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I'm looking for a proof that $X=S^1\times D^2/S_1\times S_1$ is homotopy equivalent to $S^2\vee S^3$.

I can't think of any rigorous proof and I'm not even sure if I can see why this is true. Can you help me? (If this helps, I've proved $S^n/S^k$ is homotopy equivalent to $S^n\vee S^{k+1}$).

Jules
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  • And how does identyfing a boundary of a torus look like? Should I try to imagine this? – Jules Aug 17 '16 at 14:54
  • I wouldn't, because $S^3$ sits inside of 4-dimensional space. We may as well ask to imagine $S^3$ as a quotient of the solid 3-ball by identifying the boundary 2-sphere. Which I can't. – Chris Gerig Aug 17 '16 at 15:25

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You are taking a solid torus and identifying its boundary to a point. Here is my trick:

This quotient is unchanged if we first glue in other objects "around" this solid torus and then identify them all to a point. In particular, I can glue in another solid torus to the boundary of the original solid torus (which gives a 3-sphere!) and we then identify that second solid torus to a point (which includes the boundary of the original torus). So X is equivalent to $S^3/(S^1\times D^2)$. Since the solid torus is just a thickened 1-sphere, we get the homotopy-equivalent $S^3/S^1$.

Chris Gerig
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