# Why isn't the unit interval $I=[0,1]$ the universal cover of $S^1$?

The question is very simple, i think. But i cannot see the answer. The fact is that a universal cover has a fiber which is isomorphic (under a choice of a point) to the fundamental group of the covered space. In this case $\mathbb{R}$ whit the exponential map has this property, while $I$ with the exponential map not. But actually it is a covering, is simply connected and so by definition also the universal cover.

Where is the flaw in this (paradoxal) argument?

$I$ with the exponential map is not a covering. Indeed, let $p\colon I\to S^1$ be continuous and take some small neighbourhood $U$ of $p(1)$ in $S^1$. If $U$ is small enough, then it is homeomorphic to an open interval. But $p^{-1}(U)$ contains, as a connected component, a small neighbourhood of $1$ in $I$, and this neighbourhood is homeomorphic to a half-open interval.

While $[0,1]$ is simply connected, the natural projection map $\pi \colon [0,1] \rightarrow S^1$ given by $\pi(\theta) = e^{2\pi i \theta}$ (where we think of $S^1$ as the circle in the complex plane $S^1 = \{ e^{i \theta} \, | \, \theta \in [0,2\pi] \}$) is not a covering map. For example, the fiber $\pi^{-1}(1) = \{ 0, 1 \}$ contains two points while every other fiber contains only a single point which can't happen for a cover map.

• In fact, there doesn't any map $f \colon [0,1] \rightarrow S^1$ which is a covering map because the possible covering spaces of $S^1$ are $S^1$ and $\mathbb{R}$ and they are both not homeomorphic to $[0,1]$. Commented Mar 13, 2017 at 22:47
• * which can't happen for a covering map from a connected covering space, right? I believe this phenomenon is possible for disconnected covers. Commented Sep 26, 2023 at 22:51

Another way to tell that the map is not a covering map is recalling that if the base is connected, then the covering has fibers of the same cardinality, which is clearly not the case (take the fiber at $0$ and at $1/2$).

Yet another way is to use the fact that a compact base space with infinite fundamental group must have a noncompact universal cover (although rather indirect, this also shows that there cannot exist any cover whatsoever $p:I \to S^1$ and is a nice result in itself). One important note is that this does not use "uniqueness" of the universal cover, only that we have a covering map from a simply connected space.

• Do you know a reference for this result about (non)compact coverings? Or a proof sketch? Commented Oct 6, 2023 at 16:37

This question has already been well answered, but it seems a shame not to mention an additional fact about the map $p: [0,1] \to S^1, t \mapsto e^{2\pi it}$. From it we can obtain the pushout diagram of groupoids, with the top $2$ being discrete groupoids,

$$\begin{matrix} \{0,1\} & \to & \{0 \}\\ \downarrow &&\downarrow\\ \mathbb I & \xrightarrow{p} & \mathbb Z \end{matrix}$$ where $\mathbb I = \pi_1([0,1],\{0,1\})$ is the fundamental groupoid on two base points $0,1$. This groupoid has $4$ arrows, including $\iota: 0 \to 1$, and with $\iota$ is a generator for the category of groupoids, as $\mathbb Z$ with $1$ is for the category of groups. (That is, an arrow $g$ of a groupoid $G$ is completely determined by a morphism $g': \mathbb I \to G$ such that $g'(\iota) = g$.)

Fun Fact $$:$$ Let $$p : \widetilde X \longrightarrow X$$ be a covering map and $$x_0 \in X.$$ Let $$X$$ be path-connected and $$\widetilde X$$ be simply-connected. Define a map $$\Phi : p^{-1} (x_0) \times \pi_1 (X, x_0) \longrightarrow p^{-1} (x_0),\ \ (e,[\gamma]) \mapsto e \cdot [\gamma] = \widetilde \gamma_e (1)$$ where $$\widetilde \gamma_e$$ is the unique lift of $$\gamma$$ starting at $$e \in p^{-1} (x_0).$$ Then $$\Phi$$ defines a well-defined right group action of $$\pi_1 (X, x_0)$$ on $$p^{-1} (x_0)$$ which is free and transitive. Consequently, by the virtue of orbit-stabilizer theorem it follows that the degree of the covering is $$\left \lvert \pi_1 (X, x_0) \right \rvert.$$

This is a quite straightforward thing to prove. Well-definedness follows directly from the homotopy lifting property, freeness and transitivity of the action follows respectively from the facts that $$\pi_1 \left (\widetilde X \right ) = 0$$ and $$\widetilde X$$ is path-connected, both of which hold as $$\widetilde X$$ is assumed to be simply-connected.

How does that help here? $$:$$

Claim $$:$$ There does not exist any covering map $$p : [0,1] \longrightarrow S^1$$ of finite degree.

If it is not the case then since $$[0,1]$$ is simply-connected and $$S^1$$ is path-connected it follows from the above fact that the degree of the covering map $$p$$ is same as $$\left \lvert \pi_1 (S^1) \right \rvert = \left \lvert \mathbb Z \right \rvert = \infty,$$ a contradiction to the hypothesis.

In your case $$p$$ is the exponential map and hence $$p^{-1} (1) = \{0, 1 \},$$ which is a finite set and consequently it cannot be a covering map.

Nobody mentioned the main property of a covering space so far: A covering map is a local homeomorphism.

Assume that $$p: I=[0,1]\to S^1$$ is a covering map. Then, there is an open set $$U=[0,\epsilon)\subset I$$ such that $$p|_U:U\to p(U)$$ is homeomorphism, $$p(U)\in T_{S^1}$$. This is not possible: Let $$V=p(U)-\{p(0)\}.$$ Then $$V$$ is disconnected open set. But, $$(p|U)^{-1}(V)$$ is connected, contradicting the continuity of $$p|U$$.

• I think there is no boundary of $p(U)$ since it is open. Commented Sep 27, 2023 at 10:19
• $p$ is surjective. Commented Sep 27, 2023 at 10:48
• There is a nbd of $p(0)$, an open arc, whose preiamge is $[0,\epsilon)$ and its disoint translates. So, V is disconnected. Commented Sep 27, 2023 at 11:05
• Why $1$? There is nothing special about $1$ here. Commented Sep 27, 2023 at 11:17
• My apologies. The entire time, I thought you were discussing $p(x)=\exp(2\pi ix)$, rather than treating a general covering map. So most of my comments were irrelevant and I've retracted them. However somebody (the accepted answer) did already mention local homeomorphism. Your own argument needs more work, it's not true for all opens sets $X$ of $S^1$ that $X\setminus\{x\}$ (for some point $x$) is disconnected. Commented Sep 27, 2023 at 13:01