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Let $ X = \mathbb R^3 \setminus A$, where $A$ is a circle. I'd like to calculate $\pi_1(X)$, using van Kampen. I don't know how to approach this at all - I can't see an open/NDR pair $C,D$ such that $X = C \cup D$ and $C \cap D$ is path connected on which to use van Kampen.

Any help would be appreciated. Thanks

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Try removing an open cylinder passing through the circle. – Qiaochu Yuan May 22 '12 at 13:23
@QiaochuYuan Thanks, but I'm still not seeing it... – Martin May 22 '12 at 13:44

I am not sure whether there is a nicer choice but this is how I think about it. Intuitively the fundamental group should be $\mathbb Z$ - a path may jump through the hoop a couple of times or not. I choose the open sets to model this somewhat. One open set is the interiour of a filled torus with the circle lying on the surface. The other set is the whole of $\mathbb R^3$ with the closed disk (bounded by the circle) removed. Then the first set contracts to a circle, the second set contracts to a sphere and the intersection is contractible.

Edit: To make the sets more precise: $$U=\mathbb R^3-D^2\simeq S^2$$ such that $$A=\partial D^2\subseteq D^2$$ and $$V=int(S^1\times D^2)\simeq S^1$$ such that $$A=\ast\times \partial D^2\subseteq S^1\times D^2.$$ Then $$U\cap V=int(S^1\times D^2-\ast\times D^2)\cong int(I\times D^2)\simeq\ast$$

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Sorry, I'm not seeing how the intersection is contractible? Isn't the intersection an annulus? – Neal May 22 '12 at 14:12
@Neal The intersection is a filled open cylinder (a filled open torus which has a 'snip' through it), which is contractible – Daniel Freedman May 22 '12 at 14:20
And of course by "it is an annulus with a slice missing" I mean the intersection and not the second set. – Simon Markett May 22 '12 at 14:28
Oh, I see. The filled torus has the circle as a meridian, not a longitude. – Neal May 22 '12 at 14:53
Maybe clarify that in your answer? – Neal May 22 '12 at 14:57

You can add a point to get $S^3-A$, without changing the fundamental group (this follows from van Kampen's theorem). Now $S^3$ minus any point is homeomorphic to $\mathbb{R}^3$, so choose this any point to lie on $A$! This gives a new space, still with the same fundamental group, but now you've got $\mathbb{R}^3-B$, where $B$ is a line (say the x axis). Think about what would happen if you had a solid ball minus a line segment; that should give you what you need to deformation retract to a solid torus.

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You should definitely check out the Hatcher's Algebraic Topology book page 46.

It was very hard for me to imagine at first but $\mathbb{R}^3 - S^1$ deformation retracts onto $S^1 \wedge S^2$ so just choose $S^1$ and $S^2$ for C and D respectively, since the space is formed as wedge product of two spaces, the intersection is going to be a point only(by definition) whose fundamental group is trivial for sure. Similarly $\pi_1(S^2)$ is also trivial then $\pi_1(\mathbb{R}^3 - S^1)$ is isomorphic to the fundamental group of the circle which is $\mathbb{Z}$.

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