# Computing $\pi_1$ of subset $X\subseteq \mathbb R^2$, $X$ union of 3 simp.conn. subspaces w/ simp.conn. pairwise intersection but empty intersection

Let $$X$$ be a subset of $$\mathbb R^2$$, and suppose that $$X$$ is equal to the union of open and simply connected subspaces $$V_1,V_2,V_3$$. Moreover, assume that the pairwise intersections $$V_i \cap V_j, \forall i,j$$ are simply connected, but $$\bigcap_{i=1}^3 V_i = \emptyset$$. Find $$\pi_1 (X)$$.

Here's my sketch of a proof of this:

1. Suppose for a similar space $$Y$$ that $$V_1 \cap V_2 \cap V_3 \neq \emptyset$$. Then since each pairwise intersection is simply-connected, $$V_1 \cap V_2 \cap V_3$$ will also be simply connected.
2. This implies that $$Y$$ would be the union of open, simply connected subspaces with connected intersection, so $$Y$$ is contractible.
3. However, since the intersection is empty, therefore we must have $$X=Y\setminus S$$ for some simply-connected subspace $$S\subset Y$$.
4. In $$X$$, let $$x_i \in V_i \cap V_j$$ and we can make three paths $$\gamma_i$$ from $$x_1 \to x_2, x_2 \to x_3, x_3 \to x_1$$.
5. The composition $$\Phi = \gamma_3 \circ \gamma_2 \circ \gamma_1$$ is a loop in $$X$$ that will enclose $$S$$, and in particular $$\Phi$$ is homotopic to $$S^1$$.
6. Since each $$V_i$$ is simply connected, the homotopy class of $$\Phi$$ will be unchanged by choosing different component paths $$\gamma_i$$.
7. Therefore $$\pi_1 (X) \cong \mathbb Z$$.

This is what I drew for geometric intuition:

Edit: One other thought I had for proving this was somehow showing that $$X$$ is homotopic to an annulus? I think this is probably true? Not sure how I would prove that, though.

I think the easiest way to calculate $$\pi_1(X)\cong\mathbb Z$$ is to construct the universal cover. Consider infinitely many disjoint copies of each $$V_i$$, denoted $$\{V_i(k)\}_{k\in \mathbb Z}$$. Then glue $$V_1(k)$$ to $$V_2(k)$$ along the copy of the intersection $$V_1\cap V_2$$. Similarly glue $$V_2(k)$$ to $$V_3(k)$$. Finally glue $$V_3(k)$$ to $$V_1(k+1)$$. Now verify that this space is simply connected and that the group of deck transformations is $$\mathbb Z$$.
Edit: Here are more details on why the glued space is simply connected. Since a loop is compact, it will land in a finite union of the lifted sets $$V_i(k)$$, so by induction it suffices to show that the union of two open simply connected spaces glued along a simply connected subspace is simply connected. This follows from van Kampen's theorem, but actually it follows from a much weaker and easier to prove lemma:
Lemma: Let $$X=U\cup V$$ where $$U,V$$ are path connected open sets, and suppose that $$U\cap V$$ is also path connected. Let $$x_0\in U\cap V$$. Then any loop in $$X$$ based at $$x_0$$ is homotopic rel boundary to a product of loops completely contained in either $$U$$ or $$V$$. In fancy language, there is a surjection $$\pi_1(U)*\pi_1(V)\twoheadrightarrow \pi_1(X)$$.
To prove this, one uses the Lebesgue number lemma to break up a loop into segments each contained in either $$U$$ or $$V$$, enlarges the segments so that their endpoints lie in the intersection $$U\cap V$$, and then adds hairs that connect the ends back to the basepoint. See for example Lemma 1.15 of Allen Hatcher's Algebraic Topology.