Let $f: \mathbb{R}^{n+1}\setminus{\{0\}}\rightarrow{S^{n}}$ and $g: S^{n}\rightarrow{\mathbb{R}^{n+1}}\setminus{\{0\}}$ be functions given by $f(x)=\displaystyle\frac{x}{||x||}$ and $g(x)=x$, where $|| x ||$ is the module of x.
Are $ f $ and $ g $ continuous functions?
Does anyone have any suggestions to demonstrate the continuity of f and g?
On
I will do the argument for $g: S^n \rightarrow \mathbb{R}^{n+1}\setminus\{0\}$ where $g(x) = x$. This is easily done by the definition of continuity. (Note that we take the topologies on $S^n$ and $\mathbb{R}^{n+1}\setminus\{0\}$ to be the subspace topologies in $\mathbb{R}^{n+1}$.) Let $U$ be open in $\mathbb{R}^{n+1}\setminus\{0\}$. Then $U= B\cap \left(\mathbb{R}^{n+1}\setminus\{0\}\right)$ for some open $B\in\mathbb{R}^{n+1}$. Therefore, \begin{align*}g^{-1}(U) &= \left\{x\in S^n: g(x)\in U\right\} \\&= \left\{x\in S^n: x\in B\cap \left(\mathbb{R}^{n+1}\setminus\{0\}\right)\right\} \\&= S^n\cap \left[B\cap \left(\mathbb{R}^{n+1}\setminus\{0\}\right)\right].\end{align*} Note that $\mathbb{R}^{n+1}\setminus\{0\}$ is open in $\mathbb{R}^{n+1}$ (for example, it is the complement of the closed set $\{0\}$; singletons are closed in Hausdorff spaces, which $\mathbb{R}^{n+1}$ is in its standard topology). Therefore, $B\cap \left(\mathbb{R}^{n+1}\setminus\{0\}\right)$ is open in $\mathbb{R}^{n+1}$ (finite intersection of opens sets is open). Thus, by definition of the topology on $S^n$, the set $S^n\cap \left[B\cap \left(\mathbb{R}^{n+1}\setminus\{0\}\right)\right]$ is open in $S^n$. We conclude that $g$ is continuous.
It is similar to show that $f: \mathbb{R}^{n+1}\setminus\{0\}\rightarrow S^n$ given by $f(x) = \frac{x}{\|x\|}$ is continuous. Here is the outline of the proof. Let $U$ be open in $S^n$. Then $U= B\cap S^n$ for some open $B$ in $\mathbb{R}^{n+1}$. To prove that $f$ is continuous, you need to show that $$f^{-1}(U) = \left\{x\in\mathbb{R}^{n+1}\setminus \{0\}: \frac{x}{\|x\|}\in B\cap S^n\right\}$$ is open in the topology of $\mathbb{R}^{n+1}\setminus \{0\}$. That is, you want to show that $f^{-1}(U) = B^{\prime} \cap \left(\mathbb{R}^{n+1}\setminus \{0\}\right)$ for some open $B^{\prime}$ in $\mathbb{R}^{n+1}$. If you draw the picture out in $\mathbb{R}^2$, it is fairly easy to see why $f^{-1}(U)$ is open. In general, you just need to work though the definitions, and is slightly more tedious. By the way, since the collection of open balls $\left\{B_{\epsilon}(x): x\in\mathbb{R}^{n+1},\,\epsilon>0\right\}$ is a basis for the topology on $\mathbb{R}^{n+1}$, it suffices to show that $f^{-1}(U)$ is open for all balls $B_{\epsilon}(x)$.
The continuity of $g$ is straight forward. The map $g$ is simply the inclusion map. Fix $x_0 \in S^n$, $\varepsilon > 0$, and let $\delta = \varepsilon$. Then $$\|x - x_0\| < \delta \implies \|g(x) - g(x_0)\| = \|x - x_0\| < \delta = \varepsilon.$$ The continuity of $f$ is less straight forward. Fix $x_0 \in \Bbb{R}^n \setminus \{0\}$, and $\varepsilon > 0$. We wish to find a $\delta > 0$ such that $$\|x - x_0\| < \delta \implies \|f(x) - f(x_0)\| = \left\|\frac{x}{\|x\|} - \frac{y}{\|y\|}\right\| < \varepsilon.$$ In order to figure out such a $\delta$, it is typically prudent to start with the $\|f(x) - f(x_0)\| < \varepsilon$, and work backwards.
Remember: we are always allowed to substitute $\|f(x) - f(x_0)\|$ for something larger. If we can make this larger quantity smaller than $\varepsilon$, then $\|f(x) - f(x_0)\|$ will be smaller than $\varepsilon$ too.
