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Let $X$ be the union of the unit circle centered at 0 and the line segment between the points $(1,0)$ and $(2,0)$. What is the universal covering of $X$? Compute the fundamental group of $X$.

My Attempt:

The fundamental group seems easy enough. Since the line segment between the two points has the point $(1,0)$ as its deformation retract, the fundamental group of $X$ will simply be the fundamental group of the circle which is $\mathbb{Z}$.

As for the universal cover, can we take the usual covering of the circle which is $\mathbb{R}$? Is there a way to formally show this?

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  • $\begingroup$ The universal cover will look like $A \cup B$ where $A$ is the $y$-axis and $B$ is the set of $\{(t,n) : t \in \Bbb R, n \in \Bbb N\}$. $\endgroup$
    – Nicolas Hemelsoet
    Dec 18, 2017 at 5:13
  • $\begingroup$ @NicolasHemelsoet can you explain? $\endgroup$
    – Nicholas Roberts
    Dec 18, 2017 at 5:19
  • $\begingroup$ Oh sorry I was thinking to a different space (I was thinking to the union of a circle with a tangent line). But are you sure about the fundamental group ? If seems to me that $X$ retracts on a wedge of two circles. $\endgroup$
    – Nicolas Hemelsoet
    Dec 18, 2017 at 5:22
  • $\begingroup$ @NicolasHemelsoet How do you figure? I thought any line was null-homotopic, so it can be homotoped to a point. So all we are left with is the circle. Is this bad logic? $\endgroup$
    – Nicholas Roberts
    Dec 18, 2017 at 5:26
  • $\begingroup$ You are right, but when you do this there is still a little portion of circle which will become a full circle. For example move the line so that $X$ becomes $S^1$ with the vertical segment from $(0,-1)$ to $(0,1)$. Now, you can retracts this segment and you see that you have indeed two circles intersecting in one point. This gives $\pi_1(X)$ is the free group on two generators. $\endgroup$
    – Nicolas Hemelsoet
    Dec 18, 2017 at 5:29

1 Answer 1

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Let $\widetilde{X}$ denote the universal covering of $X$. Then we can write $\widetilde{X}$ as a subset of $\mathbb{R}^2$: $$\widetilde{X} = \{(x,0)\in\Bbb{R}^2~:~x\in\Bbb{R}\} \cup \{(n,t)\in\Bbb{R}^2~:~n\in\mathbb{Z}, \; 0\leq t \leq 1\}.$$

A picture of $X$ and $\widetilde{X}$ is below. One should be able to construct the covering map from these descriptions.

enter image description here

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  • $\begingroup$ Thank you Adam. This makes sense. How about the fundamental group. Am I right in saying it is $\mathbb{Z}$? $\endgroup$
    – Nicholas Roberts
    Dec 18, 2017 at 20:04
  • $\begingroup$ Yes, the fundamental group is isomorphic to $\Bbb{Z}$. $\endgroup$
    – Adam Lowrance
    Dec 18, 2017 at 20:08
  • $\begingroup$ @adam lowrance wouldn’t The real line work as well? Since this spiky space can be retracted to it $\endgroup$
    – Zee
    Dec 19, 2017 at 4:20
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    $\begingroup$ The real line is not the universal cover of $X$ because a covering map must be a local homeomorphism. Informally, the point $(1,0)$ in $X$ locally has a "Y" shape and so the same must be true of its preimage in any covering space. On the other hand, one can prove that $X$ deformation retracts onto the circle $S^1$, and so $\pi_1(X)\cong \mathbb{Z}$. A major part of the proof that $\pi_1(X)\cong\mathbb{Z}$ involves the universal cover of $S^1$, which is the real line $\mathbb{R}$. $\endgroup$
    – Adam Lowrance
    Dec 19, 2017 at 4:33
  • $\begingroup$ +1. Was trying to figure out how to say what @AdamLowrance posted but it's already a great answer. $\endgroup$
    – Randall
    Dec 20, 2017 at 3:55

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