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If you have for example an exam question like: Determine the one point compactification of the punctured plane?

Is it enough to say: The punctured plane is homeomorphic with the sphere without the north- and south pole (stereographic projection). Now we have a sphere with two holes in it, so if we want to make this compact, we have to identify those two points with one point, so we bring those two points together in that point. If you take for example the center of the sphere, you get an torus with inner radius zero. If the point is outside the sphere, you get some torus where you identify the points on one circle of the torus.

But you all consider this in the subspace topology from $\mathbb{R}^3$, you have also to prove then, that the one point compactfication (where the opens are $Y \backslash C$ and opens of $X$) is in fact the same? So that you have to prove $Y = X \cup {\infty}$ with that topology is the same as the sphere with two points identified. Because I have now idea how to do that? Can anyone explain this?

($Y$ is denoted as the one point compactfication of $X$)

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    Do you mean the subspace topology from $ℝ^3$? – 2017-01-30
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    yes indeed.. :) – 2017-01-30

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One thing you can do is to consider a sphere and making the compactification by identifying its poles (and doing this abstractly as a quotient space rather than embedding it to $ℝ^3$). You know that the sphere without poles is the punctured plane, so the identified sphere is a one-point compactification.

But still, to know that this is the space you are imagining in $ℝ^3$ you would need a proof. So maybe it would be easier to actually write down the map from the (punctured) plane to the torus using polar coordinates of plane and translating them to the standard parametriztion of the torus.