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Well, I'm trying to prove that you can express the 3-dimensional sphere $S^{3}$ as the union of two solid tori. I tried first use that a solid tori is homeomorphic to $S^{1}$$\times$$D^{2}$ and use this to obtain some quotient space which would be homeomorphic to $S^{3}$, but I couldn't go any further. Is this the way or I need a little more to prove that? Thanks in advance.

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    i think the tag would be general topology2012-10-15
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    I am not good in topology but I guess you would get an answer more easily if you make a more specific title, maybe 'represent S^3 as union of solid tori'?2012-10-15
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    I'd like to see something more rigorous but the following is a standard way to visualize this. Take a chunk out of one torus, so that you have two tubes. Use one tube to fill in the hole in the other torus. Now you have two solid 3-balls. Rejoining the broken off piece and then identifying the boundaries of the tori amount to identifying the two solid 3-balls along their boundary. The result of this is always $S^3$.2012-10-15
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    This might help: http://math.stackexchange.com/questions/202534/s3-cong-d2-times-s1-bigsqcup-s1-times-s1-s1-times-d2/2012-10-15
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    (I'd add that you have to be careful about *how* you glue the two solid tori together. The resulting space has a trivial fundamental group, while the solid tori both have fundamental group isomorphic to $\mathbb{Z}$. If you're careless about the gluing, you'll end up with a lens space.)2012-10-15
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    We discussed this http://math.stackexchange.com/questions/199891/the-join-of-two-copies-of-s1 and http://math.stackexchange.com/questions/195848/homeomorphism-question-relating-to-the-topological-3-sphere2012-10-15
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    @MarianoSuárez-Alvarez, evidently so, but I like to think that I add my own special zest.2012-10-15
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    This might be helpful 2013-04-21

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