Good evening, I need to compute $\pi_1(\mathbb{R}^2 \setminus (\mathbb{Z} \times \{0\}))$. I see that the set $\mathbb{R}^2 \setminus \mathbb{Z} \times \{0\}$ has a countable set of points less than whole $\mathbb{R}^2$, so at each point we can retract the space into circles, and finally the space can retract to a wedge of infinite countable circles. Aplying Seifert-van Kampen theorem we can conclude that the fundamental group is free product of $\mathbb{Z}$ with itself countable times. Is this idea correct? Thanks a lot.
Fundamental group of $\mathbb{R}^2 \setminus (\mathbb{Z} \times \{0\})$
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algebraic-topology
2 Answers
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Here is a solution using the notion of the fundamental groupoid on a set $C$ of base points, which I introduced in 1967, and is discussed at this mathoverflow question. This method is developed in the book now available as Topology and Groupoids (T&G) (first edition published as "Elements of Modern Topology", 1968).
Let $X$ be the space in the question. We can write $X$ as the union of two open sets $U,V$ where $U$ is the top half of $X$ plus a little bit and $V$ is the bottom half plus a little bit, so that $U,V$ are contractible and $W= U \cap V$ is a countable union of contractible pieces. Let $C= \{(n+1/2,0): n \in \mathbb Z\}$. By 6.7.2 of T&G, the following square is a pushout of groupoids.
$$ \begin{matrix} \pi_1(W,C) & \to & \pi_1(V,C)\\ \downarrow&& \downarrow\\ \pi_1(U,C) & \to & \pi_1(X,C). \end{matrix}$$ It is worth noting that if you use the proof by verifying the universal property, then you do not need to know in advance that pushouts of groupoids exist, nor how to compute them.
We now use the general groupoid methods developed in Philip Higgins' Categories and Groupoids and in T&G Chapter 9 to say that as $\pi_1(W,C)$ is a discrete groupoid, and $\pi_1(U,C),\pi_1(V,C)$ are connected groupoids with trivial vertex groups, then $\pi_1(X,C)$ is a connected, free groupoid with vertex groups as required.
An advantage of this method is it applies to other analogous situations. For example, let $Y$ be the (non Hausdorff) space obtained from $\mathbb R \times \{1,-1\}$ by identifying all points $(x,1), (x,-1)$ for $ x \in \mathbb R$ except for $x \in \mathbb Z$. Then we get the same kind of conclusion. Here is a picture of part of this identification: the red and blue lines are identified, apart from certain points, to give the black line:
The use of groupoids adds a spatial component to group theory, which has proved powerful. It allows for example the construction of a new groupoid say $U_f(G)$ for each groupoid $G$ and each function $f: Ob(G) \to Y$ to a set $Y$ such that the following diagram, in which $D(Y)$ is the discrete groupoid on the set $Y$, is a pushout of groupoids:
$$\begin{matrix} D(Ob(G)) & \xrightarrow{f} & D(Y) \\ \downarrow& & \downarrow \\ G & \to & U_f(G). \end{matrix}$$ This construction includes that of free groups, free groupoids, and free products of groups.
One also needs to develop some general theorems, for example that if $G$ is a free groupoid, so also is $U_f(G)$, and that the vertex groups of a free groupoid are free groups.
Further, colimits of groupoids are constructed by using colimits of sets. This process is described in general in Appendix B of the book partially titled Nonabelian Algebraic Topology.
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Yes, this is the right idea. For more thoughts on this, Hatcher's algebraic topology text, in particular, chapter 1.2, has some similar questions. For example, can you show that the complement of a finite set of points in $\mathbb{R}^n$ is simply connected for $n \geq 3$?