# Intersection of Simply-Connected Sets

Let $U,V$ be two simply connected subsets of a topological space.

Prove or disprove: $U \cap V$ is simply connected.

• HINT: Find two simply connected subsets of the plane whose intersection isn’t even connected; a pair of kidneys will work, if you orient them properly. ;-) Jul 20, 2012 at 22:52
• It is perhaps worth noting that this is essentially the only way in which $U\cap V$ can fail to be simply connected; if $U\cap V$ is path-connected then it is simply connected by Mayer-Vietoris. Jul 20, 2012 at 22:58
• Thanks, although it sounds like you're killing a fly with a steamroller... Jul 20, 2012 at 23:03
• @AlexBecker That is not true. Consider $U$ as the upper closed hemisphere of $S^2$ and $V$ as the lower one. Mar 14, 2017 at 0:41
• @Aloizio Indeed, this is only true in the plane. Nov 1, 2018 at 15:27

Let $S^1$ be the circle in $\mathbb R^2$, $U=\{(x,y)\in S^1: x\geq 0\}$ and $V=\{(x,y)\in S^1: x\leq 0\}$. Then $U$ is the right half of a circle and $V$ is the left half, both of which are simply connected. What is $U\cap V$?
• I was tempted to post a one-character hint: $\between$ Jul 20, 2012 at 23:04