Real Projective Space & Quotient of a Sphere

Question

I have been trying to solve the Exercise 7.11 of Loring Tu's An Introduction to Manifolds (Second Edition, page no. 77). It reads as follows.

The way I proceeded to solve it, it has been discussed partly here. But I also wanted to understand the solution supplied by the author given below.

I have some questions about this solution.

My Questions

Here $\pi_2$ is continuous because it is a projection map. How do we deduce that $f: \mathbb{R}^{n+1} - \lbrace 0 \rbrace \to S^n, f(x) = \frac{x}{||x||},$ is continuous so that $\pi_2 \circ f$ is continuous?
In the second part of the proof, Tu applies the same technique (which is used to prove $\bar{f}$ continuous) to prove $\bar{g}$ continuous. What is the motivation for defining $g: S^n \to \mathbb{R}^{n+1} - \lbrace 0 \rbrace$ by $g(x) = x$? In other words, how do we know that $g$ needs to be defined in this way?

For 1, one approach is to consider the map $\iota_{S^n} \circ f : \mathbb{R}^n\smallsetminus \{0\} \to \mathbb{R}^n$ which is just $f$ with codomain $\mathbb{R}^n$. This map is continous, since its components are continous, and therefore by characteristic property of subspace topology, the map $f$ is continous. For 2, i think the motivation comes from the observation by playing around with $f$ until you know how to define $g$ so that $\tilde{g}$ is the inverse of $\tilde{f}$. — Kelvin Lois, Commented Aug 5, 2020 at 1:33

Mathematician.... · Accepted Answer · 2023-08-27 16:35:54Z

To deduce that $f$ is continuous it is enough to note that x is continuous because it is a polynomial and $\lVert x \rVert$ is continuous because it is the norm. The composition of $\frac{x}{\lVert x \rVert}$ is continuous because the domain does not contain 0 (the denominator will never be 0). Therefore, the composition $\pi \circ f$ is continuous.
The motivation for defining g that way is $\bar{g} \circ \bar{f} = \left[\frac{x}{\lVert x \rVert}\right] = [x]$ and $\bar{f} \circ \bar{g} = \bar{f}([x]) = \left[\frac{x}{\lVert x \rVert}\right] =\left[\frac{x}{1}\right] = [x]$ which results (checking the other composition) in these functions being inverses of each other.

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