgroup$ – hmakholm left over Monica Nov 30 '12 at 22:42
  • $\begingroup$ I think so. There's a universal $S_n$-bundle $ES_n \to BS_n$, and it should have an associated $n$-fold cover given by a construction analogous to an associated bundle (or just constructed directly from covering space theory). $\endgroup$ – Qiaochu Yuan Nov 30 '12 at 22:46
  • $\begingroup$ Ok, my objection is as follows: any covering $X\to B$ gives indeed a $S_n$-valued cocycle, and thus a principal $S_n$-bundle $P\to B$. Maybe I don't understand this part, but I fail to see how you can recover the bundle from there. Because you'd need a $S_n$-action $\rho$ on $\lbrace 1,\dots,n\rbrace$ to recover $X\to B$ as an associated bundle via the usual construction $$\lbrace 1,\dots, n\rbrace\times_{\rho} P,$$ so my contention is that I don't think that any covering is actually an associated bundle to the principal $S_n$-bundle $P$ it defines. $\endgroup$ – Olivier Bégassat Nov 30 '12 at 23:26
  • $\begingroup$ Could somebody tell me wether my contention is founded? $\endgroup$ – Olivier Bégassat Nov 30 '12 at 23:58
  • $\begingroup$ I can't say I'm familiar with that construction – I was thinking more along the lines of the correspondence between vector bundles of rank $n$ and principal $\textrm{GL}_n$-bundles. $\endgroup$ – Zhen Lin Nov 30 '12 at 23:59

The answer to both your questions is yes, and Qiaochu gave the basic idea. The base space is $BS_n$ and the fiber is $ES_n$. You can make this concrete (very analogous to Grassmannians) by using the model $BS_n \equiv C_n(\mathbb R^\infty) / S_n$ and $ES_n = C_n(\mathbb R^\infty)$ where $C_n$ indicates the configuration space of $n$ labelled points in $\mathbb R^\infty$. i.e. $C_n (\mathbb R^\infty) = Emb(\{ 1,2,\cdots, n\}, \mathbb R^\infty)$.

edit: this is a response to Zhen Lin's 2nd comment:

The theory of classifying spaces (or looking at it another way, obstruction theory). For simplicity, assume $X$ is connected. Give $X$ a CW-structure with one $0$-cell, then a map $X \to BS_n$ when restricted to the $1$-skeleton gives a homomorphism $\pi_1 X \to S_n$, this is the action of $\pi_1$ on $S_n$ described in most intro algebraic topology courses. Now ask, can you extend the map on the $1$-skeleton $X^1 \to BS_n$ to the $2$-skeleton $X^2 \to BS_n$ ? The obstructions (if any) would be elements of $\pi_1 BS_n$, corresponding to the action on the fiber along a $2$-cell attachment. But these are trivial since the covering space pulls-back to a cover of $D^2$, and covering spaces over discs are trivial. Similarly, the obstruction to extending to $X^3$ are elements of $\pi_2 BS_n = *$.

  • $\begingroup$ I guess we need some kind of local embedding theorem for covering maps? $\endgroup$ – Zhen Lin Dec 31 '12 at 2:32
  • $\begingroup$ What are you talking about by "need"? What are your purposes, and what are you assuming? The proofs follow immediately, say if your covering space is over a CW-complex, for example. But if you don't say exactly what you're looking for it's impossible for me to say what you need. $\endgroup$ – Ryan Budney Dec 31 '12 at 2:53
  • $\begingroup$ Well, if I recall correctly, one way of proving that the Grassmannian is a $\mathrm{B} O(n)$ is to show that every vector bundle can be locally embedded in a trivial vector bundle, or something like that. What is the general theory you are referring to? $\endgroup$ – Zhen Lin Dec 31 '12 at 3:08
  • $\begingroup$ My response is a little too long for a comment so it appears below the "edit" line in my answer. $\endgroup$ – Ryan Budney Dec 31 '12 at 3:49

I can see why this question is asked, but feel the better way of looking at covering maps of $X$, see here, is to say that if $X$ is "locally nice" then the fundamental groupoid functor $\pi_1$ determines an equivalence of categories

$$TopCov(X) \to GpdCov(\pi_1 X), $$

from covering maps of $X$ to covering morphisms of $\pi_1 X$. A covering morphism $p: Q \to G$ of groupoids satisfies for $x \in Ob Q$ and $g \in G$ starting at $px$ there is a unique $h \in Q$ starting at $x$ and such that $p(h)=g$.

One then shows that the category $GpdCov(G)$ is equivalent to the category of actions of $G$ on sets. If $G$ acts on a set $S$ on the left then there is an action groupoid which one can write $S \rtimes G$ and the projection $S \rtimes G \to G$ is a covering morphism.

  • $\begingroup$ Interesting how a month old unanswered question got two replies within less than half an hour. $\endgroup$ – Olivier Bégassat Dec 30 '12 at 22:03
  • $\begingroup$ Yes, I was already aware of this. Unfortunately by replacing spaces by groupoids in this question, things take the flavour of "make the homotopy hypothesis true by defining spaces to be Kan complexes and defining $\infty$-groupoids to be Kan complexes". $\endgroup$ – Zhen Lin Dec 31 '12 at 2:30

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