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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.

lucar
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  • As mentioned in the answer below the map actually seems to be a covering. Although sometimes people will include connectedness as a prerequisite for a covering map. This might be why the article states that it is not a covering map. – Niek de Kleijn Jul 16 '18 at 11:07

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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 $V = \varphi((0,2\pi)) = S^1 \backslash \{ 1 \}$.

Paul Frost
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  • I think $p$ is a covering too (I had essentially the same idea of you). What makes me doubt is that the author says "one easily sees it is NOT". – lucar Jul 15 '18 at 16:20
  • Any sequence of coverings $p_n : X_n \to X$ produces a continuous $p : \bigcup_{n=1}^\infty X_n \times { n } \to X$. I think in general it will be no covering because we will not able to find a neighborhood $U$ of a point $x$ which is evenly covered by all $p_n$. I don't know an example, but the above map $p$ is certainly not. – Paul Frost Jul 16 '18 at 10:50
  • Jeremy Brazas (the author of the article you quoted in your question) has answered https://math.stackexchange.com/q/989083. The example of a semicovering map which as not a covering map presented there is a different one than the above map $p$. – Paul Frost Jul 23 '18 at 12:50