fundamental group of the union of two subsets 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
 A: 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.
