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Is the 2-sphere $S^2$ with $k$ points removed homeomorphic to a topological group for

a) k=1

b) k=2

c) k=3

d) k=0

For (a), the sphere with a point deleted is isomorphic to $R^2$ with no points deleted, correct? And since $R^2$ is a topological group, then the sphere with a point deleted is a topological group.

For (c), let $X=S^2$ with 3 points deleted. Then $X$ is isomorphic to $R^2$ with 2 points deleted. So, the fundamental group of $X$ is the free group on 2 generators, which is not abelian. We have that a topological group always has an abelian fundamental group. Hence, $X$ is not a topological group.

As for (b) and (d), I am a little stuck.

For (b), I have that the sphere with 2 points deleted is the plane with 1 point deleted. So it has fundamental group the integers which are abelian. However, the theorem I used early was $X$ is a topological group implies the fundamental group of $X$ is abelian. If this was an if and only if statement I would be done, but I don't believe it is. So, how do I prove the sphere with 2 points deleted is or is not a topological group?

Also, for (d) I had the same type of problem. The sphere with no points deleted is just the sphere. The fundamental group is 0. And that would be abelian correct? So again, I can't really use that theorem.

Please help.

Mark
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    This paper is of interest with regards to part (d): http://www.raczar.es/webracz/ImageServlet?mod=publicaciones&subMod=revistas&car=revista62&archivo=p075.pdf PDF warning. You can identify the two points removed with $\Bbb C^*$ (by sending one point to infinity and the other to zero) which is a topological group. – Cameron Williams May 01 '16 at 01:00
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    d) $S^2$ is not a topological group. See here. For b) $\mathbb{R}^2\setminus{0,0}$ is homeomorphic to $\mathbb{R}\times S^1$, so you just need to prove the product of topological groups is a topological group in the obvious way. – Moya May 01 '16 at 01:00
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    Is the case with two points deleted $\mathbb{C}^*$? – Michael Burr May 01 '16 at 01:01
  • Yes, it is... I made it too complicated in my answer - I just separated the argument from the modulus. –  May 01 '16 at 01:07

1 Answers1

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"Is" a topological group is not exactly right, since you don't have a group structure (yet). In some places you ask the question correctly - is the sphere with points deleted homeomorphic to (the topological space underlying) a topological group.

For two points deleted: that is isomorphic to the plane with the origin deleted, which in turn is isomorphic to the Cartesian product of the circle $S^1$ and the open half-line $(0,+\infty)$. $S^1$ has a topological group structure, for example as $\mathbb R / \mathbb Z$, where $\bar x$ maps to the point on the circle with angle $2x\pi$, and $(0,+\infty)$ is a topological group under multiplication.

The other case is covered in the comments.

Edit: Or, as noted in Michael Burr's comment to the original question, this is also the group $\mathbb C^*$ under multiplication (which is pretty much what I showed - think about argument and modulus of complex numbers, and how they behave under complex number multiplication).