Compute the fundamental group of $\mathbb{R}^4\backslash \mathbb{R}^2$ [closed]

How would I compute $\pi_1(\mathbb{R}^4\backslash \mathbb{R}^2)$? I'm having trouble writing down an explicit homeomorphism or deformation retraction between $\mathbb{R}^4\backslash \mathbb{R}^2$ and another space that would be easier to compute the fundamental group of.

Solution 1:

$\mathbb{R}^4\setminus \mathbb{R}^2$ is hard to imagine, while $\mathbb{R}^3\setminus \mathbb{R}^1$ is much easier. It has $\mathbb{R}^2\setminus \{0\}$ as deformation retract through projection.

So similarly, define projection $(x_1,x_2,x_3,x_4)\mapsto (x_3,x_4)$ to show that it's a deformation retraction.

Solution 2:

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, $\Bbb R^n\backslash \mathbf V\cong \mathbf V\times(\mathbf W\backslash \mathbf 0)$.

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. $\square$

So, $\Bbb R^4\backslash \Bbb R^2$ is homeomorphic to $\Bbb R^2\times (\Bbb R^2\backslash \{\mathbf 0\})$. Therefore, $$\pi_1(\Bbb R^4\backslash \Bbb R^2)\cong \pi_1(\Bbb R^2)\times \pi_1(\Bbb R^2\backslash \{\mathbf 0\})\cong \pi_1(\Bbb R^2\backslash \{\mathbf 0\})\cong\pi_1(\Bbb S^1)\cong \Bbb Z. $$ Note that punctured plane is strong deformation retract onto circle, that's why $\pi_1(\Bbb R^2\backslash \{\mathbf 0\})\cong\pi_1(\Bbb S^1)$.