Calculating the fundamental group of $\mathbb{R}^3$ without a line

Question

I’d like to show that the fundamental group of $\mathbb{R}^3 \setminus \Delta$, where $\Delta$ is the line parametered by $t \rightarrow (0,0,t)$, is $\mathbb{Z}$.

To do so, I was thinking of using the map $\phi(\theta,r,z)=(r \cos(\theta),r \sin(\theta), z)$, which gives an homeomorphism (I think) between the topological space $U=S^1 \times ]0,+\infty[ \times \mathbb{R}$ and $\phi(U)=\mathbb{R}^3$. I was trying to calculate $\phi^{-1}(\mathbb{R}^3 \setminus \Delta)=U \setminus \phi^{-1}(\Delta)$ in the hopes of finding something whose fundamental group is $\mathbb{Z}$, but I couldn’t manage to do it yet. Could this approach work…?

Answer 1

Define $H:\Bbb{R}^{3}\setminus\Delta\times[0,1]\to\Bbb{R}^{3}\setminus\Delta$ given by $H((x,y,z),t)=(x,y,(1-t)z)$ then it is a deformation retract from $\Bbb{R}^{3}\setminus\Delta$ to $\Bbb{R}^{2}\setminus\{0\}$.

And we all know that $\Bbb{R}^{2}\setminus\{0\}$ is homotopy equivalent to the circle using the map $G:(\Bbb{R}^{2}\setminus\{0\})\times[0,1]\to \Bbb{R}^{2}\setminus\{0\} $ such that $G(x,t)=\frac{x}{1-t+t||x||}$ . Then this is a deformation retract to the unit circle.

This proves that $\Pi_{1}(A)\cong\Pi_{1}(D^{2}\setminus\{0\})\cong\Pi_{1}(S^{1})\cong\Bbb{Z}$ .

To see more visually, you can identify $\Bbb{R}^{3}$ with the open unit ball and use the same maps to deform it to $D^{2}\setminus\{0\}$ which deformation retracts to $S^{1}$.

Answer 2

Here is a more general fact.

Theorem: Let $n>m$ be positive integers. Let $\mathbf V$ be an $m$-dimensional vector subspace of $\Bbb R^n$, and $\mathbf W$ be it's complimentary subspace, i.e., $\Bbb R^n=\mathbf V\oplus \mathbf W$. Then, $\Bbb R^n\backslash \mathbf V$ is homeomorphic to $ \mathbf V\times(\mathbf W\backslash \mathbf 0)$.

Proof. Recall that $\Bbb R^n=\mathbf V\oplus \mathbf W$ means that every point $x\in \Bbb R^n$ can be written as $x=v+w$ for some some $v\in \mathbf V, w\in \mathbf W$; further if $x=v'+w'$ for some other $v'\in \mathbf V, w'\in \mathbf W$ then $v=v',\ w'=w$. So, Consider the continuous map $$\Phi\colon \Bbb R^n\backslash \mathbf V\ni v\oplus w\longmapsto (v,w)\in \mathbf V\times(\mathbf W\backslash \mathbf 0).\ $$ with its continuous inverse $$\Phi^{-1}\colon \mathbf V\times(\mathbf W\backslash \mathbf 0)\ni(v,w) \longmapsto v\oplus w\in \Bbb R^n\backslash \mathbf V.\ $$ $\Phi$ is continuous as it is the restriction of a linear map (induced by projections) $\Bbb R^n\to \mathbf V\times \mathbf W$. Similarly, $\Phi^{-1}$ is also continuous as it is the restriction of a linear map (induced by vector-addition operation of $\Bbb R^n$). So, we are done. $\square$

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So, $\Bbb R^3\backslash \Delta$ is homeomorphic to $\Delta\times (\Bbb R^2\backslash \{\mathbf 0\})$, which is homeomorphic to $\Bbb R\times (\Bbb R^2\backslash \{\mathbf 0\})$. Therefore, $$\pi_1(\Bbb R^3\backslash \Delta)\cong \pi_1(\Bbb R)\times \pi_1(\Bbb R^2\backslash \{\mathbf 0\})\cong \pi_1(\Bbb R^2\backslash \{\mathbf 0\})\cong\pi_1(\Bbb S^1)\cong \Bbb Z. $$ Note that punctured plane is strong deformation retract onto circle, that's why $\pi_1(\Bbb R^2\backslash \{\mathbf 0\})\cong\pi_1(\Bbb S^1)$.