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.

• Hint. It may be easier than you think. Can you show that the norm function is a continuous function to the reals? Then its reciprocal will be, except at the origin. Mar 28, 2017 at 16:12
• @EthanBolker Thanks for the hint. I edited my post with a proof of this fact. Would you mind critiquing it when you get a chance? Mar 28, 2017 at 17:03
• For the norm business, just note $||x|-|y||\le |x-y|.$
– zhw.
Mar 28, 2017 at 17:31
• @user193319 You end with "clearly, then, the function defined is continuous". However, you use the result that the quotient of real functions is continuous. $f(x)=x$ is not a real function. You then either have to argue with components and use the fact that projections are continuous (this will not generalize well for normed spaces in general) or change your argument. Mar 29, 2017 at 4:13
• Anyway, to prove that $g$ is continuous is a standard argument of "composition/multiplication/product/restriction/etc" of continuous maps. Writing down what those are is just a matter of straightforward (and a bit patient) checking. Mar 29, 2017 at 4:16

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 $$v.$$
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.
You want to show that $||g(x)-g(y)||$ is small when $||x-y||$ is small. For this use the following, $$g(x)-g(y)= \frac{(x-y)||y|| + y (||y||-||x||)}{||x|| \ ||y||}$$