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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. –  hilary Dec 9 '11 at 8:15

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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? –  Chris Eagle Dec 9 '11 at 14:54
    
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. –  Paul Dec 10 '11 at 1:15
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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. –  Chris Eagle Dec 10 '11 at 1:28
    
@Chris: Oh yes, you are right. I should be more careful. See my edited answer. –  Paul Dec 10 '11 at 5:55
    
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 pages.bangor.ac.uk/r.brown/gpdsweb.html –  Ronnie Brown Apr 26 '12 at 21:11

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