[Math] Union of 2-sphere with line segment in $\mathbb{R}^3$ removing one point homotopy equivalence.

algebraic-topologygeneral-topologyhomotopy-theory

I am working on a problem from Lee's Introduction to Topological Manifolds where one is asked to compute fundamental groups using Van Kampen's theorem. I know how to use Van-Kampen's theorem but I cannot find a homotopy equivalence between one of my subsets and a space whose fundamental group I know.

Consider the union of the unit 2-sphere in $\mathbb{R}^3$ with the line segment $[0]\times[0]\times[-1,1]\in \mathbb{R}^3$. If I remove a point, say at $(1,0,0)$ then I have the unit 2-sphere with a line down the middle and one point on the side removed. I am told that this is homotopy equivalent to $\mathbb{S}^1$. Can someone explain to me what deformations must be made to make this space $\mathbb{S}^1$?

Best Answer

First make the hole larger, so you have a sphere with a big hole in the side and a line down the middle. Then stretch things so that you have a 2-dimensional disk with a curved line segment sticking out and connecting two points on the disk. (So you get a disk with a 'handle'.) So far this a homeomorphism.

Now draw a line segment between the two points on the disk the handle connects, and deformation retract the disk to that line segment. This leaves you with something homeomorphic to $S^1$. Since a deformation retraction is a homotopy equivalence, you are done.

The above is how I found the equivalence, but you can shorten it considerably if you are writing it up. Take the original space and orient it so that the line in the middle connects the north and south poles. Draw a longitudinal line connecting the poles, and note the union of this line and the line in the middle is an $S^1$, and that the space retracts to this $S^1$.