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So I know how the universal cover of the line with two origins is(looks like a comb going in both directions with alternating midline points corresponding to the different origins). I was wondering why the higher homotopy groups vanish in the universal cover, which would imply that the higher homotopy groups of the main space vanishes.

Enigma
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2 Answers2

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Let $E$ denote the universal cover of the line with two origins, the "two-sided comb" as you call it. I claim that $E$ is weak equivalent to a point. Since $E$ is simply connected, to show this it suffices to show that $E$ has trivial homology. To show $E$ has trivial homology, it suffices to show that the union $E_n$ of $n$ consecutive "teeth" of the comb (including the midpoints in between these teeth but not the midpoints at the ends of the first and last teeth) has trivial homology for each $n$, since every compact subset of $E$ is contained in such a subspace $E_n$.

We can now prove $E_n$ has trivial homology by an easy induction using Mayer-Vietoris. In the base case, $E_1$ is just homeomorphic to $\mathbb{R}$. For the induction step, note that $E_{n+1}=E_n\cup U$ where $E_n$ and $U$ are both open in $E_{n+1}$, $U\cong\mathbb{R}$, and $E_n\cap U\cong \mathbb{R}$ (here $U$ is the new tooth together with the new midpoint where it attaches and the old tooth that shares that midpoint, so $E_n\cap U$ is just the latter tooth). An easy Mayer-Vietoris computation then shows that if $E_n$ has trivial homology, so does $E_{n+1}$.

Eric Wofsey
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Let $L$ denote the line with two origins and for lack of better notation, $p:E \rightarrow L$ be the universal cover space with $p$ the projective map. As cover spaces are fibre bundle, here the fibre being $\mathbb Z$, we have the long exact sequence of homotopy groups: $$\cdots \rightarrow \pi_n(\mathbb Z) \rightarrow \pi_n(E) \rightarrow \pi_n(L) \rightarrow \pi_{n-1}(\mathbb Z) \rightarrow \cdots $$ As $E$ is contracible it has the homotopy type of a point and for $n\geq 2$ we have, $\pi_n(E) \cong \pi_{n-1}(\mathbb Z) \cong 0$ giving the short exact sequence $$0 \rightarrow \pi_n(L) \rightarrow 0 $$ And the higher homotopy groups must vanish.

Birch Bryant
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  • But how do you know $E$ is contractible? I thought that's what the question was asking about. – Eric Wofsey Jun 05 '17 at 05:44
  • In fact, I don't think $E$ is contractible, though I believe it has the weak homotopy type of a point. – Eric Wofsey Jun 05 '17 at 05:50
  • Yeah, I don't think E is contractible either. That was the issue I was having proving this result in the first place. It would be helpful if the user could expand on why the universal covering space is aspherical. – Enigma Jun 05 '17 at 05:52
  • All of the points $(0,n)$ by the identification imposed on $\mathbb R \times \mathbb Z$. Collapse those down to the origin and now view them as a countable rayed star centered at the origin in $\mathbb R^2$ and use the typical contraction of $\mathbb R^2$ to the origin – Birch Bryant Jun 05 '17 at 06:01
  • But how do you know collapsing all those points together doesn't change the homotopy type? – Eric Wofsey Jun 05 '17 at 06:04
  • @EricWofsey I concede there's something fishy going on around the end points of the rays maybe collapse the rays first, anyway I say screw it your answer works +1 – Birch Bryant Jun 05 '17 at 06:17