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I apologise in advance for the vagueness of this question but I have not been able to find very much info on the topic and have made very little progress on my own.

I am trying to understand why the knot group $\pi_1 (S^3 - K)$ of the trefoil is isomorphic to Artin's 3-strand braid group $B_3$. I know that the Wirtinger presentation for $\pi_1 (S^3 - K)$ gives Artin's presentation for $B_3$ directly but I was hoping someone could paint a more topological picture which takes homotopy classes directly to braids (or vice-versa) without using group presentations as the middle man. Thanks in advance.

Edit: Thanks for the replies guys. I should have stated that $S^3-K$ is diffeomorphic to the space $SL(2,\mathbb{R}) / SL(2,\mathbb{Z})$ (John Baez says so in his blog). It is possible that the disk with 3 holes is a deformation retract of $SL(2,\mathbb{R}) / SL(2,\mathbb{Z})$ but I don't know much about this space. I'll update if I find the answer myself.

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    $B_3$ is the fundamental group of the configuration space of unordered triplets of points in $\mathbb{C}$. This space might be homotopy equivalent to $S^3 \setminus K$...?2012-05-11
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    This may be useful to you: http://mathoverflow.net/questions/96423/fundamental-group-of-the-configuration-space-in-a-plane-closed2012-05-11
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    Speaking of John Baez, have you seen this?: http://math.ucr.edu/home/baez/week261.html It explains it very clearly!2012-05-11
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    To answer the point made by @QiaochuYuan , it's well known that knot complements are aspherical spaces, and so $S^3\setminus K$ is a $K(B_3,1)$. So is the configuration space of unordered triples in $\mathbb{C}$ - by considering the long exact sequence in homotopy of the Fadell-Neuwirth fibration on the ordered space of points in $\mathbb{C}$ with $m$ puncture points, and then taking a covering space projection on to the space of unordered points. Both of these spaces are homotopy equivalent to CW-complexes, and so are weakly homotopy equivalent by the classification of $K(G,n)$s.2013-05-27
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    Obviously, this uses the fact that $S^3\setminus K$ and the configuration space of unordered triples have isomorphic fundamental group, so doesn't answer the OP's question, but it does suggest looking for an explicit homotopy equivalence between the spaces.2013-05-27
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    http://math.stackexchange.com/questions/1774198/fundamental-group-of-mathbbr3-minus-trefoil-knot?rq=1 this more recent post seems to answer the question2016-12-14

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