Define the unit sphere $S^{n-1}$ in $\Bbb{R}^n$ by the equation $S^{n-1}=\{ x | \| x\|=1\}$. If $n\gt 1$, it is path connected.
To be honest I don’t understand the proof, even after assuming $g$ is continuous and subjective. The Unit Sphere $S^{n-1}$ is Path-Connected. Claim: If $h:X\to Y$ is continuous and $X$ is path connected, then $h(X)$ is path connected. Proof: By theorem 18.2 $f:X\to f(X)$ defined by $f(x)=h(x) ,\forall x\in X$, is continuous. $h(X)=f(X)$. let $f(x),f(y)\in f(X)$. Since $X$ is path connected and $x,y\in X$, $\exists g:[0,1]\to X$ such that $g$ is continuous, $g(0)=x$ and $g(1)=y$. By theorem 18.2, $f\circ g:[0,1]\to f(X)$ is continuous. So $f\circ g(0)=f(x)$ and $f\circ g(1)=f(y)$. Thus $f\circ g$ is a path from $f(x)$ to $f(y)$. Hence $h(X)=f(X)$ is path connected. Is this proof correct? How to use this result to show $S^{n-1}$ is path connected? I don’t even known(explicitly) what is the path from $x$ to $y$,in order to show $ \Bbb{R}^n -\{0\}$(punctured euclidean space) is path connected. This is example 4.
Edit: Better version of proof of above claim is to consider surjective map i.e. If $f:X\to Y$ is continuous and surjective/onto, $X$ is path connected, then $Y$ is path connected.
Map $g:\mathbb{R}^n\backslash\{0\}\to S^{n-1}$ defined by $g(v)=\frac{1}{\lVert v\rVert}\cdot v$ is continuous and surjective. So $g(\mathbb{R}^n\backslash\{0\})= S^{n-1}$, because $g$ is surjective. Since image of a path connected space under continuous map is path connected, $g(\mathbb{R}^n\backslash\{0\})= S^{n-1}$ is path connected. Our desired result.
Best Answer
The proof is correct. You use this result by considering $g:\mathbb{R}^n\backslash\{0\}\to S^{n-1}$ given by $g(v)=\frac{1}{\lVert v\rVert}\cdot v$. This function is continuous as a composition of well known continuous functions: $v\mapsto \lVert v\rVert$, $x\mapsto\frac{1}{x}$ and $(r,v)\mapsto r\cdot v$ (vector by scalar multiplication). It is also well defined because
$$\lVert g(v)\rVert=\bigg\lVert \frac{1}{\lVert v\rVert}\cdot v\bigg\rVert=\frac{1}{\lVert v\rVert}\cdot \lVert v\rVert=1$$
Finally it is surjective, because if $v\in S^{n-1}$, which is a subset of $\mathbb{R}^n\backslash\{0\}$, then $$g(v)=\frac{1}{\lVert v\rVert}\cdot v=\frac{1}{1}\cdot v=v$$
and so $v$ belongs to the image of $g$ (in fact $g$ maps $v$ to itself).
So all you need to show is that $\mathbb{R}^n\backslash\{0\}$ is path connected. Given two vectors $v,w\in\mathbb{R}^n\backslash\{0\}$ we can consider the following function
$$\alpha:[0,1]\to\mathbb{R}^n\backslash\{0\}$$ $$\alpha(t)=t\cdot w+(1-t)\cdot v$$
It is a simple, straight line connecting $v$ and $w$. It is continuous (as a composition of well known continuous functions: addition, subtraction, scalar multiplication and vector addition) and $\alpha(0)=v$ while $\alpha(1)=w$. The only problem with that function is that if $0$ is exactly between $v$ and $w$ (this happens for example when $w=-v$ and $t=\frac{1}{2}$) then $\alpha$ goes through it, meaning its image is not necessarily in $\mathbb{R}\backslash\{0\}$. But if $v,w$ are linearly independent then $\alpha$ is a well defined path. Because being linearly independent implies that $t\cdot w+(1-t)\cdot v=0$ equation has no solution, and so $0$ is not in the image of $\alpha$.
So what to do when $v,w$ are linearly dependent? It is not a problem, because there always is some $u\in\mathbb{R}^n\backslash\{0\}$ such that $\{v,u\}$ are linearly independent and $\{w,u\}$ are linearly independent. This is precisely where we need $n>1$. Because every linearly independent set, in this case singleton, can be extended to a basis which has to have at least two elements. Finally we compose our path (see: path composition) by first connecting $v$ to $u$ and then $u$ to $w$ (which we already know how to do by previous reasoning).