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: enter image description here

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.


1 Answer 1


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 $\mathbb Z$.

Alternatively, you could use Leray's Nerve Theorem (https://en.wikipedia.org/wiki/Nerve_complex#Nerve_theorems) to show that the space is homotopy equivalent to a circle.

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.

  • $\begingroup$ Perhaps you should explain how to show that the "glued space" is simply connected. $\endgroup$ May 13 at 15:03
  • $\begingroup$ @KritikerderElche I added some details. $\endgroup$ May 14 at 3:20

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