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Ahoy Mathematicians,

We are preparing for qualifying exams and ran across this question. There were two proposed methods of approaching it.

  1. Recognize that the thrice punctured plane is homotopic to the wedge of 3 circles. Following the example in Hatcher for the 2-fold coverings of the wedge of 2 circles, we construct the 3 connected covering spaces (up to relabeling) of the wedge of three circles. There is also the disconnected covering space of two copies of itself.
  2. Use the lifting correspondence to show that the 2 fold covering spaces of the thrice punctured disk are in one-to-one correspondence with subgroups H of $\mathbb{Z} * \mathbb{Z} * \mathbb{Z}$ of index 2. There are 3 such groups, giving 3 connected covering spaces.

You'll notice that the ideas in these two approaches agree. The question is one of rigor. Which answer is more accurate? Does Approach 1 actually give us the correct covering spaces, or do we need to think about crossing it with something to cover the surface rather than the 1-mfld? If we use Approach 2, does this answer the question, or do we need to give more of a classification than just the fundamental groups?

Thanks!

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Ahoy OP!

Ultimately, I think it comes down to how the answer is written and how much wiggle room your grader is willing to give.


I see the first argument as

“We have a bijection between connected two-fold covers and index two subgroups of the fundamental group. Since the fundamental group is invariant under homotopy equivalence, we can find two-fold covers of the three-petaled rose. Here they are, and here are their fundamental groups. Thus, we know that there are exactly three connected covers of the surface, classified by their fundamental group.

The key aspect here is the bolded statement. A loose grader might understand that you’re implicitly relating the covers of the graph with the covers of the surface via their fundamental groups, but an answer I feel comfortable with would include some form of the bolded statement.


The second proof feels less solid to me. The argument seems to presuppose that we already know what the index two subgroups of $F_3$ are. Off the top of my head, I couldn’t rattle them off for you. The way I would prove that we know that there are three such subgroups would be to use your first argument. I’m sure there’s a nice group theoretic answer as well, but that is one that you would have to spell out.

That said, were you to spell out that group theoretic answer, then it would be a fine proof. I see nothing wrong with that direction inherently.


TL;DR If you give all the details and prove your statements, both options seem perfectly viable.

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  • $\begingroup$ Also, note that if the question asked you to “give” or “draw” the covering spaces and not to “classify” them, then both answers would fall short. $\endgroup$ – Santana Afton Jun 25 at 4:41
  • $\begingroup$ Thanks for this thorough analysis! I appreciate it. $\endgroup$ – Beccah M. Jun 25 at 17:16
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I'd like to propose another method of "classification", for completeness perhaps.

Let $G$ be a group. It is a result (tom Dieck, p. 345) that the set of isomorphism classes of principal $G$-bundles $p : \widetilde{X} \rightarrow X$, denoted $\mathcal{B}(X, G)$ is in bijective correspondence with the set $[ X, BG]$ of homotopy classes $X \rightarrow BG$ where $BG$ is the classifying space of $G$, i.e. we have $$\mathcal{B}(X, G) \cong [X, BG].$$

Letting $G = S_n$ the symmetric group on $n$ letters, an exercise in tom Dieck, p. 351 shows that principal $S_n$-bundles $p : \widetilde{X} \rightarrow X$ correspond to $n$-fold covering spaces $p_n : \widetilde{X} \times_{S_n} \{ 1, \dots, n \} \rightarrow X$ where $\widetilde{X} \times_{S_n} \{ 1, \dots, n \} = \widetilde{X} \times \{ 1, \dots, n \} / g(x, n) = (xg^{-1}, gn) $.

Then if we let $X = S^1 \vee S^1 \vee S^1 \simeq \mathbb{R}^2 \setminus \{ x_1, x_2, x_3 \}$ and $G=S_2$, $\mathcal{B}(X, G)$ becomes the set of covering spaces we would like to "classify" by the second result, and by the first result we see that these covering spaces correspond to the set $[X, BG]$ of homotopy classes $S^1 \vee S^1 \vee S^1 \rightarrow BS_2$. Finally, it is a general result that $$ BS_n \cong \mathrm{UConf}_n (\mathbb{R}^\infty) $$ the unordered configuration space of $\mathbb{R}^\infty$ on $n$-tuples. In our case, $BS_2$ can be described as $$ BS_2 \cong \frac{\{ f \mid f : \{1, 2 \} \rightarrow \mathbb{R}^\infty \} }{S_2} = \frac{\{ f \mid f : \{1, 2 \} \rightarrow \mathbb{R}^\infty \} }{\{\mathrm{id}, (1 \, 2 )\} }. $$

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  • $\begingroup$ Thanks for the response! Some of the machinery in this response is a big beyond the pay grade of the class, but it's definitely a nice solution. Thanks for adding it to round out the discussion. :) $\endgroup$ – Beccah M. Jun 25 at 17:19
  • $\begingroup$ You're welcome! :) $\endgroup$ – mathphys Jun 26 at 12:17
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There is a very computable approach, also using classifying spaces, using the facts that 2-fold coverings are the same as principal $\mathbb{Z}/2$ bundles and $B\mathbb{Z}/2 \simeq \mathbb{RP}^\infty \simeq K(\mathbb{Z}/2,1)$, the Eilenberg-Maclane space representing $H^1(-;\mathbb{Z}/2)$. Then we can compute isomorphism classes of 2-fold covering spaces over your punctured plane $P$ by computing

$$ H^1(P;\mathbb{Z}/2) \cong H^1(\vee^3S^1; \mathbb{Z}/2) \cong(\mathbb{Z}/2)^3 $$

so in fact there are 8 of them (up to isomorphism over $id_P$), corresponding to the different choices of wether they are trivial over each circle. There is only one disconnected double cover: the trivial one. Moreover since $\mathbb{Z}/2 \cong O(1)$ we can say that the bundles are classified by their first Steifel-Whitney class.

I think I can explain why I get 7 connected double covers where you get 3: my classification result is up to isomorphism of covering spaces over $P$, but I think your "up to relabeling" corresponds to isomorphisms of the bundle over some homeomorphism of $P$, which when considered as $\vee^3 S^1$ corresponds (up to homotopy) to permuting the wedge summands, which in $H^1$ corresponds to permuting the factors. Then, up to permutation, there are only three covering spaces that are not trivial: $(1,0,0),\ (1,1,0)$ and $(1,1,1)$.

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  • $\begingroup$ I agree with your analysis of why you got 7 and we got 3 connected covers. Thanks for this response! $\endgroup$ – Beccah M. Jun 25 at 17:18

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