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I have problems understanding these two spaces: 1) the complement to a plane in $\mathbb{R}^4$ 2) the complement to the circle $S^1$ in $S^3$. what are they homotopic to?

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The first is easy - pick the plane $P=(x,y,0,0)$. Then $\mathbb R^4-P$ is the same as $\mathbb R^2\times (\mathbb R^2-(0,0))$. Since $\mathbb R^2$ is trivial, homotopy-wise, we get $\mathbb R^2-(0,0)\cong S^1$ –  Thomas Andrews Jan 20 '12 at 21:29
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The second is almost as easy, if you consider $S^3$ to be the one-point compactification of $\mathbb R^3$, and then make your circle pass through the point of compactification, so you get $\mathbb R^3-C$ for some $C$ an image of $\mathbb R$ with conditions on the behavior at $\pm \infty$ –  Thomas Andrews Jan 20 '12 at 21:33
    
I assumed we were taking some "obvious" embedding of $S^1$ in $S^3$, but, as Jim points out, it is much more complicated if the embedding can be arbitrary. –  Thomas Andrews Jan 20 '12 at 21:38
    
The homology of $S^3\setminus S^1$ (independent of the imbedding of $S^1$ in $S^3$) is straightforward to compute using Alexander duality, for instance. We can compute that $\tilde{H}_i(S^3\setminus S^1)\cong H^{2-i}(S^1)$ which you should be able to compute fairly readily. (More precisely, we are using the Cech cohomology theory but this is isomorphic to the simplicial cohomology theory on the class of triangulable pairs.) Of course, this computation does not answer your question completely but does give you spaces to which $S^3\setminus S^1$ cannot be homotopy equivalent. –  Amitesh Datta Jan 21 '12 at 3:07

2 Answers 2

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Think of the plane in $\mathbb R^4$ as being spanned by the first two coordinate axes: $\{(x,y,0,0):x,y\in\mathbb R\}$. Then projection onto the second two coordinates is a homotopy equivalence $(x,y,z,t)\mapsto (z,t)$ onto $\mathbb R^2\setminus\{(0,0)\}$. Now $\mathbb R^2\setminus\{(0,0)\}$ is homotopy equivalent to a circle.

For $S^1$ in $S^3$, recall that $S^3$ is a union of two solid tori glued along their boundary in a certain way. Think of your circle $S^1$ as being the core of one of these tori. Then removing it will be homotopy equivalent to the other solid torus, which is homotopy equivalent to a circle $S^1$. If your circle sits in $S^3$ as a knot, then this won't work, and in fact the homotopy type of knot complements is a very rich set.

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To add to the other two answers, it is also worthwhile to see that the homotopy of $\mathbb{R}^4$ minus a plane to $S^1$ can be made to factor through $S^3$ minus the unknot. That is, $\mathbb{R}^4 \setminus \{(x,y,0,0)\}$ deformation retracts to $\{(x,y,z,w)\mid x^2+y^2+z^2+w^2=1 \} \setminus \{(x,y,0,0)\mid x^2+y^2 =1\}$ (i.e.$S^3 \setminus \text{planar unknot}$) by the homotopy $H(t,\vec{v}) = (1-t)\vec{v} + t \vec{v}/\|\vec{v}\|$

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