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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.

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    @Qiaochu, That would be a fundamental thing I'm missing.2011-06-02

2 Answers 2

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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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    Okay, that makes sense.2011-06-02
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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).