I read the following statement in the old question "Intersection of Simply-Connected Sets" (Intersection of Simply-Connected Sets):
If $U$ and $V$ are simply connected and $U \cap V$ is path connected, then $U \cap V$ is simply connected.
The author said, that this follows by Mayer-Vietoris, but I dont see why this is the case since this is a statement about homology. Can anyone give me a hint?
I thought about using the fact, that the first homology group is the abelization of the fundamental group, but I couldnt derive a proper argumentation out of this.
If this statement is false, is there any general condition?
Thanks a lot !
Best Answer
This isn't true. Another counter-example (similar to Dan Rust's) is simply taking the closed upper hemisphere of $S^2$ as $U$ and the lower as $V$. If you want $U,V$ open in $\mathbb{R}^3$, just inflate the hemispheres a little.
What is true is the following:
Given two simply-connected open regions $U,V$ on the plane such that $U \cap V$ is path-connected, then $U \cap V$ is simply connected (instead of requiring $U \cap V$ to be path-connected, one could simply conclude that $U \cap V$ has all its path-connected components simply connected).
This follows from Mayer-Vietoris, since we have the following fragment:
$$\cdots \to H_2(U \cup V) \to H_1(U \cap V) \to H_1(U) \oplus H_1(V) \to \cdots $$ We now have that $H_2(U \cup V)=0$ since $U \cup V$ is a non-compact $2$-manifold. We also have that $H_1(U)=0=H_1(V)$, therefore we can conclude that $H_1(U \cap V)=0$.
It follows now from this MO post that $\pi_1(U \cap V)$ must be free. In particular, if it were non-trivial it would not be perfect, leading to a contradiction. Funny thing is that you can even use Hurewicz's theorem to prove this, as is shown here.