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I am trying to compute $\pi_1(X)$ where $X$ is the following collection of circles:

$$ X = \left[\bigcup_{(m,n) \in \mathbb{Z}^2} (m+ in) + S^1 \right] \cup \left[\bigcup_{(m,n) \in \mathbb{Z}} \big((m + \tfrac{1}{2}) + i (n + \tfrac{1}{2}) \big) + (1 - \sqrt{2})S^1 \right]$$

I have computed the formula to make the circle tangent patterns to be what it looks like. This is possibly related to the truncated square tiling.

There is an action $\mathbb{Z}^2 \times X \to X$ leaving this space invariant, so this is a covering space $X \to Y$. So perhaps it's sufficient to compute $\pi_1(X/\mathbb{Z}^2)$.

enter image description here

Eric Wofsey
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cactus314
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    this is a 1-dimensional CW complex with a countably infinite number of cells. So the fundamental group is free of countable rank. – Rylee Lyman Jun 27 '19 at 02:17

1 Answers1

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This space is just a graph (consider the intersection points of the circles as vertices and the arcs connecting them as edges). Modding out a spanning tree, it is thus homotopy equivalent to a wedge of circles. It is clear there must be countably infinitely many circles, so $\pi_1(X)$ is a free group on countably infinitely many generators.

Eric Wofsey
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