This questions is asked in Armstrong's Topology book, and I am totally stuck.... I could really use a major hint:
Think of $S^n \subset \mathbb{E}^{n+1}$. Given a loop $\alpha \in \pi_1 (S^n , x_0)$, for $n \geq 2$, show that there is a loop $\beta \subset \mathbb{E}^{n+1}$ such that $\beta(0) = \beta(1) = x_0$ for which it is made up of a finite number of straight line segments, and satisfies $||\alpha(s) - \beta(s)|| < 1$ for all $s \in [0,1]$. Deduce that $S^n$, $n \geq 2$ is simply connected.
My main question is, how does it help to have $\beta$ composed of straight lines, and how would this imply the space is simply connected?
It also says to explain why $n = 1$ doesn't work; I fell like after I understand the former part of this proof, or the intuition, this would make sense. Or maybe, if someone could give me intuition for a reason why this case fails, I could gain an insight to why the other method should work.
Any help would be great, big thanks ahead of time.
To show that $\alpha$ is trivial in $\pi_{1}$ it suffices to show that it is freely homotopic to a point. By "freely homotopic" I mean homotopic through loops that need not be based at $x_{0}$. Suppose we could find a point $x$ on the sphere for which the antipodal point $-x$ does not lie on $\alpha$. We could homotope $\alpha$ into the point $x$ via the "straight-line homotopy", projected back on to the sphere. The only way this could fail is if the straight lines passes through the origin (how do you project $0$ back on the sphere?), but that would only happen if $\alpha$ contained the antipodal point, since the line containing antipodal points on the sphere passes through the origin.
Anyhow, $\alpha$ might be this hideous, continuous, space-filling monstrosity. We cannot assume it misses any points. But if it is homotopic to something simple, like the $\beta$ described above, then we apply the above proof to $\beta$, since a finite union of straight segments cannot fill the sphere.