A trivial fundamental group

Question

I am reading fundamental groups from Munkres book.

As stated in the definition, I understand a fundamental group relative to a base point $x_0$ includes all the loops based at point $x_0$.

Later they give example of a fundamental group in $\mathbb{R}^n$. This group, $\pi_1(X,x_0)$ is a trivial fundamental group as it consists of the identity alone.

I am having trouble understanding the definition of a trivial fundamental group. Here, by identity do they refer to $e_{x_0}$ as $e_{x_0}$ includes the paths originating and ending at $x_0$. Also, how does a unit ball in $\mathbb{R}^n$ have a trivial fundamental group?

Answer 1

Remember the fundamental group includes not the loops themselves, but the homotopy classes of loops.

So stating that the fundamental group $\pi_1(X,x_0)$ is trivial is equivalent to stating that every loop going from and to $x_0$ are homotopic. This is true for $\mathbb{R}^n$. Up to a translation, we can assume $x_0$ is $0$. Now, let $\gamma : [0,1] \to X$ be a loop based at $0$. Then we have an homotopy $\Gamma : [0,1]^2 \to \mathbb{R}^n$ between $\gamma$ and the constant loop $t \mapsto 0$, given by :

\begin{equation} \Gamma_s(t) = s \cdot \gamma(t) \end{equation}

Indeed, $\Gamma_0(t) = 0 \cdot \gamma(t) = 0$ and $\gamma_1(t) = 1 \cdot \gamma(t) = \gamma(t)$. $\Gamma$ is moreover obviously continuous, so it is a valid homotopy.

Now if you have two paths, $\gamma_1$, and $\gamma_2$, their must be homotopic since both are homotopic to $t \mapsto 0$ and the "is homotopic" relation is an equivalence relation (certainly proved somewhere in your book).

The homotopy used to prove the same result in a ball is identical.

Answer 2

Every loop in these spaces can be shrunk to a point within that space.

Answer 3

Notice that a ball in $\mathbb{R}^n$ is contractible, that is, it is homotopic to a point. So all loops at $x_{o}$ in the ball are null-homotopic, which is to say, homotopic to the constant loop at $x_{o}$. Constant loop at $x_{o}$ means that the loop doesn't go anywhere, ie. the loop stays at $x_{o}$. Hence $[x_{o}] = id$ and $\pi (X, x_{o})$ is trivial.