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Let $A$ be two distinct points in $R^3$. How would I go about showing that $R^3\backslash A$ is homotopy equivalent to the one-point union $S^2\vee S^2$?

Any help appreciated

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The two spaces are not homeomorphic: one is compact and the other isn't. – Mariano Suárez-Alvarez Nov 20 '11 at 16:50
You seem to be confused about the difference between "homotopy equivalent" and "homeomorphic". None of the pairs of spaces you have mentioned are homeomorphic, and $\mathbb{R}^3$ with two points removed is certainly not homeomorphic to $S^2\vee S^2$. – Jim Belk Nov 20 '11 at 16:55
You seem to be unsure about what "homeomorphic" and "homotopy equivalent" mean. I think you really meant to write "homotopy equivalent" everywhere. – Rudy the Reindeer Nov 20 '11 at 16:56
yeah, sorry I do mean homotopy equivalent. i'm being an idiot. – al1023 Nov 20 '11 at 17:34
@al1023: If you can see the geometric reason why $\mathbb{R}^3 \setminus \{ x \}$ is homotopy-equivalent to $S^2$, then you should be able to see why $\mathbb{R}^3 \setminus \{ x, y \}$ is homotopy-equivalent to $S^2 \vee S^2$. It's more or less the same, though the formula is more tricky to write down. – Zhen Lin Nov 20 '11 at 17:39

You probably want an explicit formula. Suppose the points are $(-1,0,0)$ and $(1,0,0)$; take the two spheres with center at these points, with radius $1$. First consider a deformation retraction of the interiors of the balls onto the spheres: just move along the radii (by multiplying the distance from the sphere by $1-t$, $t\in[0,1]$). Then you can deformation retract the exterior onto the two spheres and the axis $x$: move along lines orthogonal to the $x$ axis (again by multiplying by $1-t$ the distance along the line). Finally, deformation retract the two pieces of the $x$ axis. From this you can get a formula for a deformation retraction.

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