$\bf{S^3}$ as 2 point compactification of $\bf{S^2} \times \bf{R}$

Question

I'm reading a paper where the authors are regarding the three sphere as a 2 point compactification of $\bf{S^2} \times \bf{R}$ and I've haven't seen that construction of $\bf{S^3}$ before and I'm not sure it makes total sense to me. I know that we can view $\bf{S^3}$ as the one point compactification of $\bf{R^3}$, a union of 2 solid tori, the boundary of a 4-ball or a union of two 3-balls. I was thinking that since $\bf{S^3} = \partial(D^4)$, then we get the following: $\bf{S^3} = \partial(D^3 \times D^1) = (\partial D^3 \times D^1) \bigcup (D^3 \times \partial D^1) = (\bf{S^2} \times D^1) \bigcup (D^3 \times S^0) $, but I'm not sure if this line of thought takes me anywhere. It would be appreciated if someone could help me view $\bf{S^3}$ as this particular compactification.

Answer 1

To see this use the fact that S$^2$×R is homeomorphic to S$^2$×(-1,1).
Add the boundaries to the cylinder.
Now contract S$^2$×{1} to a point n and
contract S$^{-1}$×{-1} to another point s.
The compactifying points n and s become the north and south poles of a sphere.
The cylinder is like a round Mercator projection of the world,