How to show this function $h:\Bbb{R}^n\setminus{\{0\}} \to S^{n-1} \times \Bbb{R}$ is continuous? Is the function $h:\Bbb{R}^n\setminus{\{0\}} \to S^{n-1} \times \Bbb{R}$ defined by $x=(x_1, \dots, x_n) \mapsto \left(\frac{x_1}{\|x\|}, \dots, \frac{x_n}{\|x\|}, \log\|x\|\right)$ continuous?
I tried to do it using open sets. If $U \in S^{n-1}\times\Bbb{R}$ is open then we can write $U$ as a union of open sets (in the product topology) of the form $\Bbb{B}\cap S^{n-1} \times (a,b)$ where $\Bbb{B}$ is an open ball in $\Bbb{R}^n$ and $(a,b)$ is an open interval in $\Bbb{R}$. So we only need to show $h^{-1}(\Bbb{B}\cap S^{n-1} \times (a,b))$ is open for all such sets.
However $h^{-1}(\Bbb{B}\cap S^{n-1} \times (a,b)) = \{\exp(z)(y_1,\dots,y_n):(y_1,\dots,y_n)\in\Bbb{B}\cap S^{n-1}, z\in(a,b)\}$ and I am unsure how to show that this is open.

To prove that a map $x : f(x)=(f_1(x), \dots, ,f_p(x))$ is continuous where $x \in \mathbb R^n$, it is sufficient to prove that each $x \mapsto f_i(x)$ is continuous.
$x \mapsto \Vert x \Vert$ is continuous as it is the square root of a polynomial map which is continuous.
$x \mapsto \frac{x_i}{\Vert x \Vert}$ is continuous as the ratio of two continuous map with the denominator not vanishing.
Finally $x \mapsto \log \Vert x \Vert$ is continuous as the composition of two continuous maps.
This allows to conclude to the continuity of $h$.
