Fundamental Group Calculation in $\mathbb{R}^3$

Question

We let $X$ be the following subset of $\mathbb{R}^3$: $$ X=\left \{x \in \mathbb{R}^3 \vert 1 \leq \vert x \rvert \leq 2 \right \} \subseteq \mathbb{R}^3.$$ Now we note that $X$ has two boundary components, $S_1$ and $S_2$. Here $ S_1=\left \{x \in \mathbb{R}^3 \vert \vert x \rvert =1 \right \} \subseteq \mathbb{R}^3 $ and $ S_2=\left \{x \in \mathbb{R}^3 \vert \vert x \rvert =2 \right \} \subseteq \mathbb{R}^3 $. Moreover, we can generate an equivalence relation $\sim$ on $X$ by setting $x \sim y$ if $x \in S_1,y \in S_2$ and $y=2x$. Now consider the quotient topological space $X/ \sim$, for any point $x_0 \in X/ \sim$, we want to compute the fundamental group $\pi_1(X/\sim,x_0)$? I have almost no idea on this question. I guess maybe the wanted fundamental group is the same as the fundamental group of torus, so the answer is $\mathbb{Z} \times \mathbb{Z}$. But I do not know how to prove it.

Answer 1

Think of $X$ as a collection of spheres centered at the origin with radii ranging from $1$ to $2$. Topologically speaking, it is homeomorphic to $[0,1]\times S^2$. The quotient has the effect of identifying $0\times S^2$ with $1\times S^2$, so $X$ is homeomorphic to $S^1\times S^2$.

Then, $\pi_1(X)\cong \pi_1(S^1\times S^2)\cong \pi_1(S^1)\times \pi_1(S^2)\cong\mathbb{Z}\times 1\cong \mathbb{Z}$.

Answer 2

Notice that we have a copy of $S^2$ inside our quotient space (which I will call $Y$) with radius $\frac{3}{2}$. Furthermore, at each point on this sphere, we get a circle by traveling perpendicular to the point, so $Y\cong S^2\times S^1$. Then it follows that $$\pi_1(Y)=\mathbb{Z}$$

Answer 3

We can calculate the fundamental group using universal cover.

Instead of looking at your quotient space, we start with punctured $\Bbb R^3$, i.e. the space $E=\Bbb R^3\setminus\{(0,0,0)\}$. Define an equivalence relation on $E$ by $x\sim y$ if $x=2^ny$ for some integer $n\in\Bbb Z$. Hopefully you can see that the quotient space $E/\sim$ is homeomorphic to your quotient space. We also have that the canonical projection $\pi:E\to E/\sim$ is a covering map.

The covering automorphisms/deck transformations, i.e. the homeomorphisms $\varphi:E\to E$ having the property $\pi=\pi\circ\varphi$, are precisely those of the form $\varphi_n(x)=2^nx$ for an integer $n$. So the covering automorphism group (the group of all covering automorphisms with binary operation being function composition) is isomorphic to $\Bbb Z$. Since $E$ is simply-connected, the fundamental group of $E/\sim$ is isomorphic to the covering automorphism group and hence $\Bbb Z$.

(Note: To see that covering automorphisms $\varphi$ must be of that form, first observe that $\varphi_n$ described above ($\varphi_n(x)=2^nx$) are covering automorphisms. Then the condition $\pi=\pi\circ\varphi$ on covering automorphisms gives $\pi\circ\varphi(x)=\pi(x)$, i.e. we have $x\sim\varphi(x)$, and hence $\varphi(x)=2^nx$ for some integer $n$. This integer $n$ may depend on the point $x$, but it happens that it does not because of the theorem that two covering automorphisms being equal at some point must be equal everywhere, so that we compare the two maps $\varphi$, $\varphi_n$ at $x$ to see $\varphi=\varphi_n$ everywhere, i.e. $\varphi_n$ are all of the covering automorphisms.)