Why is $\pi_1(\Bbb{R}^n,x_0)$ the trivial group in $\Bbb{R}^n$?

Question

My Algebraic Topology book says

Let $\Bbb{R}^n$ denote Euclidean n-space. Then $\pi_1(\Bbb{R}^n,x_0)$ is the trivial subgroup (the group consisting of the identity alone).

I wonder why that is. I can imagine infinite continuous "loops" in $\Bbb{R}^3$ that start and end at $x_0$.

Thanks in advance!

Answer 1

$\mathbb{R}^n$ is contractible (homotopy equivalent to a point). Hence its fundamental group is the same as the fundamental group of a singleton, i.e. trivial.

Answer 2

The problem with your last sentence is that $\pi_1(X,x_0)$ is not the set of loops based on $x_0$, but of homotopy classes of loops based on $x_0$.

Can you see why every loop in $\mathbb R^n$ based on $x_0\in \mathbb R^n$ is homotopic to the constant map based in $x_0$?

Answer 3

Let $\gamma\colon[0,1]\to\mathbb R^n$ be a loop in $x_0$, then \begin{align} H\colon [0,1]\times [0,1] &\longrightarrow \mathbb R^n\\ (s, t) &\longrightarrow (1-t)\cdot\gamma(s)+t\cdot x_0 \end{align} is a continous map such that

1. $H(0,t) = H(1,t) = x_0$ for all $t\in[0,1]$,
2. $H(s,0) = \gamma(s)$ and $H(s,1) = x_0$ for all $s\in[0,1]$.

These are the properties of a loop homotopy between the loop $\gamma$ and the constant loop in $x_0$, i.e. $c_{x_0}\colon [0,1]\to \mathbb R^n, s\mapsto x_0$. It is a continous deformation from $\gamma$ to $c_{x_0}$ while keeping the start and end point fixed.

Therefore $\gamma$ and $c_{x_0}$ are loop homotopic and the homotopy classes $[\gamma]$ and $[c_{x_0}]$ are equal. We conclude $$ \pi_1(\mathbb R^n, x_0) = \left\{\, [\gamma] \ \big|\ \text{$\gamma\colon[0,1]\to\mathbb R^n$ a loop in $x_0$}\,\right\} = \left\{ [c_{x_0}] \right\}. $$