$S^3$ $\cong$ $D^2\times S^1\bigsqcup_{S^1\times S^1} S^1\times D^2$? 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?
 A: 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.)
A: An argument that avoids coordinates goes like this.  The $4$-ball $D^4$ is homeomorphic to $D^2 \times D^2$.  Since $\partial D^4 = S^3$, 
$$S^3 \simeq \partial (D^2 \times D^2) = S^1 \times D^2 \cup D^2 \times S^1$$
