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My overarching question is why in the Artin presentation of the (geometrically defined) braid group there are no more than two relations. I've learned one way to prove this is using the fact that the group $Br_n$ is the $\pi_1$ of configuration space $Conf_n=(\mathbb{C}^n\backslash \{z_i=z_j, i\neq j\})/S_n$ together with the fact that when computing the $\pi_1$ of a CW complex you can restrict your attention to its 2-skeleton.

Now I'm wondering if there is an explicit description of a cell decomposition for the configuration space (in particular of the 1- and 2-cells in it).

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    $\begingroup$ CW structures on configuration spaces are often not very practical to study their homotopical properties, as there will be a lot of extraneous cells accounting for the essential non-compactness of the configuration space. In general, there are three common approaches in this direction. (1) Find cell complexes homotopy equivalent to the configuration spaces: see the book "Geometry and topology of configuration spaces" by Fadell and Husseini for descriptions of such CW-complexes. (to be continued...) $\endgroup$ Oct 10, 2014 at 14:41
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    $\begingroup$ (continuation) Another important construction here is that of Salvetti complexes for complements of subspace arrangements. (2) Find cell structures on compactifications of configuration spaces: this is done for the one-point compactification of the space of configurations in the plane by Fox and Neuwirth in their article "Braid groups" (eudml.org/doc/165794). They use this to compute a presentation of the braid groups. (3) Inspired by Fox and Neuwirth, it is common nowadays to analyse "open-cell decompositions" or even more general stratifications of configuration spaces. $\endgroup$ Oct 10, 2014 at 14:41
  • $\begingroup$ Thank you very much Ricardo. You can write this as an answer as well. $\endgroup$ Oct 10, 2014 at 17:13
  • $\begingroup$ If you don't mind replacing the configuration space with another space which has the same homotopy type, there is also the "configuration complex" due to Jeff Smith. $\endgroup$
    – user43326
    Oct 11, 2014 at 5:50

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