I know by abstract results that $\mathbb R^2-\{\mathrm{pt}.\}$ has a triangulation. By how can I visualise one?
2026-03-27 10:15:49.1774606549
How does a triangulation of the punctured plan look like?
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You can think about how the complex exponential sends $\mathbb{C}$ to $\mathbb{C}-\{0\}$, where values different by $2\pi i$ are sent to the same value. This means if you take a triangulation of $\mathbb{C}$ that has this vertical symmetry, you can take its image to get a triangulation of $\mathbb{C}-\{0\}$.
Here's the sort of thing you can get: