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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 :)

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    $\begingroup$ math.stackexchange.com/questions/2745909/… gives a proof on why these $U$and $V$ does not exists. $\endgroup$
    – Marcos
    Jan 25 at 15:19
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    $\begingroup$ 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- $\endgroup$ Jan 25 at 16:35

1 Answer 1

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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.

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    $\begingroup$ Nice! It also works for path-connected $U,V$ and simply connected $U \cup V$. $\endgroup$
    – Paul Frost
    Jan 25 at 15:30
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    $\begingroup$ 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. $\endgroup$
    – Thorgott
    Jan 25 at 15:40
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    $\begingroup$ @Matematleta I was writing a proof without homology but the question was closed $\endgroup$ Jan 25 at 16:23
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    $\begingroup$ What a shame. The OP even said explicitly that he/she hasn't studied singular homology so the answer is useless to him/her. $\endgroup$ Jan 25 at 16:28
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    $\begingroup$ I don't think this can be done using van Kampen. I thought about it, but it didn't get me anywhere. $\endgroup$
    – freakish
    Jan 25 at 17:13

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