How do I show that $\mathbb{R}^n$ is simply-connected?

Question

I have shown that $\mathbb{R}$ is simply connected by reparametrization.

However, how do I show that $\mathbb{R}^n$ is general?

Answer 1

So a space is simply connected if it is path connected and the fundamental group is trivial.

Observe that $\mathbb R^n$ is convex, which means the straight line between any two points in the space is contained in the space, that is for any $\textbf x, \textbf y$, $(1 - t) \textbf x + t \textbf y \in \mathbb R^n$ for all $t \in [0, 1]$.

In particular, the convexity of $\mathbb R^n$ tells us $\mathbb R^n$ is path-connected.

Now we still need to show $\pi_1(\mathbb R^n)$ is trivial, so let $[f] \in \pi_1(\mathbb R^n, \textbf x_0)$ be arbitrary. Observe that $f$ is path homotopic to the constant map at $\textbf x_0$, $e_{x_0}$ via the straight line homotopy (which is well-defined by convexity of $\mathbb R^n$) given by $F : I \times I \to \mathbb R^n$ defined by $F(s, t) = (1 - t)f(s) + t \textbf x_0$. This shows that $[f] = [e_{x_0}]$ for arbitrary $[f]$ which means $\pi_1(\mathbb R^n)$ is trivial.