Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $S^1$.

share|cite|improve this question
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

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.

share|cite|improve this answer
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
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

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 $U_x$.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.