# Comparing different topological spaces regarding homeomorphisms and fundamental groups.

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.

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 $\Bbb E ^2\setminus \{0\}$.
• 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 $\mathbb{Z}$. – JKnecht Feb 13 '16 at 3:22
• 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. – JKnecht Feb 13 '16 at 3:23
• 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. – Nitrogen Feb 13 '16 at 3:30