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In $(\mathbb{R^2}, \tau_E)$, I would like to find some open sets $U, V$ such that $U, V$ and $ U \cup V$ are all simply connected but $U \cap V$ is disconnected. I am not sure whether it is possible or not.

I thought that I could take $U = \{x^2 + y^2 = 1, y < \frac{1}{2}\}$ and $V = \{x^2 + y^2 = 1, y > - \frac{1}{2}\}$. This way I obtain that $U \cap V$ is disconnected but the problem is that $ U \cup V$ is only connected and not simply connected (as it is $\mathbb{S}^1$).

Do you have any idea of how to either find some $U,V$ that fit or prove that it is impossible to find such sets ?

Thank you very much in advance :)

yrual
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    https://math.stackexchange.com/questions/2745909/x-a-cup-b-be-an-open-cover-of-x-if-x-a-b-are-simply-connected-then-a-c?rq=1 gives a proof on why these $U$and $V$ does not exists. – Marcos Jan 25 '22 at 15:19
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    This question should be reopened. The OP has stated explicitly that he/she has not studied the method used in the answer. Several posters have asked for a proof without singular homology- – Matematleta Jan 25 '22 at 16:35

1 Answers1

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This cannot happen. Since $U\cup V$ is simply connected, then by the Hurewicz theorem it has trivial first homology. Since they are all path-connected then they also have trivial zeroth reduced homology. Now apply the reduced Mayer-Vietoris sequence to get

$$\tilde{H}_1(U\cup V)\to \tilde{H}_0(U\cap V)\to \tilde{H}_0(U)\oplus \tilde{H}_0(V)\to \tilde{H}_0(U\cup V)$$

exact sequence. Clearly this implies that $\tilde{H}_0(U\cap V)=0$ and so $U\cap V$ is path connected.

Note that $\mathbb{R}^2$ is irrelevant here, it works for any topological space. And as pointed by Paul Frost in the comment, it is enough when only $U\cup V$ is simply connected while $U$ and $V$ are path-connected.

freakish
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    Nice! It also works for path-connected $U,V$ and simply connected $U \cup V$. – Paul Frost Jan 25 '22 at 15:30
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    If you unwind how the Mayer-Vietoris sequence works, you effectively get the following nice geometric picture: Take two points in $U\cap V$. Find a path connecting them in $U$ and a path connecting them in $V$. Following one with the inverse of the other gives a loop, which is contractible, so there's a homotopy from this path to the constant one. If you take a fine enough subdivision of the square the homotopy is defined on, you can find an edgepath in it whose image lies entirely in $U\cap V$, proving path-connectedness. – Thorgott Jan 25 '22 at 15:40
  • @freakish Unfortunately I have never heard yet of the results you used in your answer but I will explore in this direction. Thank you very much. – yrual Jan 25 '22 at 15:44
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    @Matematleta I was writing a proof without homology but the question was closed – Aitor Iribar Lopez Jan 25 '22 at 16:23
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    What a shame. The OP even said explicitly that he/she hasn't studied singular homology so the answer is useless to him/her. – Matematleta Jan 25 '22 at 16:28
  • @AitorIribarLopez You can write an answer to the question which was referred to when the present question was closed. – Paul Frost Jan 25 '22 at 16:59
  • @Matematleta Seifert - van Kampen will not be helpful because its assumption is that $U \cap V$ is path-connected. – Paul Frost Jan 25 '22 at 17:02
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    I don't think this can be done using van Kampen. I thought about it, but it didn't get me anywhere. – freakish Jan 25 '22 at 17:13
  • @ Paul Yes, right. Thank you. I will delete my (silly) comment. Can you give me a hint (just a direction, not a proof) on how to give a proof that does not use homology? – Matematleta Jan 25 '22 at 17:13
  • @freakish as Paul Frost pointed out, it's a non-starter. Ugh. – Matematleta Jan 25 '22 at 17:14
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    @AitorIribarLopez I did not know that. So perhaps the OP does not know either. Why not just keep the question open? The OP stated clearly they had not studied homology. So, the answer (although it's the way to go, very nice!) is not useful to him/her. – Matematleta Jan 25 '22 at 17:16
  • @Matematleta I think the idea given by Thorgott here in the second comment might be a good one. I didn't check details though. – freakish Jan 25 '22 at 17:22
  • @freakish looks like an interesting approach. Thanks. – Matematleta Jan 25 '22 at 19:31