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I asked this question previously on stackexchange ( but could not get a solution. Any help is appreciated.

The question is (Topology II: homotopy and homology: classical manifolds):

Show that the quotient space $X = S^2 \times S^2 / [(x_1,x_2) \sim (Rx_1,Rx_2)]$ where R is the reflection in the equatorial plane, is homeomorphic to $S^4$.

I am still in the process of learning topology and I really don't think I can prove this result. I apologize in advance if this is trivial, but it will be of great help if someone could provide a homeomorphism. I would really like to use the mapping of $X$ to $S^4$ in my work.

I did manage to get a mapping from $S^1 \times S^2 / [(x_1, x_2) \sim (Rx_1,Rx_2)]$ to $S^3$, but I am having trouble extending the result to $S^2 \times S^2$

Thank you.

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This sounds like a hw problem. – John Klein Feb 28 '11 at 18:02
As I mentioned in my post, I apologize if it is a simple hw problem. I am materials science student and I came across this problem in my research. As it happens it is an exercise problem in the book I mentioned. I tried my best (with the little knowledge I have), but could not get a mapping. – Srikanth Feb 28 '11 at 18:11
Think of it this way : $S^2$ is the collection of two $2$-disks glued along the equator. In $S^2\times S^2$, we therefore have four copies of $D_2\times D_2$. The relations you have tells you that you basically end up with two copies of $D_2\times D_2$ which have $S^3$ as a boundary and the relations (the ones left over) tell you to glue these two boundaries, thereby giving $S^4$. – Somnath Basu Feb 28 '11 at 19:09
@Somnath. Thanks. I think I understand it. I actually did the similar thing when showing $S^1×S^2/E$ to be equivalent to $S^3$. I didn't realize it though. Thanks for your help. I will try to figure out the exact mapping now. – Srikanth 0 secs ago – Srikanth Feb 28 '11 at 19:27

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