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

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

pre-kidney
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    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. ;-) – Brian M. Scott Jul 20 '12 at 22:52
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    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. – Alex Becker Jul 20 '12 at 22:58
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    Thanks, although it sounds like you're killing a fly with a steamroller... – pre-kidney Jul 20 '12 at 23:03
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    @AlexBecker That is not true. Consider $U$ as the upper closed hemisphere of $S^2$ and $V$ as the lower one. – Aloizio Macedo Mar 14 '17 at 00:41
  • @Aloizio Indeed, this is only true in the plane. – domotorp Nov 01 '18 at 15:27

1 Answers1

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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 $U\cap V$?

Alex Becker
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