Consider the annulus $$A:=\{(x,y)\in\mathbb{R}^{2}\mid 1\leq x^{2}+y^{2}\leq 4\}$$ I have to find all (connected) covering up to isomorphism.

My proof so far:

First of all, we have a homeomorphism $A\cong S^{1}\times [1,2]$. Therefore, we find $\pi_{1}(A)\cong \pi_{1}(S^{1}\times [1,2])\cong \pi_{1}(S^{1})=\mathbb{Z}$. All the subgroups of $\mathbb{Z}$ are of the form $\mathbb{Z}n$ for $n\in\mathbb{N}$ and since $\mathbb{Z}$ is abelian, all of them define distinct conjugacy classes of subgroups.

Now there is a theorem, which says that,if $A$ admits an universl cover, the then there are for each $n$ a covering $(\widetilde{A},p)$ such that $p_{\ast}(\pi_{1}(\widetilde{A}))\cong n\mathbb{Z}$ and this are all coverings up to isomorphism.

Now, I know that for $S^{1}$, we can define the maps $f_{n}:z\mapsto z^{n}$, which define covering maps of $S^{1}$, such that $(f_{n})_{\ast}(\pi_{1}(S^{1}))\cong n\mathbb{Z}$. So therefore, my idea was to define the maps $g_{n}:S^{1}\times [1,2]\to A\cong S^{1}\times [1,2]$ exactly by $g_{n}(z,\lambda):=(f_{n}(z),\lambda)$. This should then be a covering with the claimed property $(g_{n})_{\ast}(\pi_{1}(S^{1}\times [1,2]))\cong n\mathbb{Z}$.

Now to my question: Does this look right so far? and furthermore, the annulus $A$ is not simply connected, therefore all of my constructed coverings above are not universal. So I still have to find a universal cover, in order to show that I can apply the theorem above.

  • 3
    $\begingroup$ You know the universal cover of $S^1$, and you observed that $A$ is homotopy equivalent to $S^1$, so the universal cover of $A$ should be similar... $\endgroup$
    – kamills
    Jan 5, 2021 at 15:32
  • 1
    $\begingroup$ Consider $p:\Bbb R\times [1,2]\longrightarrow A$ defined as $p(\theta,r)=re^{2\pi i\theta}$ $\endgroup$
    – Sumanta
    Jan 5, 2021 at 15:40
  • 1
    $\begingroup$ Ahh I see thanks.... And the rest of my idea is right? $\endgroup$
    – B.Hueber
    Jan 5, 2021 at 15:41

1 Answer 1


Thanks to the comments, I think I can answer it by myself.

All the coverings of the annulus are (up to isomorphism):

(1) The universal cover $$p:\mathbb{R}\times [1,2] \to A, \\ (t,\lambda)\mapsto (\lambda \cos(2\pi t),\lambda\sin(2\pi t))$$

(2) For each $n\in\mathbb{N}$ the cover $$g_{n}:S^{1}\times [1,2] \to A, \\ (e^{i\theta},\lambda)\mapsto (\lambda \cos(\theta n ),\lambda\sin(\theta n ))$$

  • 1
    $\begingroup$ I would suggest writing (2) as $(e^{i\theta},\lambda)\mapsto \lambda e^{in\theta}$ where we identify $\mathbb R^2 = \mathbb C$, just because otherwise it reads as "select a point on the unit circle with coordinate $z = e^{i\theta}$, extract the angle $\theta$, and feed the angle to trig functions" which is messy. Otherwise excellent! $\endgroup$ Jan 6, 2021 at 2:11

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