I want to prove that the fundamental group of $S^n$ is trivial.
I found this demonstration by Nersés Aramian which can be found in this pdf.
It is essentially demonstrated that every path in $S^n$ is homotopic to a path that is not surjective.
I would like to understand this demonstration. My problem with this proof is at the end of it.
Why is the comment made about the $\sigma$ and $\tilde{\sigma}$ maps?
And the last paragraph was very confusing for me.
The bit about $\sigma$ and $\tilde{\sigma}$ is claiming that any continuous function $\sigma: J \to D^n$ such that the endpoints of $J$ are not mapped to the center of $D^n$ is homotopic to a continuous function $\tilde{\sigma}: J \to D^n$ which does not go through the center at all. You can picture in your head what such a $\tilde{\sigma}$ could look like: Either a straight line between the endpoints of the path, or if that happens to go through the center, then a curved path between them. Perhaps a semicircle. Or any continuous path you care to imagine that avoids the center.
Aramian simply says that "Clearly $\sigma \simeq \tilde{\sigma}$" without giving an explicit proof, but it's actually true that all paths with the same endpoints are homotopic to each other in $D^n$. After all, following one path and then the other in reverse gives a loop, and we know that $\pi_1(D^n) = 0$, so the loop must be null homotopic, implying that the two paths are homotopic.
In the last paragraph, Aramian constructs a homotopy of $f$ to a new path $g$. To construct $g$, they consider each point in which $f$ crosses $*$ and bend $f$ locally to avoid $*$. They show that this is done a finite number of times (once within each open interval $I_n$) and use the argument from the previous paragraph to show that there is a homotopy from $f$ to $g$ within each of these $I_n$. Since $f$ and $g$ agree outside of these intervals, we get a homotopy from $f$ to $g$ in total.