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The following is a qualifying exam problem. I am having great difficulty and have no clue where to start. I believe it is a legitimately difficult problem. Any help is greatly appreciated. Thank you very much.

Let $n$ be a positive integer. Let $X$ be the space of all $n$-tuples $(x_1, ..., x_n)$ of points of $\mathbb{R}^4$ such that for $i \neq j$, $x_i \neq x_j$ , with the topology induced by the inclusion into $\mathbb{R}^{4n}$ . Let $\sim$ be the equivalence relation on $X$ given by $(x_1, ..., x_n) \sim (x_{\sigma(1)}, ..., x_{\sigma(n)})$ for every permutation of $1, ..., n$. Compute $\pi_1(X)$ and $\pi_{1}(X/\sim )$.

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I'm not really sure what tools you're allowed to use, but here goes. I'm going to use the notation $X_n$ instead of $X$.

Claim 1: $\pi_1(X_n)$ is trivial.

Proof: By induction on $n$. When $n=1$, $X_1 = \mathbb{R}^4$ so is simply connected. Now, assume $X_{n-1}$ is simply connected.

Let $\pi:X_n\rightarrow X_{n-1}$ be the projection onto the first $n-1$ coordinates. Then $\pi$ gives $X_n$ the structure of a fiber bundle with fiber $\mathbb{R}^4\setminus\{x_1,...,x_n\}$. From this we get a long exact sequence in homotopy groups $$\ldots\rightarrow \pi_1\left(\mathbb{R}^4\setminus\{x_1,...,x_n\}\right)\rightarrow \pi_1(X_n)\rightarrow \pi_1(X_{n-1})\rightarrow\pi_0\left(\mathbb{R}^4\setminus\{x_1,...,x_n\}\right)\ldots $$

Note that $\mathbb{R}^4\setminus\{x_1,...,x_n\}$ is both connected and simply connected and further, by assumption $\pi_1(X_{n-1}) = 0$. Then, exactness implies $\pi_1(X_n) = 0$ so $X_n$ is simply connected as well.

Claim 2: The space $X_n/\sim$ is nothing but the orbit space $X_n/S_n$ where $S_n$, the symmetric group, acts freely on $X_n$ by permuting coordinates. In particular, $X_n\rightarrow X_n/S_n$ is a covering and so $\pi_1(X/\sim) = \ldots$

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    $\begingroup$ You posted this in the middle of my writing it myself! I'll just drop a couple of keywords. The space $X_n$ is what is known as the complement of an arrangement of subspaces (the subspaces in question in this case being the $\binom{n}{2}$ subspaces you removed from $\mathbb R^{4n}$ to construct $X_n$) There is an amazingly rich and beautiful and useful theory about these objects, in particular about their topology. The arrangement of subspaces which gives $X_n$ is what is called a fiber type arrangement: this means more or less that the argument Jason used to compute its $\pi_1$ (...) $\endgroup$ – Mariano Suárez-Álvarez Apr 17 '13 at 5:19
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    $\begingroup$ (...) works; and there are many other examples of this situation. A great introduction to all this is the book by Peter Orlik and Hiroaki Terao on arrangements of hyperplanes (which deals only with, well, hyperplanes, not general subspaces, but oh well) and some lecture notes on the subject written by Richard Stanley (and available on his webpage) $\endgroup$ – Mariano Suárez-Álvarez Apr 17 '13 at 5:20
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    $\begingroup$ I guess I'll add another keyword to the list: The spaces $X_n$ are called the configuration spaces of $n$ points in $\mathbb{R}^4$. $\endgroup$ – Jason DeVito Apr 17 '13 at 12:12

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