If $A$ and $B$ are bounded simply connected open subsets of $\mathbb{R}^2$ and if $ A \cap B \neq \emptyset$ then a connected component $C$ of $A\cap B$ is simply connected?

I think it must be true intuitively but how can I prove it?


1 Answer 1


Suppose for a contradiction that there is a connected component $C \subset A \cap B$ with $\pi_{1}(C) \neq \{id\}$.

It is not hard to see that there is a simple closed curve $c$ representing a non-trivial element $ [c] \in \pi_{1}(C)$ (using the fact that it is open if neccesary). By the Jordan curve theorem, $\mathbb{R}^{2} \setminus c$ has two connected components, one bounded say $U_{1}$ and one unbounded say $U_{2}$.

Now, atleast one of $A,B$ does not contain $U_{1}$ entirely, since if they did then clearly $c$ would be contractible inside $C \subset A \cap B$.

Without loss of generality, suppose there exists $p \in U_{1} \setminus A$, then clearly $id \neq [c] \in \pi_{1}(A)$, which contradicts the fact that $A$ is simply connected.

  • $\begingroup$ I think you mean $p\in U_1\setminus A$, if I am understanding correctly. $\endgroup$ May 14, 2018 at 19:31
  • $\begingroup$ Yes, that was a typo. Thanks for pointing it out. $\endgroup$
    – Nick L
    May 14, 2018 at 19:40

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