We have \begin{align*} \|f(x) - f(x_0)\| &= \left\|\frac{x}{\|x\|} - \frac{x_0}{\|x_0\|}\right\| \\ &= \left\|\frac{x}{\|x\|} - \frac{x}{\|x_0\|} + \frac{x}{\|x_0\|} - \frac{x_0}{\|x_0\|}\right\| \\ &\le \left\|\frac{x}{\|x\|} - \frac{x}{\|x_0\|}\right\| + \left\|\frac{x}{\|x_0\|} - \frac{x_0}{\|x_0\|}\right\| \\ &\le \left\|\frac{x}{\|x\|} - \frac{x}{\|x_0\|}\right\| + \left\|\frac{x}{\|x_0\|} - \frac{x_0}{\|x_0\|}\right\| \\ &= \left|\frac{1}{\|x\|} - \frac{1}{\|x_0\|}\right|\|x\| + \frac{1}{\|x_0\|}\|x - x_0\|. \end{align*} In order to make this less than $\varepsilon$, we can make each of the terms in the sum less than $\varepsilon / 2$. The right term is easy; if we force $\delta \le \frac{\varepsilon \|x_0\|}{2}$, then $$\|x - x_0\| < \delta \implies \|x - x_0\| < \frac{\varepsilon \|x_0\|}{2} \implies \frac{1}{\|x_0\|} \|x - x_0\| < \frac{\varepsilon}{2}$$ as needed.
The other term is more tricky for a couple of reasons. Firstly, the $\|x\|$ term is not constant; we can't just treat it the same way we did with the $\|x_0\|$ term. It's not even bounded, meaning that $\|x\|$ could be arbitrarily large! That is, unless we limit $\delta$. If we force, say $\delta \le \frac{\|x_0\|}{2}$, then we can say for sure that $$\|x - x_0\| < \delta \implies \|x - x_0\| < \frac{\|x_0\|}{2} \implies \|x\| \le \|x - x_0\| + \|x_0\| < \frac{3\|x_0\|}{2}.$$ Note that I could have replaced $\frac{\|x_0\|}{2}$ with any positive number (e.g. $1$). I've chosen $\frac{\|x_0\|}{2}$ specifically because I know I'll want (very soon) for $x$ to be bounded away from $0$. The closer to $0$ that $x$ becomes, the larger $\frac{1}{\|x\|}$ becomes, and the larger $\left|\frac{1}{\|x\|} - \frac{1}{\|x_0\|}\right|$ becomes. By bounding $\delta$ by $\frac{\|x_0\|}{2}$, I ensure that
\begin{align*} \|x - x_0\| < \delta &\implies \|x_0\| - \|x\| \le \|x - x_0\| < \frac{\|x_0\|}{2} \\ &\implies \|x\| > \|x_0\| - \frac{\|x_0\|}{2} = \frac{\|x_0\|}{2} \\ &\implies \frac{1}{\|x\|} < \frac{2}{\|x_0\|}. \end{align*}
Under this assumption of $\delta \le \frac{\|x_0\|}{2}$, note that $$\left|\frac{1}{\|x\|} - \frac{1}{\|x_0\|}\right| = \frac{1}{\|x\| \cdot \|x_0\|} \cdot \Big|\|x\| - \|x_0\|\Big| \le \frac{2}{\|x_0\|^2}\|x - x_0\|.$$ From the working above, we further have, under this same assumption $$\left|\frac{1}{\|x\|} - \frac{1}{\|x_0\|}\right| \|x\| \le \frac{2}{\|x_0\|^2}\|x - x_0\| \cdot \frac{3\|x_0\|}{2} = \frac{3}{\|x_0\|}\|x - x_0\|.$$ We can make this less than $\frac{\varepsilon}{2}$ by making $\delta \le \frac{\varepsilon\|x_0\|}{6}$.
So, in summary, in order for our $\delta$ to work, it must be less than or equal to $\frac{\|x_0\|}{2}$, $\frac{\varepsilon\|x_0\|}{6}$, and $\frac{\varepsilon\|x_0\|}{2}$. Note that the latter is redundant, so our choice for $\delta$ is (at long last) $$\delta = \min \left\{\frac{\|x_0\|}{2}, \frac{\varepsilon\|x_0\|}{6}\right\}.$$
This is not the proof. To write out the proof, you now have to write all of the above out in a sensible order. You start by assuming that $\|x - x_0\| < \delta$, and you must conclude (using the above working) that $\|f(x) - f(x_0)\| < \varepsilon$. I'm going to leave it to you.