Line with two origins and higher homotopy groups 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.
 A: 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 $E_{n+1}$.
A: 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.
