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?

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 $g'(\iota) = g$.) 
