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?
 A: $I$ 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.
A: 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.
A: 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.
A: 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 $g'(\iota) = g$.) 
