If $X$ is path connected, locally path connected and semi-locally simply connected topological space and $x_0\in X$, consider the set $\chi =\{(X_{\alpha},x_\alpha) \} $ of covering spaces of $X$ with covering map $p_\alpha \colon X_\alpha \rightarrow X$, $p_\alpha (x_\alpha)=x_0$. (Clearly, this is a non-empty collection, $(X,x_0)\in \chi$ )

Put a partial ordering relation on this collection: $(X_\alpha,x_\alpha)\geq (X_\beta,x_\beta)$, if there is a covering map $q\colon X_\alpha \rightarrow X_\beta$ such that $p_\beta \circ q =p_\alpha$.

Can we use Zorn's lemma to prove existence of universal covering space of $X$ as maximal element of this collection?

If yes, are the conditions stated (path connected, etc.) in hypothesis are necessary (they are necessary when we prove existence, in usual topological way)?


1 Answer 1


One cannot prove this using Zorn's lemma, at least not in an obvious way. The problem is that I don't see how to show that if $(X_1,x_1) < (X_2,x_2) < \ldots$ is a totally ordered sequence, then there exists some upper bound $(Y,y)$ with $(X_i,x_i) \leq (Y,y)$ for all $i$. You have to come up with a space somehow, and I don't think you'll be able to bypass the usual construction.

It should be noted that the conditions "locally path connected" and "semilocally simply connected" are actually necessary for your space to have a universal cover (exercise : prove this!).

  • $\begingroup$ @Adma: the conditions are necessary to hold lifting properties and for covering map to be "covering". $\endgroup$
    – Beginner
    Sep 12, 2011 at 6:59
  • $\begingroup$ Well, the natural candidate for that space would be the limit $Y = \varprojlim X_n$, wouldn't it? I haven't checked but it seems to me that the natural maps from $Y$ should be covering maps. $\endgroup$
    – t.b.
    Sep 12, 2011 at 9:23
  • 1
    $\begingroup$ Dear @Theo, your idea is (as usual) quite interesting and elegant, but unfortunately I don't think it works. Consider the unit circle $S \subset \mathbb C$ and the system of morphisms of coverings of that circle $q_n:S_n=S \to S_{n-1}=S:z\mapsto z^n$ , where the $n$-th covering is given by $p_n:S\to S:z \mapsto z^{n!}$ . The projective limit is a compact (Tychonov!) solenoid, which is not a covering of $S$ and even less its universal cover. $\endgroup$ Sep 12, 2011 at 11:23
  • $\begingroup$ @Georges: Dear Georges, thank you. You're right, of course. No need to make it sound better than it was, I should have thought at least a few minutes before posting. $\endgroup$
    – t.b.
    Sep 12, 2011 at 11:31
  • $\begingroup$ @Theo: Dear Theo, I stand by my judgment that your answers usually are interesting and elegant! $\endgroup$ Sep 12, 2011 at 12:07

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