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What is 1st homology group of $X = \text{plane} - \bigcup_{n}\{(1/n, 0)\}\cup\{(0,0)\}$?

My guess is since it is $\operatorname{Ab}(\Pi_1(X))$.

It is a subgroup of $\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}\dotsb$ countably many times, generated by all vectors which are $0$ except at finitely many terms and the diagonal group ie. all vectors of form $[n , n , n , n ,\dotsc]$ where $n$ is any integer. As any loop either loops about finitely many of pts $(1/n ,0 )$ or it loops origin in which case it loops around all but finitely many of them.

Am I right? If yes what is a rigorous proof, what is a good reference?

mdp
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rohit
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  • Intuitively, I'd say it is homotopy equivalent to $(I\times{0,1})\cup({0}\cup{1/n\mid n\in\mathbb N}\times I)$. Its homology can be computed efficiently via Mayer-Vietoris sequence. – Stefan Hamcke May 02 '13 at 11:47
  • Let $Y=[-1,2]\times{-1,1}\cup{-1,2}\times[-1,1]\cup{\frac1{2n}+\frac1{2(n+1)} \mid n\in\mathbb N}\times[-1,1]$. I think one can show that $X$ deformation retracts onto $Y$. Maybe I'll make an answer of it if I find the time. – Stefan Hamcke May 02 '13 at 12:13
  • i have no clue what you are thinking of, can you please elaborate, because the space X is path connected and your spaces like (I×{0,1})∪({0}∪{1/n∣n∈N}×I) is it connected?? – rohit May 02 '13 at 12:53
  • The first space Stefan described is essentially a comb space (http://en.wikipedia.org/wiki/Comb_space) with another interval laid 'on top'. – Dan Rust May 02 '13 at 13:04
  • Yes it is connected, but I'm not sure anymore about the deformation retraction. P.S.: You really should learn how to write in latex on this site. – Stefan Hamcke May 02 '13 at 13:53

1 Answers1

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Think of the plane as $\mathbb C$ and apply the homeomorphism $z\mapsto 1/z$. Then your space is homeomorphic to $\mathbb C\setminus\mathbb N$. Now it's very easy to see what's happening since there the accumulation point has been pushed out to infinity. The result is then homotopy equivalent to a wedge of countably many circles. (Countably many circles identified at a single point.) The first homology of a countable wedge of circles is similar to what you described in your post. It is a direct sum of countably many copies of $\mathbb Z$: in notation, $\bigoplus_{n=1}^\infty \mathbb Z$. This is different from the direct product in that all but finitely many coordinates are required to be $0$. You might ask, what about the diagonal elements $(n,n,...)$ that you seemed to be seeing? The idea is that if a loop passes around $0$ in the original picture, you instead think of it as passing around $\infty$, in which case it only loops around finitely many points (including $\infty$), so you don't see the diagonal terms you were talking about.

Cheerful Parsnip
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    Why are you being a grumpy parsnip Jim? – Dan Rust May 02 '13 at 12:58
  • @DanielRust: I noticed a trend of not using real names, so I hopped on that bandwagon. :) – Cheerful Parsnip May 02 '13 at 13:15
  • Nice answer! thanx – rohit May 02 '13 at 13:27
  • just for curiosity if the origin was included in X, then what wouuld have been the answer? – rohit May 02 '13 at 13:41
  • @rohit: If the origin is included, then you get a deformation retract onto the Hawaiian earring space. The first homology of the Hawaiian earring is surprisingly complicated. See http://jlms.oxfordjournals.org/content/62/1/305. It looks like it contains a countable direct sum of $\mathbb Z$s plus other stuff, including copies of the rationals. – Cheerful Parsnip May 02 '13 at 14:24