$X=A\cup B$ be an open cover of $X$. If $X,A,B$ are simply connected , then $A\cap B$ path-connected?

Question

I'm trying to prove or find counterexample to the following :

Let $X=A\cup B$ be an open cover of $X$. Assume that $X,A,B$ are simply connected , then $A\cap B$ must be path-connected.

I tried a proof by contradiction : Assume that $A\cap B$ not path-connected , and pick loop which start at one of the connected components, say $C_1$ , travelling through all the other connected components , then returning to $C_1$.

I think the fact this loop is nullhomotopic will imply that A∩B is connected, but I haven't found such a proof.

(I'm not sure if that true)

Answer 1

You can apply the Mayer-Vietoris sequence:

$$H_1(X)\to \widetilde{H}_0(A \cap B) \to \widetilde{H}_0(A) \oplus \widetilde{H}_0(B)$$

Since $X$ is simply connected, $H_1(X) = 0$. Since $A$ and $B$ are simply connected, they are path connected and so their reduced 0-th homology groups are zero. Therefore, $\widetilde{H}_0(A \cap B) = 0$ and so $A \cap B$ is path connected. Here I am using singular homology.

Answer 2

Here is an interesting example to think about that doesn't prove or disprove the question.

Let $X$ be the quasi-circle shown in the figure, a closed subspace of $\mathbb R^2$ consisting of a portion of the graph of $y = sin(1/x)$, the segment $[−1, 1]$ in the $y$ axis, and an arc connecting these two pieces.

The important point about this example is that $X$ is simply connected even though it somewhat resembles a circle.

Now if you pick $A$ to be an open piece in the top part and $B$ to be an open piece in the bottom part, we get $B,X$ is simply connected, and $A$ is almost simply connected, i.e. all loops are null-homotopic ($A$ fails to be simply connected because it isn't path-connected). Lastly observe that $A\cap B$ is not path-connected!

Edit: $A$ is the intersection of the pink region with the quasi-circle and $B$ is the intersection of the purple region with the quasi-circle.