How would I prove that the $n$-sphere $S^n$ is path connected for $n\geq 1$?
I have seen this question. I do not really understand this section:
For b) The application $\varphi$ from $\mathbb{R}^n \setminus \{ 0 \}$ to $S^{n-1}$ defined by $x \to x/\Vert x \Vert_2$ is continuous. If $x,y$ are on the sphere and not on a diameter, the path $\varphi([x,y])$ is continuous and join $x$ to $y$ on the sphere. If $x$, $y$ are on a diameter, $\varphi([x,z][z,y])$ is a solution.
What is $\varphi$? What is $x/||x||_2$?
This is my proof attempt:
Proof: $\newcommand{\R}{\mathbb{R}}$
We want to show that the $n$-sphere is path connected for $n\geq 1$.
Let $S^n$ be an $n$-sphere where $n\geq 1$. Then $S^n\subseteq\R^{n+1}$.
We know $\R^{n+1}$ is path connected, as is every open and every closed ball in $\R^n$.
Let $f:\R^{n+1}\to S^n$ be a function ???. (What would this be?)
Then by Theorem 6.29, $f(\R^{n+1}) = S^n$ is a path connected subspace of $S^n$
Because $S^n$ is path connected in the subspace topology that $S^n$ inherits from $S^n$, we have that $S^n$ is path connected in $S^n$.
In other words, the $n$-sphere $S^n$ is path connected for $n\geq 1$.
$\square$
Thank you for any help you can provide.
EDIT: Removed textbook specific definition and theorem references.
$X=\mathbb{R}^{n+1}\setminus\{0\}$ is path-connected: take straight line segments between any two points in $X$ and if that line segment would go through $0$, take a path via a third non-origin point so that both line segments mis $0$.
$f: X \to \mathbb{S}^n$ given by $$f(x)=\frac{1}{\sqrt{\sum_{i+1}^{n+1} x_i^2}} \cdot x$$
(or $f(x)=\frac{x}{\|x\|}$ more succinctly) is a well-defined continuous map whose image is exactly $\Bbb S^n$ and the continuous image of a path-connected space is path-connected.