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In Fundamental group of $\mathbb{R}^2 \setminus (\mathbb{Z} \times \{0\})$ it is said because $Y=\mathbb{R}^2-(\mathbb{Z}\times \{0\})$ has a countable points less than whole $\mathbb{R}^2$, then at each point we can retract the space into circles, and finally the space can retract to a wedge of infinite countable circles.
Could anyone elaborate this more? Thank you.

Laurence PW
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  • What is the fundamental group of $\mathbb{R}^2 \setminus {0}$? Use this intuition for the example. – Alvin Jin Oct 14 '21 at 18:55
  • What more do you need? Enlarge each point into a circle of radius 1/2. Now you have a long line of tangent circles (a chain of beads). Now slide each tangent circle down to the same point of tangency on each. (This cannot be done in the plane.) – Randall Oct 14 '21 at 18:55
  • @AlvinJin Intuitively I know the fundamental group, but it is difficult for me to do it formally – Laurence PW Oct 14 '21 at 19:01
  • @Randall Yes I know the intuition, but I have difficulty to write it formally – Laurence PW Oct 14 '21 at 19:02
  • Write what formally? Do you want to write down an explicit homotopy equivalence? – Randall Oct 14 '21 at 19:02
  • @Randall yes the retraction – Laurence PW Oct 14 '21 at 19:04
  • Fun fact: Every non-compact orientable surface is homotopy equivalent to a wedge of at most countably many circles. – Sumanta Oct 16 '21 at 03:32

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