5
$\begingroup$

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

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

  • 6
    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. ;-)2012-07-20
  • 2
    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.2012-07-20
  • 3
    Thanks, although it sounds like you're killing a fly with a steamroller...2012-07-20
  • 1
    @AlexBecker That is not true. Consider $U$ as the upper closed hemisphere of $S^2$ and $V$ as the lower one.2017-03-14
  • 0
    @Aloizio Indeed, this is only true in the plane.2018-11-01

1 Answers 1

8

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 $U\cap V$?

  • 2
    You've got some skinny kidneys :)2012-07-20
  • 0
    I was about to post left and right hemispheres intersecting on a circle, but this is even simpler! +1.2012-07-20
  • 9
    I was tempted to post a one-character hint: $\between$2012-07-20
  • 0
    But then you couldn't make your kidney pun...2012-07-20
  • 0
    @BrianM.Scott I like it! But that is probably (just a little) more work to describe formally than this.2012-07-20
  • 0
    @BrianM.Scott You should have, maybe you would get [200 upvotes](http://math.stackexchange.com/a/74383/14657)!2012-07-20
  • 0
    @Ragib: Oh, well. I’ve never been able to resist a pun.2012-07-20