TITLE: $X=A\cup B$ be an open cover of $X$. If $X,A,B$ are simply connected , then $A\cap B$ path-connected?
QUESTION [6 upvotes]: 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)

REPLY [3 votes]: 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.