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Let $X$ be a topological space and $U,V \subset X$ two open subsets such that $U \cap V$ and $U \cup V$ are both simply connected. How can i show that $U$ and $V$ are simply connected? Thanks in advance.

hilary

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    yes i know. but i tried with van kampen's. but i have no idea how to apply it.2011-12-09

1 Answers 1

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Ok. Let me expand my answer. By Seifert–van Kampen theorem, we know that the fundamental group $\pi_1(U\cup V)$ of $U\cup V$ is the free product of the fundamental groups of $U$ and $V$ with amalgamation of $\pi_1(U\cap V)$. Since $U\cap V$ is simply connected by assumption, i.e. $\pi_1(U\cap V)=0$, $\pi_1(U\cup V)$ is the free product of $\pi_1(U)$ and $\pi_1(V)$. Since $U\cup V$ is simply connected by assumption, i.e. $\pi_1(U\cup V)=0$, we must have $\pi_1(U)=0$ and $\pi_1(V)=0$; otherwise, if $\pi_1(U)\neq 0$ or $\pi_1(V)\neq 0$, the free product of $\pi_1(U)$ and $\pi_1(V)$ must be non-trivial. For the above facts about free product, you can refer to here.

Note added: As Chris said, I should prove that $U$ and $V$ are path-connected before I can apply Seifert-van Kampen theorem. Here is the proof: note that $U\cup V$ and $U\cap V$ are simply-connected by assumption, which implies that $U\cup V$ and $U\cap V$ are connected. Now by the Mayer-Vietoris sequence, we have $$H_0(U\cap V)\rightarrow H_0(U)\oplus H_0(V)\rightarrow H_0(U\cup V)\rightarrow 0.$$ Since the rank of the zero homology $H_0$ is equal to the number of connected components, by the above exact sequence $H_0(U)$ and $H_0(V)$ has rank 1, which implies that $U$ and $V$ are connected. Since $U$ and $V$ are open by assumption, $U$ and $V$ must be path-connected.

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    Surely we need to show that $U$ and $V$ are path-connected before we can apply the theorem?2011-12-09
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    Right, but I think this is standard: if $U$ and $V$ are simply connected, then of course they are connected. Moreover, they are open by assumption. And it can be proved that open connected set must be path connected.2011-12-10
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    But we *need* $U$ and $V$ to be path-connected *before* we can use van Kampen to conclude they are simply-connected. It's not hard to show that if $U$ and $V$ are open and both $U \cup V$ and $U \cap V$ are path-connected, then $U$ and $V$ are path-connected, but it has to be done.2011-12-10
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    @Chris: Oh yes, you are right. I should be more careful. See my edited answer.2011-12-10
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    Please do not overlook the fact that there is a Seifert-van Kampen theorem for the fundamental groupoid on a set of base points, which allows for the computation of the fundamental group of the circle, as well as many other examples. For the use of groupoids, see http://pages.bangor.ac.uk/r.brown/gpdsweb.html2012-04-26
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    It is possible to extract from the usual proof of the Seifert-van Kampen Theorem a proof that the union of two open simply-connected sets with path connected intersection is simply connected, without using the notion of group or groupoid. But you get more information by having these concepts around. The same applies in higher dimensions, using higher order versions of the fundamental groupoid.2012-10-07