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