Fundamental group of torus knot without thickening

The calculation of the fundamental group of a $$(m, n)$$ torus knot $$K$$ is usually done using Seifert-Van Kampen theorem, splitting $$\mathbb{R}^3\backslash K$$ into a open solid torus (with fundamental group $$\mathbb{Z}$$) and its complementary (with fundamental group $$\mathbb{Z}$$). To use Seifert-Van Kampen properly, usually the knot is thickened so that the two open sets overlap in a path-connected open set with fundamental group $$\mathbb{Z}$$. And the resulting fundamental group is $$\langle x, y \ | \ x^m = y^n\rangle$$.

I find the thickening part a bit artificial. My question is: can we save this proof avoiding the thickening? For example, does it exist a version of Seifert-Van Kampen for which we can use the closed solid torus (minus the knot) and the closure of its complementary?

To be a knot in the first place, it essentially needs a well-defined regular neighborhood, sort of as proof of its tameness. If you think of a knot as being in the one-skeleton of a triangulation of $$\mathbb{R}^3$$, then if you want a triangulation of the complement without too many more simplices, you can remove a regular neighborhood of the knot -- after all, this is a deformation retract of $$\mathbb{R}^3-K$$. It is also possible to triangulate $$\mathbb{R}^3-K$$, but it takes infinitely many more simplices!
There is a cell structure of the torus minus $$K$$ with infinitely many cells, and one can extend this to a cell structure of $$\mathbb{R}^3-K$$. If $$X_1,X_2$$ are closed sets with $$X_1\cup X_2=\mathbb{R}^3$$ and $$X_1\cap X_2$$ being the torus, then $$X_1-K$$ and $$X_2-K$$ inherit the cell structure, and the van Kampen theorem for complexes applies.
(Note: if you took the solid torus minus $$K$$ and the closure of the complement of this, as you suggest, their union would be all of $$\mathbb{R}^3$$! This is like the example of covering $$S^1$$ by $$[0,1/2]$$ and $$(1/2,1]$$ through the quotient map $$[0,1]\to S^1$$. These are two simply connected subsets of $$S^1$$ that intersect at a point, but van Kampen, if it were to apply, would give $$\pi_1(S^1)=1$$. This illustrates what open sets are meant to handle, but also $$(1/2,1]$$ is not a subcomplex of $$S^1$$.)
• I mean the closure in $\mathbb{R}^3\backslash K$. Could this work? Do you have a reference for the version of van Kampen that you mentioned? – Marco All-in Nervo Feb 26 at 17:33
• @MarcoAll-inNervo Yes, that closure gives $X_1$ and $X_2$. As for the version of van Kampen I cited, I think this might be Seifert's version (which I haven't read myself), but Hatcher Prop A.5 shows that subcomplexes have deformation retract neighborhoods. It's also possible to adapt the idea of the proof of Prop 1.26 by replacing the cell structure of $X_1\cap X_2$ with $(X_1\cap X_2)\times [0,1]$, which replaces $\mathbb{R}^3-K$ with a homotopy equivalent space, but $(X_1\cap X_2)\times (0,1)$ can be the shared neighborhood. Notice: these sort of hide a thickening behind a theorem! – Kyle Miller Feb 26 at 20:39
• Another approach (though possibly similar to Seifert's) is to use the fact that every map is homotopic to a simplicial map, if you subdivide the domain enough. That is, each element of $\pi_1$ can be thought of as a path along the $1$-skeleton. This sequence of edges can be used to decompose such an element as an element of the amalgamated product, basically by following the usual proof of the van Kampen theorem. The open sets, after all, are used to cover $I$ with the preimages through a map from $I$, and compactness implies the path can be decomposed into finitely many pieces. – Kyle Miller Feb 26 at 20:43