# Showing two topological spaces are homotopy equivalent

Could someone explain, like just a geometric description, how the space $\mathbb{R}^3\backslash A$, where $A$ is the unit circle in the $xy$-plane, is homotopy equivalent to $S^2\vee S^1$, the one point union of a 2-sphere and a circle. I can't see it.

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## 1 Answer

Sure. Firstly, let's not consider $\mathbb{R}^3$, because it's hard to visualize. Convince yourself that you can consider a solid sphere or radius say... 10, instead ($D^3$). Remove A. Now just as a punctured disk is homotopy eq. to $S^1$, the solid sphere with out that inner circle is homotopy eq. to the solid sphere without an inner torus.

Draw a picture.

Keep pushing out the torus. Eventually, it will meet itself or the outer wall of the sphere. Then all that's left is a spherical shell and a single line between the poles, i.e. a sphere with 2 points identified. Contract a meridian to get the two points together.

Then you have a wedge of a 2-sphere and a circle.

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thank you! I get it now :) –  908979 Dec 7 '11 at 18:41