I want to show that if $\mathbb{R}^n$ has a division algebra structure, then $S^{n-1}$ is parallelizable. I know the approach, but I'm struggling with one detail.
Write $S^{n-1} = \{x \in \mathbb{R}^n : \|x\| = 1 \}$. Fix $e = (1,0,\dots, 0) \in S^{n-1}$, and let $f(x) = \frac{e \cdot x}{\|e \cdot x\|}$, for all $x \in S^{n-1}$, where $e \cdot x$ is the multiplication of the divison algebra, not the dot product. I already have that $f$ is a smooth bijection. How can I show $f^{-1}$ is smooth, which would show that $f$ is a diffeomorphism? I tried looking at $f$ with respect to charts on $S^{n-1}$, but since there is very little given about the division algebra structure (we have no unit, commutativity, or associativity), this makes it very hard to look at the expression for $f$ in certain coordinates, compute the Jacobian, and show that it is invertible.
The charts I was using were defining $U_i^+ = \{x \in S^{n-1} \mid x_i > 0 \}$ with $U_i^-$ defined similarly, and $\phi_i^{\pm} = (x_1, \dots, \hat{x_i}, \dots, x_n)$.
If your algebra product has no zero divisors, then the linear map $m_e(x) = ex$ is injective, hence has an inverse $m_e^{-1}$, which is smooth because it's linear. So $$f^{-1}(x) = \frac{m_e^{-1}(x)}{\|m_e^{-1}(x)\|}$$ is smooth.