I've found this assertion (here , pag 5) about covering spaces: the map $p\colon\mathbb{S}^{1}\times\mathbb{N}\to\mathbb{S}^{1}$ given by $\left(z,n\right)\mapsto z^{n}$ is NOT a covering map of the 1-sphere (on $\mathbb{N}$ the discrete topology, as usual).
Well, to show it I should prove that each point of the 1-sphere doesn't admit any fundamental neighborhood. For example take the point $1=e^{2i\pi}$ and a small connected neighborhood $U$ of this point. The preimage of $U$ is clearly a disjoint union of open connected sets of the form $U_{j}\times\left\{ j\right\} $ where $U_{j}$ contains an n-th root of 1. But it seems also clear that $U_{j}\times\left\{ j\right\} $ is homeomorphic through $p$ to the initial set $U$. This is exactely the proof to show that the point admits a fundamental neighborhood. Where is the mistake?
It should also be remarked that $p\colon\mathbb{S}^{1}\times\left\{ 1,...n\right\} \to\mathbb{S}^{1}$ defined in the same way is a covering since it has finite fiber, so somewhere I should exploit the fact that the fiber is infinite.
I think $p$ is a covering.
Let $\varphi : \mathbb{R} \to S^1, \varphi(t) = e^{it}$. The set $U = \varphi((-\pi,\pi)) = S^1 \backslash \{ -1 \}$ is an open neighborhood of $1$ in $S^1$. We have $p^{-1}(U) \cap S^1 \times \{ n \} = \bigcup_{k=0}^{n-1} U(k,n) \times \{ n \} $ where $U(k,n) = \varphi((\frac{2\pi k}{n} - \frac{\pi}{n},\frac{2\pi k}{n} + \frac{\pi}{n}))$. The intervals $J(k,n) = (\frac{2\pi k}{n} - \frac{\pi}{n},\frac{2\pi k}{n} + \frac{\pi}{n})$ are pairwise disjoint and contained in $(-\frac{\pi}{n}, \frac{2\pi (n-1)}{n} + \frac{\pi}{n}) = (-\frac{\pi}{n}, 2\pi - \frac{\pi}{n})$. This shows that also the $U(k,n)$ are pairwise disjoint. In fact, the $U(k,n)$ are the components of $S^1 \backslash \{ \eta_1, .... \eta_n \}$, where the $\eta_k$ are the $n$-th complex roots of $-1$.
Each $U(k,n) \times \{ n \}$ is mapped by $p$ homeomorphically onto $U$. Therefore $U$ is evenly covered. A similar argument works for $V = \varphi((0,2\pi)) = S^1 \backslash \{ 1 \}$.