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I was trying to understand this problem:

Construct a simply-connected covering space of the space $X \subset \mathbb R^3$ that is the union of a sphere and a diameter.

And my idea was to only use two spheres separated by a line, but when I looked here Covering spaces need big help Hatcher I found that there has to be infinite number of spheres, is this because we want a universal cover because we want a simply connected cover and also because we want to have loops (start and end at the same time)? Would it break the homomorphism condition because it would not be locally homeomorphic ? If so, why the local homeomorphic condition will be broken?

Any explanations will be greatly appreciated!

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2 Answers 2

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Let me use two spheres then. Image the line $[-1,1]\times\{(0,0)\}$ embedded in $\Bbb R^3$, the unit spheres positioned with centres $(-2,0,0)$ and $(2,0,0)$. Use the obvious kind of projection map from here onto $X$ which is the union of the same line with the unit sphere centred at the origin.

Consider the part of the sphere at $(3,0,0)$. This point maps to $(-1,0,0)$. Any neighbourhood of that point in $X$ catches some small segment of the line; however the preimage of that neighbourhood in our "covering" space is just some sphere-neighbourhood of $(3,0,0)$ with no line presented. Local homeomorphy is lost. You need to attach a further line from $3\to 5$ to make this work. Oh, but then you need to attach a sphere which touches the line at $(5,0,0)$ for the same sorts of reason. Oh but then you have to attached another line. And so on...

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    $\begingroup$ I am trying to draw what you are explaining and then I will try to rationalize it and let you know. $\endgroup$
    – Emptymind
    Commented Apr 8 at 16:18
  • $\begingroup$ So your line is along the x axis, right? $\endgroup$
    – Emptymind
    Commented Apr 8 at 16:23
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    $\begingroup$ @Emptymind All covering spaces have the possibility to lift a loop to either a loop or a non-closed path. The simply connected covering spaces only lift loops to loops if the loop is nullhomotopic. $\endgroup$
    – FShrike
    Commented Apr 8 at 19:03
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    $\begingroup$ @Emptymind You are mentioning loops but ... which loops do you mean? I never talked about specific loops and you never talked about loops in your question. The space you described is simply connected and the space I describe is also simply connected, so all loops are nullhomotopic. As for "folding", you've got two spheres connected by a stick. The "covering" map (which fails to be a covering map) which seems obvious to try, since it's essentially how the real universal covering map works, would rotate each sphere about the endpoint they're connected to. It looks like a fold - not important. $\endgroup$
    – FShrike
    Commented Apr 8 at 20:05
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    $\begingroup$ @Emptymind I've got a function $p$, a point $x$, and I mentioned what $p(x)$ is. There is no loop there $\endgroup$
    – FShrike
    Commented Apr 8 at 21:41
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I am not sure what space you had in mind, @FShrike interpreted it as two spheres connected by a line segment, which is simply connected but doesn't cover $X$ (there is no map from it to $X$ that is a local homeomorphism).

I on the other hand thought that what you had in mind was the space consisting of two spheres arranged in a circular parttern and connected by two line segments, which would be a special case of $n$ spheres in a circular pattern, with all adjacent spheres connected by a line segment like a string of beads.

These spaces, let's denote them by $X_n$, do cover $X$ (a covering map being the quotient map after rotation over $2\pi/n$), but they are not simply connected: intuitively it is clear that a curve passing through all line segments and taking any path from the north pole to the south pole (or rather, the points at which the line segments are attached) is not null-homotopic.

To prove that the spaces $X_n$ are not simply connected rather than proving that this particular curve is not null-homotopic, you can also directly use the theory of covering spaces: any connected covering of a simply connected space is a homeomorphism. However, each $X_n$ is covered by $X_{mn}$ for any natural number $m$, and this is an $m$-fold covering, so not a homeomorphism for $m > 1$. It follows that $X_n$ is not simply connected for any $n$.

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    $\begingroup$ I assume you mean the one of FShrike (because $X_2$ isn't). Note that this space being connected, we get the same fundamental group for any base point, so we can take the base point to be on the line segment. A loops that passes through one sphere has to pass through the base point every time it visits the other sphere, so each loop can be written as a product (i.e. concatenation) of loops that visit only one sphere. Since this one sphere together with the segment up to the base point is homotopy equivalent to just the sphere, which is simply connected, each of these loops is trivial $\endgroup$
    – doetoe
    Commented Apr 8 at 17:39
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    $\begingroup$ "what is the equivalence relation of the quotient?" In general, when a group $G$ acts on a space $S$, there is an equivalence relation where $s\sim t$ if $gs = t$ for some $g\in G$. The quotient is called the orbit space. In our case, the cyclic group of order $n$ acts on $X_n$ in the obvious way $\endgroup$
    – doetoe
    Commented Apr 9 at 7:32
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    $\begingroup$ "yes it has only one hole in it as we are like string of beads, that is why it is not nullhomotopic, am I correct?" - intuitively yes, to make it rigorous will require some work (but should be quite doable) $\endgroup$
    – doetoe
    Commented Apr 9 at 7:34
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    $\begingroup$ "why the m-fold covering is not a homeomorphism?" Because it has not continuous inverse (it is not a bijection) $\endgroup$
    – doetoe
    Commented Apr 9 at 7:40
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    $\begingroup$ Your answer was amazing also and so beneficial for me :) thank you very very much! $\endgroup$
    – Emptymind
    Commented Apr 9 at 15:46

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