TITLE: Open sets $U, V$ such that $U, V$ and $ U \cup V$ are all simply connected but $U \cap V$ is disconnected
QUESTION [5 upvotes]: In $(\mathbb{R^2}, \tau_E)$, I would like to find some open sets $U, V$ such that $U, V$ and $ U \cup V$ are all simply connected but $U \cap V$ is disconnected. I am not sure whether it is possible or not.
I thought that I could take $U = \{x^2 + y^2 = 1, y < \frac{1}{2}\}$ and $V = \{x^2 + y^2 = 1, y > - \frac{1}{2}\}$. This way I obtain that $U \cap V$ is disconnected but the problem is that $ U \cup V$ is only connected and not simply connected (as it is $\mathbb{S}^1$).
Do you have any idea of how to either find some $U,V$ that fit or prove that it is impossible to find such sets ?
Thank you very much in advance :)

REPLY [7 votes]: This cannot happen. Since $U\cup V$ is simply connected, then by the Hurewicz theorem it has trivial first homology. Since they are all path-connected then they also have trivial zeroth reduced homology. Now apply the reduced Mayer-Vietoris sequence to get
$$\tilde{H}_1(U\cup V)\to \tilde{H}_0(U\cap V)\to \tilde{H}_0(U)\oplus \tilde{H}_0(V)\to \tilde{H}_0(U\cup V)$$
exact sequence. Clearly this implies that $\tilde{H}_0(U\cap V)=0$ and so $U\cap V$ is path connected.
Note that $\mathbb{R}^2$ is irrelevant here, it works for any topological space. And as pointed by Paul Frost in the comment, it is enough when only $U\cup V$ is simply connected while $U$ and $V$ are path-connected.