Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
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.) –  Qiaochu Yuan Jun 2 '11 at 20:01
    
@Qiaochu, That would be a fundamental thing I'm missing. –  JSchlather Jun 2 '11 at 20:03
add comment

2 Answers

up vote 17 down vote accepted

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.

share|improve this answer
    
Okay, that makes sense. –  JSchlather Jun 2 '11 at 20:07
add comment

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).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.