The main question for me is what is meant by an embedding of the cylinder $S^1\times[0,1]$ into $\mathbb{R}^2$. I know that an embedding is a continuous function $f:X\to Y$ between two topological spaces that is injective from $X$ to its image $f(X)$ and is a homeomorphism. What I tried is writing down cylinder coordinates, but that's a function to $\mathbb{R}^3$ which should not be (?). Also, the complement seems to be locally compact. I would say that's true because $\mathbb{R}^2$ is a Hausdorff space and the complement also lives in the Hausdorff space which is always locally compact (?). And I have no clue how to find the one-point compactification of the complement. I'm really stuck with this embeddings and compactnesses in topology. All help is appreciated!

  • $\begingroup$ Try showing that a cylinder and an annulus are homeomorphic. It should be a little easier to proceed after that. $\endgroup$
    – Rocket Man
    Apr 27, 2018 at 11:55

1 Answer 1


Hint 1: If $S^1\subseteq\mathbb{R}^2$ is the standard sphere then try this:

$$f:S^1\times [0,1]\to \mathbb{R^2}$$ $$f(v, t)=(t+1/2)v$$

It is easy to visualize it: you just take the circle and make it thicker. The translation by $1/2$ (or any other number greater than $0$) is necessary because otherwise $f$ wouldn't be injective.

Hint 2: As for one-point compactification: as you can see the complement of the image is a disjoint union of an open ball and a complement of a closed ball. As an open subset of $\mathbb{R}^2$ it is locally compact, indeed.

The one-point compactification of an open ball in $\mathbb{R}^2$ is the two-dimensional sphere $S^2$.

The one-point compactification of the complement of a closed ball is $S^2/\{x,-x\}$ (so it is like a torus with "glued" inner hole).

The one-point compactification of a disjoint union is the wedge sum of one-point compactifications of each component.

All in all the result is

$$S^2\vee \big(S^2/\{x,-x\}\big)$$

  • $\begingroup$ Is ${x,-x}$ a set of explicit points? Thanks for pointing out the torus-thing, I can't believe I did not come up with that. $\endgroup$
    – Algebear
    Apr 27, 2018 at 12:57
  • $\begingroup$ @GuusPalmer $x$ is a fixed point on the sphere. The choice doesn't matter. So you glue a point and its antipode. Actually you can glue any two points, but it's perhaps easier to visualize antipodes. $\endgroup$
    – freakish
    Apr 27, 2018 at 12:58

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