Cross-posted on MO

Let $\mathsf{CW}$ be the category of CW-complexes and $\mathsf{CW}_*$ that of pointed CW-complexes (possibly disconnected, one basepoint in each component). I would like to know whether there is a functor $\mathsf{CW} \to \mathsf{CW}$ mapping a CW-complex to its universal cover. In the pointed case this works, i.e. there is a functor $\mathsf{CW}_* \to \mathsf{CW}_*$ mapping a pointed CW-complex $(X,x_0)$ to its universal cover $$ \widetilde X(x_0) \colon = \{[\alpha \colon I \to X] \mid \alpha(0)= x_0 \} $$ where square brackets denote homotopy relative endpoints, and the covering projection is $[\alpha] \mapsto \alpha(1)$. The basepoint of $\widetilde X(x_0)$ is the class of the constant map, and induced maps are given by requiring that basepoints map to basepoints, using lift uniqueness. But without the basepoint I am unable to define this as a functor.

A related question (one that I am more interested in) is: define the category $\mathsf{CW}^\text{tw}$ to be the category of pairs $(X,w)$ with $X$ a CW-complex and $w \in H^1(X;\mathbb{Z}_2)$, and maps preserving cohomology classes. Is there a functor $\mathsf{CW}^\text{tw} \to \mathsf{CW}$ which maps a $(X,w)$ to a two sheeted covering space $\overline X \to X$ whose first Stiefel-Whitney class is $w$? Again, this is possible in the pointed setting, and a positive answer to the first question would imply a positive answer here as well. The second functor actually exists on the subcategory $\mathsf{MFD} < \mathsf{CW}^\text{tw}$ of smooth (actually also topological) manifolds with their first Stiefel-Whitney classes: take the sphere bundle of the top exterior algebra of the tangent bundle. This makes me think it would not be an easy task to show that they don't exist over all CW-complexes, i.e. the statement is not that implausible after all.

I would be very grateful for any help, including references and hints. Thank you for reading :)


1 Answer 1


With regard to the first question, the point you are rightly making is that fundamental groups and universal covers are defined on pointed spaces and there is no getting away from this. Grothendieck wrote to me in 1983 in a different context: " . .... both the choice of a base point, and the $0$-connectedness assumption, however innocuous they may seem at first sight, seem to me of a very essential nature. To make an analogy, it would be just impossible to work at ease with algebraic varieties, say, if sticking from the outset (as had been customary for a long time) to varieties which are supposed to be connected. Fixing one point, in this respect (which wouldn't have occurred in the context of algebraic geometry) looks still worse, as far as limiting elbow-freedom goes!"

What is true is that under the usual local conditions for a universal cover we can make the fundamental groupoid $\pi_1 X$ into a topological groupoid, see this paper, and the source map $s: \pi_1 X \to X$ has fibre at $x$ the universal cover of $X$ at $x$. This puts together as a bundle the standard construction of the universal cover of $X$ at $x$.

I also see that this page lists related questions.

Can this help in your second question?

  • 1
    $\begingroup$ Thank you for your answer. I added the "groupoids" tag because, as you mention, the fundamental groupoid substitutes the fundamental group in the unpointed setting. More to the point of the question, one can define a universal covering functor $\Pi(X) \to \mathsf{CW}$, $x \mapsto \widetilde X(x)$; but then we can't take the colimit because there are no canonical identifications of the $\widetilde X(x)$'s. I also have the intuition that my questions have a negative answer. Do you know of a proof of this? It's not that implausible of a statement, e.g. the second functor works for manifolds. $\endgroup$ Sep 21, 2015 at 20:06

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