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I understand that the long line isn't a covering space for $S^1$, as has been discussed in other places, such as here.

However, does the projection map to $S^1$ satisfy path lifting? Intuitively it seems so? If so, how does that relate to the result that path lifting + a local homeomorphism implies a covering map (Theorem 4.19 from here)?

I should say, I am not well-versed with dealing with ordinals, and am trying to self learn covering spaces etc.

SarthakC
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    As mentioned in your first link, a map that wraps each segment of the long line around the circle is not continuous at the limit ordinals. Is this discontinuous function your "projection map"? – Dennis Feb 01 '24 at 20:09
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    The map that is discussed in your link isn't well-defined at limit ordinals. – Rob Arthan Feb 01 '24 at 20:13
  • I guess I was thinking that the long line, defined as $\omega\times[0,1)$ could be mapped to $[0,1)$ through a projection map without any issues. And that could be mapped to $S^1$. Maybe I misunderstood what was being claimed in the answer to the first link I put above. Is the projection map to [0,1) itself not well defined somehow? I should say, I am not well-versed with dealing with ordinals, and am trying to self learn covering spaces etc. – SarthakC Feb 01 '24 at 20:26
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    @SarthakC The issue with that is that the long line is $\omega\times[0,1)$ equipped with the order topology, not the product topology. – Thorgott Feb 01 '24 at 23:54
  • @Thorgott: Ah, that's what I was missing. If one does consider the product topology on the long line, then I imagine path lifting holds and it is a covering map? – SarthakC Feb 02 '24 at 00:08
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    No, lifting to a product is a component-wise question and the map $[0,1)\rightarrow S^1$ does not satisfy path-lifting. – Thorgott Feb 02 '24 at 02:07
  • Yeah, okay, that makes sense, thanks. – SarthakC Feb 02 '24 at 04:59

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