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Is $S^3$ homeomorphic to $D^2\times S^1\bigsqcup_{S^1\times S^1} S^1\times D^2$ ?

Here $D^2$ denotes the closed 2-dimensional unit disk.

If it is, how to prove it?

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I'm assuming you're gluing along the boundary using the identity map. If so, then yes, your construction is homeomorphic to $S^{3}$. This is known as a Heegaard splitting, in case you want to look up a proof in a textbook. I'm sorry, but I don't know one off the top of my head. –  jmracek Sep 26 '12 at 2:09
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@jmracek, indeed we glue the boundary using the identity map. And thanks to mention ``Heegaard splitting", I just checked the wiki, link, perhaps it is the same as the standard genus one splitting of $S^3$. –  ougao Sep 26 '12 at 2:24
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2 Answers 2

up vote 3 down vote accepted

As long as you are gluing by a homeomorphism isotopic to the identity on $\mathbb{S}^1\times\mathbb{S}^1$, then the product space will be homeomorphic to $\mathbb{S}^3$.

Here are two proof sketches.

For the first proof, just formalize the standard proof-by-picture of the genus-one Heegaard splitting of $\mathbb{S}^3$. Put toroidal coordinates on $\mathbb{S}^3$. I prefer coordinates different from those on the Wikipedia page: if $x,y,z,w$ are the standard $\mathbb{R}^4$ coordinates, then I would define coordinates $(\chi,\theta,\phi)$ by $$x = \cos{\chi}\cos{\theta},$$ $$y = \cos{\chi}\sin{\theta},$$ $$z = \sin{\chi}\cos{\phi},$$ and $$w = \sin{\chi}\sin{\phi}.$$ For fixed $\chi\in(0,\frac{\pi}{2})$, this is a torus. At the endpoints of the interval, the torii collapse to circles $x^2 + y^2 = 1$ and $z^2 + w^2 = 1$.

Now observe that $\chi^{-1}[0,\frac{\pi}{4}]\cong\mathbb{D}^2\times\mathbb{S}^1\cong\chi^{-1}[\frac{\pi}{4},\frac{\pi}{2}]$ and the two spaces are identified along the boundary $\chi^{-1}(\frac{\pi}{4})\cong\mathbb{S}^1\times\mathbb{S}^1$. The identification map is the identity.

For the second proof, note that $\pi_1(\mathbb{D}^2\times\mathbb{S}^1) = \mathbb{Z}$ (the space retracts onto a core circle). We can take a generator for $\pi_1$ of each solid torus to be a longitude on the boundary. But the boundary identification takes each longitude to a meridian of the other, so the fundamental group of the resulting (closed) three-manifold is trivial.

Now apply the Poincare conjecture :)

(Note: As Baby Dragon points out, it could well be that the proof of the Poincare conjecture uses that this gluing is homeomorphic to $\mathbb{S}^3$. Still, I think it's too cute to leave out.)

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It is imaginable that the "proof" using Poincare conjecture is circular. That said, it is a good sanity check. If only Bedlam had this sort of knowledge. –  Baby Dragon Sep 26 '12 at 3:47
    
"Bedlam"? I'm sorry, I don't understand :/ –  Neal Sep 26 '12 at 4:10
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An argument that avoids coordinates goes like this. The $4$-ball $D^4$ is homeomorphic to $D^2 \times D^2$. Since $\partial D^4 = S^3$,

$$S^3 \simeq \partial (D^2 \times D^2) = S^1 \times D^2 \cup D^2 \times S^1$$

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Bah! Much prettier! –  Neal Sep 26 '12 at 4:33
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