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 !

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    $\begingroup$ What are $U$ and $V$? If they are just subsets of some topological space $X$ then it is definitely possible for $U$ and $V$ to have non-simply connected intersection (consider two unit sphere offset at their centers by some small vector). $\endgroup$ – Dan Rust Aug 17 '13 at 15:51
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    $\begingroup$ The following statement is true: If $U$ and $V$ are open and simply connected and $U ∩ V$ is path connected, then $U ∪ V$ is simply connected. $\endgroup$ – user87690 Aug 17 '13 at 16:41
  • $\begingroup$ Yes they are. Ok you are right, I thought of a similar example that made me wondering... Thanks for making it clear. $\endgroup$ – unknownMathematician Aug 17 '13 at 16:41
  • $\begingroup$ @user87690: I think you mean that $U \cap V$ is path connected. Then your statement follows from Seifert-Van Kampen, right? $\endgroup$ – unknownMathematician Aug 17 '13 at 16:42
  • $\begingroup$ @unknownMathematician: Yes, you're right. $\endgroup$ – user87690 Aug 17 '13 at 17:04

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.


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