One point compactification for open annulus Describe the one-point compactification of the open annulus $\{(x,y):1<x^2+y^2<2\}$.
I got the answer from the class, but do not fully understand the answer.
The answer is: the one-point compactification for it is the surface of a ball and identify the South Pole with the North Pole.
The illustration is below:
the open annulus $\{(x,y):1<x^2+y^2<2\}$ is homeomorphic to the surface of ball but without the South Pole and the North Pole. The neighborhood of $\infty$ in the one point compactification space must include both of the poles. Thus we have to identify them as a point. Hence we glue the the South Pole and the North Pole together and get the one-point compactification.
Questions:


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*Why is the open annulus $\{(x,y):1<x^2+y^2<2\}$ homeomorphic to the surface of ball but without the South Pole and the North Pole?

*Why the one-point compactification is just glue something together because they always are included in the same neighborhood?

The answer to this problem is that we can think the homeomorphism as a deformation with continuity and without gluing.Is it correct to think about homeomorphisms as deformations? Then we can know the open annulus $\{(x,y):1<x^2+y^2<2\}$ homeomorphic to the surface of the ball but without the South Pole and the North Pole.
One point compactification is often viewed as a process gluing things together. You would get some sense about this by taking a look at the notes here and read through the examples given there.(p51 and section 3.6.4) They have intuitive graphs there.
http://www.math.colostate.edu/~renzo/teaching/Topology10/Notes.pdf
