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Which of the following topological spaces are homeomorphic? Which have the same fundamental group?

a) The interval $(0,1)$ and $\mathbb{E}^1$

b) The torus $\mathbb{R}^2/\mathbb{Z}^2$ and the sphere $S^2$

c) $S^2$ \ $\{0,0,1\}$ and $\mathbb{E}^2$

d) $S^1$ and $\mathbb{E}^2$ \ $\{0\}$

e) The Mobius band and the cylinder


I am particularly interested in the part in pink, but you are welcome to criticize and correct even the other parts in case there are something wrong there.


$a)$ Homeomorphic. A homeomorphism from $(0,1)$ to $\mathbb{E}^1$ is e.g. $f(x) = \tan \pi (x-1/2)$.

Thus they also have isomorphic fundamental groups: The trivial fundamental group $0$.


$b)$ Not homeomorphic. The torus has the fundamental group $\mathbb{Z}^2$ while the sphere has the trivial fundamental group.


$c)$ Homeomorphic via stereographic projection.

I would say that $S^2$ \ $\{0,0,1\}$ has the fundamental group $\mathbb{Z}$ for a similar reason that $\mathbb{E}^2$ \ $\{0\}$ has the fundamental group $\mathbb{Z}$

But this cant be true because $\mathbb{E}^2$ has the trivial fundamental group and they are homeomorphic. What is the flaw in my thinking here?


$d)$ The spaces are of the same homotopy-type and thus have isomorphic fundamental groups which is $\mathbb{Z}$.

They are though not homeomorphic; e.g. the circle is compact unlike $\mathbb{E}^2$ \ $\{0\}$.


$e)$ The spaces are of the same homotopy-type and thus have isomorphic fundamental groups which is $\mathbb{Z}$.

But they are not homeomorphic; e.g. the Mobius band has a connected boundary - unlike the cylinder which has the boundary $S^1 \times \{0, 1\}$


Thanks for any input.

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1 Answer 1

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All your proofs are correct.

About the fundamental group of $S^2\setminus \{(0,0,1)\}$: With the stereographic projection you can see that $S^2\setminus \{(0,0,1)\} \cong \Bbb{E}^2$ and $S^2\setminus \{(0,0,1), (0,0,-1)\} \cong \Bbb{E}^2\setminus \{0\}$.

So $\pi_1(S^2\setminus \{(0,0,1)\})=\star$ and $\pi_1(S^2\setminus \{(0,0,1), (0,0,-1)\})\cong \Bbb{Z}$. Maybe that's what was confusing you, you have to remove two points of $S^2$ to get something homeomorphic to $\Bbb E ^2\setminus \{0\}$.

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  • $\begingroup$ I am thinking like this: If a space has the trivial fundamental group then every loop from any base point in this space can be shrunk down to a point within that space. This is obviously true for $\mathbb{E}^2$. It is not true for the loops in $\mathbb{E}^2$ \ $\{0\}$ for those loops which loop around the point $\{0\}$ and hence $\mathbb{E}^2$ \ $\{0\}$ has the fundamental group $\mathbb{Z}$. $\endgroup$
    – JKnecht
    Feb 13, 2016 at 3:22
  • $\begingroup$ I am thinking the same for $S^2$ \ $\{0,0,1\}$. The loops that go around the point $\{0,0,1\}$ can not be shrunk down to a point because of the same reason a loop that goes around $\{0\}$ in $\mathbb{E}^2$ \ $\{0\}$ can not be shrunk down to a point. Therefore $S^2$ \ $\{0,0,1\}$ should have the same fundamental group as $\mathbb{E}^2$ \ $\{0\}$, i.e. $\mathbb{Z}$. That is the reason for my confusion here. $\endgroup$
    – JKnecht
    Feb 13, 2016 at 3:23
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    $\begingroup$ Oh I see. Well the flaw in your argument here is that loops in $S^2\setminus \{(0,0,1)\}$ can actually be shrunk to a point. Indeed, imagine $S^2$ as the one point compactification of the plane, i.e., as $\Bbb E ^2 \cup \{\infty\}$, where $\infty$ plays the role of $(0,0,1)$. Then, a loop in $S^2\setminus \{(0,0,1)\}$ is the same as a loop in $\Bbb E ^2 \cup \{\infty\}$ not going through $\infty$. These loops are exactly the loops in $\Bbb E ^2$, which can be shrunk to a point. $\endgroup$
    – Nitrogen
    Feb 13, 2016 at 3:30
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    $\begingroup$ More geometrically, if you have a loop on a ball not going through the north pole, then you can always shrink that loop to the south pole without "cutting through" the north pole. You just have to stretch every south-north meridian to the south. $\endgroup$
    – Nitrogen
    Feb 13, 2016 at 3:32
  • $\begingroup$ Yes, of course. Now its totally obvious to me. I keep making those really silly mistakes in my reasoning... :) Thanks! $\endgroup$
    – JKnecht
    Feb 13, 2016 at 3:36

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