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Let the map $f:S^1 \times \mathbb{N} \to S^1$ defined by $f(z,n):=z^n$ is continuous and onto, but f is not a covering map from $S^1 \times \mathbb{N}$ onto $S^1$.

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Hint. Consider the inverse image of $1$. –  Arturo Magidin May 20 '12 at 2:36
    
I know as n approaches to inifity, the roots of the equation z^n=1 will approach to be dense on S^1, since for each open neighbourhood U of 1, the inverse image f^-1(U) contains f^-1(1) and so f^-1(U) contains a dense subset of S1 which would map under f to S1 due to the surjectivity, is this correct? –  Harry May 20 '12 at 3:56
    
@Arturo, while the roots of $1$ are dense in $S^1$, there are finitely many $f$-preimages of $1$ in each connected component of $S^1\times\mathbb N$, no? –  Mariano Suárez-Alvarez May 20 '12 at 5:09
    
@MarianoSuárez-Alvarez: I keep thinking this problem has a map from $S^2$ to $S^1$. Sigh... Thanks. –  Arturo Magidin May 20 '12 at 5:10
    
@MarianoSuárez-Alvarez Yes ... but I don't see how that's relevant to Arturo's hint. Is there a problem with my answer below? –  Neal May 20 '12 at 14:02
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3 Answers

Rather than obey the order in the title of this "question", I shall countermand this order.

For each $n \in N$, the subset $S^1 \times n$ is a component of $S^1 \times N$, it is also an open subset of $S^1 \times N$, and the restriction of $f$ to $S^1 \times n$ is a covering map.

Now check that for any function $f : X \to Y$, if each component of $X$ is open, and if the restriction of $f$ to each component is a covering map, then $f$ is a covering map.

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Finally reason! :D –  Mariano Suárez-Alvarez May 21 '12 at 1:28
    
Thanks! I have consulted my lecturer and he gave a similar reason for f is a covering map. The domain is S^1*N which isolates the roots of unity corresponding to each n in N, so at each level n the preimage of a nbh on S1 is a finite disjoint union of open sets on S1 each of which is homeomorphic to the nbh started with. So the f in the question is indeed a covering projection. –  Harry May 22 '12 at 5:38
    
@Harry You should accept his answer. –  Neal May 22 '12 at 11:38
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Here is a counterexample: Let $C_n = \mathbb R^n \times S^1 \times S^1 \times \cdots$, and define $p_n: C_n \to S^1 \times S^1 \times \cdots = X$ by $p_n(t_1, \ldots, t_n, z_1, z_2, \ldots) = (e^{2\pi i t_1}, \ldots, e^{2\pi i t_n}, z_1, z_2, \ldots)$. Then, each $p_n$ is a covering map, but $\coprod p_n : \coprod C_n \to X$ is not a covering map. Any neighborhood of $(1,1,\ldots)$ will contain a nhood of the form $U_1\times \cdots \times U_k \times S^1 \times S^1 \times \cdots$ which will not pullback to an appropriate neighborhood on $C_i$ for $i\ge k+1$. –  Justin Young May 22 '12 at 12:03
    
The example in the OP works because there is a uniformity of the required neighborhoods as $n\to \infty$. –  Justin Young May 22 '12 at 12:06
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For the record, here is an answer, that is, a proof that $f$ is a covering map. Let $z = e^{2\pi i x} \in S^1$. Then, let $U_x = \{e^{2\pi i t}\vert t\in (x-1/2, x+1/2)\}$. Then, $f^{-1}(U_x) = \coprod U_n \times \{n\}$, and $U_n = \coprod G_k$ where $G_k = \{e^{2\pi i t} \vert t - (2\pi k/n) \in (x/n - 1/2n, x/n + 1/2n )\}$ and $0\le k \le n-1$. Finally, $f\vert_{G_k}$ is a homeomorphism onto $U_x$.

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The following is incorrect because I misapplied the definition of covering map (tsk tsk). A better definition is: for each $x\in\mathbb{S}^1$ there is some neighborhood $U\ni x$ such that $f^{-1}U \cong \coprod_i U_i$ and for each $i$, $f|_{U_i}:U_i\to U$ is a homeomorphism.


Expanding the hint of Arturo Magidin in the comments, recall that for $f$ to be a covering map, for any $x\in\mathbb{S}^1$, there must be some small neighborhood $U$ about $x$ so that for any $z\in f^{-1}(x)$, there is a small neighborhood $U_z$ so that $f|_{U_z}$ is a homeomorphism of $U_z$ onto $U$. But the preimages of $1$ are dense in $\mathbb{S}^1$, so for any small $\epsilon$-neighborhood $U_\epsilon$ of $1$, there is a preimage of $U_\epsilon$ which maps onto all of $\mathbb{S}^1$.

Comment: This is true, and my definition is correct. However, it's a non-sequitur to insist that just because there is a large enough $n$ so that $U_\epsilon\times n$ maps by $f$ onto all of $\mathbb{S}^1$, there is no neighborhood which maps homeomorphically onto $U_\epsilon$! It is easier to see the correct idea by using the "better definition" given above. Take small $\epsilon$ and examine $f^{-1}U_\epsilon$.

The idea here is to exploit that $\mathbb{S}^1\times\mathbb{N}$ has a countably infinite number of components, each with a different covering map, so we cannot (as we might think to try) find an open neighborhood about (say) $1$ by taking intersections of suitable neighborhoods of each element of $f^{-1}(1)$.

Exercise: Given $\epsilon$, find $n$ so that there is an $n^{th}$ root of unity $w\neq 1$ in $B_\epsilon(1)$.

(Solution: Take $n>(2\pi\epsilon)^{-1}$.)


The correct idea may be found in Lee Mosher's answer.

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Can you explain "But the preimages of 1 are dense in Z 1 , so for any small ϵ -neighborhood U ϵ of 1 , there is a preimage of U ϵ which maps onto all of S 1" further? What does Z 1 mean? –  Harry May 20 '12 at 3:36
    
I understand the second paragraph of your argument, but I can't see the aim of that exercise which seems linked to the first paragraph that I am still puzzled about.. I will be greatly appreciated if you can explain it in more detail! Thanks! –  Harry May 20 '12 at 3:43
    
@Harry Sorry, that was a typo. Should have been "... of $1$ are dense in $\mathbb{S}^1$, so ..." –  Neal May 20 '12 at 12:38
    
The pre-image of $1$ in $S^1$ does not make sense in the context of this question, because the domain of the function $f$ is not $S^1$. The domain is $S^1 \times N$. –  Lee Mosher May 20 '12 at 23:20
    
However, because the roots of unity are dense in $\mathbb{S}^1$, for each neighborhood $U$ of $1$ there is an $n$ for which $f|_{U\times n}$ maps onto $\mathbb{S}^1$. Unless I am mistaken in my recollection of the definition of covering map, this proves that $f$ is not a covering map. –  Neal May 21 '12 at 0:34
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