Does $S^2\times[-1,1]$ decompose as $B^3\#B^3$

Question

If we let $N=S^2\times[-1,1]$ and let $S=S^2\times\lbrace 0\rbrace$, then cutting along $S$ we obtain what appears to me like two more copies of $N$: $S^2\times[-1,-\epsilon]$ and $S^2\times[\epsilon,1]$. Gluing $B^3$ to the boundary of these components corresponding to $S$ in $N$, we (I think) obtain two more balls. Am I doing something wrong here? This doesn't seem correct.

Answer 1

A trick is: you decompose $S^2$ as $S^2=D_1\cup D_2$ into 2-disks such that the borders of them are identinfied in “equator” $\partial=D_1\cap D_2$. Then do the cartesian product $$S^2\times I=(D_1\times I)\cup_{\partial\times I}(D_2\times I).$$ So, at the level $S^2\times\{0\}$ isotoping into one of the 3-balls $$B_1=D_1\times I\quad,\quad B_2=D_2\times I,$$ in the interior of each of them, is giving you a 3-ball whose border is the isotoped sphere and serves from the connected-sum $B_1\# B_2$ claimed.