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

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?

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.
• I think you mean $p\in U_1\setminus A$, if I am understanding correctly. May 14, 2018 at 19:31