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Let $D^3 =\{(x,y,z)\in \mathbb{R}^3 :x^2+y^2+z^2 ≤1\}$. Let $A= \{a_1,a_2,...,a_n\} \subset D^3$ be a subset of distinct points in the 3-ball. Compute the fundamental group of the quotient space $\pi_1(D^3/A, b)$ where $b \in D^3/A.$

Should I use Seifert Van Kampen or is there a more visual argument? Is this similar to the fundamental group of $\mathbb{R}^3/$n-points ?

bmmcutet12
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3 Answers3

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Hint: $\pi_1(D^3-A,b)\cong\pi_1(\mathbb{R}^3-A,b)$. Consider the map $F: [0,1] \times (\mathbb{R}^3-A) \to (D^3-A)$ via $$F(\lambda,x)=\begin{cases} \begin{array}{cc} (1-\lambda)x+\lambda\frac{x}{||x||} & \mbox{ for ${||x||\geq1}$}\\ x & ||x||<1 \end{array}\end{cases}$$

One can see that $F$ is a deformation retract and therefore our two pointed spaces are homotopy equivalent.

Now, to figure out $\pi_1(\mathbb{R}^3-A,b)$, you use induction and van Kampen's theorem .

Kyle Gannon
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One way of doing this computation directly is to see the situation in a broader light, namely as part of the theory of the fundamental groupoid $\pi_1(X,C)$ on a set $C$ of base points. Some key aspects of this idea were developed by P.J. Higgins in the 1960s and are well explained in his downloadable book Categories and Groupoids. The argument is also well developed in the responses to this mathoverflow question, and which raises the question of the strange restriction to methods which work only for path connected spaces.

I got into the use of Higgins' work in the mid 1960s in order to find a version of the Seifert-van Kampen Theorem which would compute the fundamental group of the circle, which is the basic example in topology; the more general version of that theorem for this aim requires two base points. This year is the 50th anniversary of my publication on this topic!

Using groupoids has advantages also in the theory of covering spaces, and for orbit spaces of non free actions of groups. An account for all this given in the book Topology and Groupoids, the third edition of a book published in 1968.

As an example application, if $X$ is a path connected and simply connected space, and $D$ is a discrete subspace of $X$ such that the inclusion $i: D \to X$ is a closed cofibration, then the fundamental group of $X/D$ at the image of $D$ is a free group. In fact there is a description of the more general case when the points of $D$ are just identified by some function $f:D \to E$ to a discrete space $E$.

Ronnie Brown
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If $n=1$ then $D^3-A$ is homotopy equivalent to the $2$-sphere $S^2$ and hence it is simply connected. If $n>1$ then we can assume that all points of $A$ except $a_1$ are on the boundary of $D^3$. Then we get that $D^3-A$ is homotopy equivalent to $S^2-\{a_2,\dots,a_n\}$ which is homotopy equivalent to $\mathbb{R}^2-\{a_3,\dots,a_n\}$ (assuming $n\ge 3$) and that has the same homotopy type of the wedge of $(n-2)$ circles. So the fundamental group of the space is free group on $(n-2)$ generators. If $n=2$ then $D^3-A$ is homotopy equivalent to $\mathbb{R}^2$ and hence is contractible.