The Unit Sphere $S^{n-1}$ is Path-Connected I am trying to understand Munkres' proof that $S^{n-1}$ is path-connected. Below is a snippet from the book. 
It's clear to me that $g$ is surjective; and I concur with him with him when he says it is rather easy to show the continuous image of a path-coonected space is path-connected. What is giving me trouble is showing that $g$ is continuous.
I tried showing that $g : \mathbb{R}^n_0 \rightarrow \mathbb{R}^n_0$ is continuous, which would entail that the restriction to $f(\mathbb{R}^n_0) = S^{n-1}$ is continuous. This didn't seem to lead anywhere though. If I am not mistaken, a basis element of $\mathbb{R}^n_0$ is $\prod (a_i,b_i)$, where each interval $(a_i,b_i)$ does not contain $0$. The preimage under $f$ is 
$$f^{-1}\left(\prod (a_i,b_i)\right) = \left\{x \in \mathbb{R}^n_0 ~|~ \frac{x}{||x||} \in \prod (a_i, b_i) \right\} = \left\{x \in \mathbb{R}^n_0 ~|~ x \in \prod (||x||a_i,||x||b_i) \right\}$$
which I can't really make sense of; I can't determine exactly what this set looks like. 
EDIT: Following the unanamious suggestion, I will attempt to prove that the norm is a continuous function. Note that the norm on $\mathbb{R}^n$ induces the standard euclidean metric on $\mathbb{R}^n$, which in turn induces the standard topology. In a previous exercise, it was shown that the metric is continuous with respect to the topology it induces. With a little thought, we can see that $||x|| = d(x,0)$ and is nothing more than the restriction of a continuous function (i.e., the metric) to $\mathbb{R}^n \times \{0\}^n$, which means the norm is continuous. 
Having the zero vector, which is a crucial to the above proof, this proof generalizes to all normed vector spaces. I'll have to think about the case in which the normed space isn't a vector space, but I think this gets the job done.
Now, in the previous chapter of munkres, I am told that if $f,g : X \rightarrow \mathbb{R}$ are continuous functions and $g(x) \neq 0$ for any $x \in X$, then $f/g : X \rightarrow \mathbb{R}$ is continuous. Clearly, then, the function defined in the picture is continuous. 
 A: Another approach: If $u,v \in S$ are linearly independent, define $f:[0,1]\to S$ by
$$f(t) = \frac{(1-t)u + tv}{|(1-t)u + tv|}.$$
Then $f$ is a path from $u$ to $v$ within $S.$  If $u,v$ are linearly dependent, we can choose $w\in S$ so that $u,w$ are linearly independent. Note $w,v$ are also linearly independent. As above, we can connect $u$ to $w$ and then connect $w$ to $v.$
A: The function $n: \mathbb{R}^n \to \mathbb{R}$ given by $n(x)=\Vert x \Vert$ is continuous. The function $i: \mathbb{R} -\{0\} \to \mathbb{R}$ given by $i(x)=\frac{1}{x}$ is continuous. The function $m: \mathbb{R} \times X \to X$ given by $(k,x)\to kx$ is continuous.
Your function $g$ is $\big(m \circ \big(\big(i \circ (n|_{\mathbb{R}^n-\{0\}}^{\mathbb{R}-\{0\}})\big)\times \mathrm{Id}_{\mathbb{R}^n-\{0\}}\big) \big)^{S^{n-1}}$, where upper index means restriction of image and lower index means restriction of domain. Since restrictions are continuous and the "product" $f_1 \times f_2$ of continuous maps $f_1,f_2$ is continuous, the result follows.
A: You want to show that $||g(x)-g(y)||$ is small when $||x-y||$ is small. For this use the following,
$$g(x)-g(y)= \frac{(x-y)||y|| + y (||y||-||x||)}{||x|| \ ||y||}$$
