I know of several reasons why the long line can't be a covering space for the circle, but I'm more curious in what exactly goes wrong with the following covering map.

Let $L$ be the long line and define $p: L \rightarrow \mathbb S^1$ by wrapping each segment of the line around the unit circle once, essentially in the same manner as with $\mathbb R$. Clearly we have that for $x \in \mathbb S^1$ the cardinality of the fiber $p^{-1}(0)$ is uncountable, which isn't possible since the fundamental group of the circle is countable. But it's not clear to me why $p$ is not a covering map. It's certainly surjective and seems to be continuous and a local homeomorphism. Yet one of these conditions must fail. Is the map not as well-defined as I originally thought.

I assume I'm misunderstanding the long line in a fundamental way.

  • 3
    $\begingroup$ The problem seems to lie in the phrase "essentially in the same manner as with $\mathbb{R}$." As kahen suggests you haven't proven that the naive interpretation of what this means is continuous. (The long line does not have the colimit topology with respect to the intervals that make it up; if it did, it would be disconnected.) $\endgroup$ Jun 2, 2011 at 20:01
  • $\begingroup$ @Qiaochu, That would be a fundamental thing I'm missing. $\endgroup$
    – JSchlather
    Jun 2, 2011 at 20:03

2 Answers 2


The "obvious" function from the long line to the circle is not continuous at the limit ordinals. As you approach a limit ordinal from the left, the map revolves around the circle infinitely many times, and therefore does not converge to a limiting value on the circle.

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    $\begingroup$ Okay, that makes sense. $\endgroup$
    – JSchlather
    Jun 2, 2011 at 20:07

Recall that every continuous map from the long line to the real numbers is eventually constant. This gives a contradiction with continuity of $p$ (as far as I can tell from a cursory glance at least).


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