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?

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.
