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Let $X=\mathbb{R}\times (-1,1)-\mathbb{Z}\times \{0\}$. I want to compute the fundamental group of that space.

I would be tempted to do some division like the following to apply Van Kampen, but to do it all the sets must intersect in the basepoint of $\pi(X,x)$, so I can't really do it. enter image description here If I incorrectly use Van Kampen, I will get to the correct result that it's free in a countable number of generators, though.

Is there any other way to do that, I mean, with the basic theory of Hatcher's of Massey's book (It's an exercise from Massey's book)?

Nell
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  • Not exactly. The problem is the use of the ideas from Topology and Groupoids, which I don't really know about, and this is just an exercise from Massey's book. – Nell Sep 05 '18 at 22:43
  • See also https://math.stackexchange.com/questions/978851/is-the-plane-minus-the-integer-lattice-homeomorphic-to-the-plane-minus-the-integ – mr_e_man Sep 05 '18 at 22:53
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    Why not just add $\mathbb R\times(\frac12,1)$ to each of your $B_i$s and declare the basepoint to be $(0,\frac34)$? – hmakholm left over Monica Sep 05 '18 at 22:59
  • It is curious to me that standard algebraic topology with its choice of a single base point is willing to take elaborate steps to avoid dealing with connected spaces. I refer you to comments of Grothendieck given in Engllsh at https://mathoverflow.net/questions/220561/descent-theorems-for-fundamental-groups-and-groupoids . Why give students a version of the van Kampen theorem which does not compute the fundamental group of the circle? – Ronnie Brown Sep 06 '18 at 21:07
  • My previous comment should replace "connected" by "nonconnected"! – Ronnie Brown Sep 07 '18 at 10:05

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I suggest to proceed as follows. Define $Y = \bigcup_{n \in \mathbb{Z}} S(n)$, where $S(n) = \{ x \in \mathbb{R}^2 \mid \lVert x - n \rVert = \frac{1}{2}) \}$. This is an infinite chain of circles. It is easy to see that $Y$ is a strong deformation retract of $X$. Now $Y$ has the structure of a CW-complex, $0$-cells being the sets $\{n - \frac{1}{2} \}$ and $1$-cells being the upper and lowers half circles of the $S(n)$. Then $Y_0 = Y \cap \mathbb{R} \times (1,0]$ is subcomplex of $Y$ which is homeomorphic to $\mathbb{R}$. In particular, $Y_0$ is contractible. It is known that then $p : Y \to Y/Y_0$ is a homotopy equivalence. But $Y/Y_0$ is nothing else than a countably infinite wedge of circles and your arguments apply.

Paul Frost
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  • Yes. I was just thinking about something similar, trying to collapse the maximal tree of that graph, which translates in what you did. – Nell Sep 05 '18 at 22:52
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Use Van Kampen inductively to prove that for all $n$ the set $X_n = B_{-n} \cup \cdots \cup B_n$ is free of rank $2n+1$, and that the inclusion $X_n \to X_{n+1}$ takes the rank $2n+1$ free group $\pi_1(X_n)$ to a free factor of the rank $2n+3$ free group $\pi_1(X_{n+1})$. Then use a direct limit argument to conclude that $\pi_1(\cup_{n=1}^\infty X_n)$ is free of infinite rank.

Lee Mosher
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I want to say a rose of countably infinitely many petals; but I'm not sure it's correct. At least it seems like it might deformation retract onto it...