TITLE: $S^3$ $\cong$ $D^2\times S^1\bigsqcup_{S^1\times S^1} S^1\times D^2$?
QUESTION [7 upvotes]: 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?

REPLY [10 votes]: An argument that avoids coordinates goes like this.  The $4$-ball $D^4$ is homeomorphic to $D^2 \times D^2$.  Since $\partial D^4 = S^3$, 
$$S^3 \simeq \partial (D^2 \times D^2) = S^1 \times D^2 \cup D^2 \times S^1$$