I have to show that if $A$ and $B$ are compact connected subsets in the plane such that $A\cap B$ is not connected (and not empty), then $\Bbb R^2\setminus(A\cup B)$ is not connected.
The tool I must use is the coboundary map. I set $U=\Bbb R^2\setminus A$ and $V=\Bbb R^2\setminus B$. So my coboundary map is $$\delta:H^0(U\cap V)=H^0(\Bbb R^2\setminus (A\cup B))\to H^1(U\cup V)=H^1(\Bbb R^2\setminus(A\cap B)).$$ I think I should prove that its kernel and its image have dimension at least one, so that $H^0(\Bbb R^2\setminus (A\cup B))$ has dimension at least two.
I know that a function is in the kernel if and only if it can be written as the (restriction of the) difference of two locally constant functions, one on $U$ and one on $V$; and that $[\omega]$ is in the image if and only if the restrictions of $\omega$ to $U$ and to $V$ are both exact.
How can I prove the kernel and the image have dimension at least one?