# Fundamental group of intersection trivial

Let $U,V$ be open sets in $\mathbb{R}^n$ such that $U,V,U\cap V$ are path connected, and $U\cup V=\mathbb{R}^n$. Let $x_0\in U\cap V$. Show that if $\pi_1(U,x_0)\ncong\{1\}$ then $\pi_1(U\cap V,x_0)\ncong\{1\}$.

My attempt: Assume $\pi_1(U\cap V,x_0)\cong\{1\}$. Then $\pi_1(\mathbb{R}^n,x_0)\cong\operatorname{colim}(\pi_1(U)\leftarrow\{1\}\rightarrow\pi_1(V))\cong\pi_1(U)\ast\pi_1(V)$ by van Kampen's theorem, where $\ast$ represents the free product of groups.

I also know that if $\pi_1(U)\cong\{1\}\cong\pi_1(V)$, then $\pi_1(U\cup V)\cong\{1\}$, but I'm not sure if that helps here. I feel like I'm close, but I'm not sure how $\pi_1(\mathbb{R}^n,x_0)\cong\pi_1(U)\ast\pi_1(V)$ implies that $\pi_1(U,x_0)\cong\{1\}$ (if I'm taking the right approach).

If $\pi_1(U,x_0)\not\cong \{1\}$ but $\pi_1(U\cap V,x_0)\cong\{1\}$ then as you said you have $\pi_1(\mathbb{R}^n,x_0)=\pi_1(U,x_0)*\pi_1(V,x_0)$ which is a contradiction since $\mathbb{R}^n$ has trivial fundamental group whereas the righthand side cannot be trivial since the fundamental group of $U$ is assumed to be not trivial.
• Right. So is it correct to say that this implies $\pi_1(U)\cong\pi_1(V)\cong\{1\}$? – mrose Mar 15 '18 at 23:56
• No. It is possible that the fundamental groups are not trivial but the amalgamated product is trivial. Set $U$ to be whole plane without the origin and set $V$ to be a disc containing the origin. Then the fundamental group of $U$ is not trivial but the fundamental group of $\mathbb{R}^2$ is of course trivial. – Levent Mar 16 '18 at 10:47
Well, $\pi_1(\Bbb R^n, x_0)\cong\{1\}$ and we have embeddings $A\hookrightarrow A*B\hookleftarrow B$ for groups $A$ and $B$.
In this case it would mean that both $\pi_1(U)$ and $\pi_1(V)$ are subgroups of the trivial group, i.e. are themselves trivial.