TITLE: Does this arrangement of polygons necessarily contain a hole?
QUESTION [6 upvotes]: Consider an arrangement of finitely-many open polygons in the plane (not necessarily convex) such that each polygon intersects at least two other non-intersecting polygons.
Is there always a sub-arrangement such that the union of all polygons in the sub-arrangement contains a hole?
In the illustration below, the blue arrangement on the left satisfies the requirements and indeed it contains a hole; the brown arrangement on the right does not satisfy the requirements (each polygon intersects two other polygons, but these other polygons intersect), and it does not contain a hole.

REPLY [2 votes]: I believe the following approach should work for all polygons (I assumed that (1) the polygons are open, (2) existence of holes refers to a non-trivial first homology group and (3) the condition means that all polygons intersecting a fixed polygon are mutually disjoint):
Let $\mathcal{P}=\{P_1,\ldots,P_n\}$ be the polygons. We first show that the union of some subsets of $\mathcal{P}$ has a hole.
By the assumption, re-ordering if necessary, we can assume that for some $(4\le) k\le n$ and for all $1\le i<j\le k$, $P_i\cap P_j\neq\emptyset $ iff $j=i+1$ or $(i,j)=(1,k)$. Using the Mayer-Vietoris sequence (for reduced homology) and the fact that polygons are contractible, we have:
If $P_i\cap P_{i+1}$ is not connected for some $i\in\{1,\ldots,k-2\}$, take the minimum such $i$, then $H_1(P_1\cup \cdots \cup P_{i+1})\simeq\tilde{H}_0(P_i\cap P_{i+1})$ is non-trivial. Otherwise, $H_1(P_1\cup \cdots \cup P_k)\simeq \tilde{H}_0((P_1\cup P_{k-1})\cap P_k)$, which is non-trivial by construction.
Suppose $H_1(\bigcup \mathcal{P})=0$, let $m$ be the smallest integer such that $H_1(P_1\cup\cdots\cup P_m)=0$. Since $(P_1\cup\cdots\cup P_{m-1})\cap P_m$ is a disjoint union of open polygons, it has a trivial first homology. Once again by Mayer-Vietoris, $0=H_1(P_1\cup\cdots\cup P_{m-1})\oplus H_1(P_m)\simeq H_1(P_1\cup\cdots\cup P_{m-1})$, a contradiction.
Edit: A later edit to the question has clarified that my assumption (3) is not what was intended. In any case, I will leave this answer as it is, because the question under assumption (3) is also a very interesting one.