How this function defines a homeomorphism from $\mathbb B^n$ to itself?

Question

How the function $F_s(x)=|x|^{s-1}x$ defines a homeomorphism from $\mathbb B^n$ to itself for $s>0$? I find this on Lee's smooth manifold. For $0<s<1$, this function is not defined on $0$. For $s>1$, the inverse of this function is not defined on $0$.
However, the $\lim_{x\to 0}F_s(x)=0$, so I think if I modify by $$F_s(x)= \begin{cases} |x|^{s-1}x& \text{if $x \neq 0$}\\ 0 & \text{if x=0}, \end{cases}$$ then this function is homeomorphism. But how do I know if this modified $F_s$ a diffeomorphism or not for $s\neq 1$?

Answer 1

You're absolutely right that the function $F_s(x)$ as given is not well defined at $x = 0$. However, your modified function is still not a diffeomorphism of the $n$ ball to itself for $s \neq 1$. To see this, apply the definition of the derivative to $F_s(x)$ at $x = 0$ (and note that $0 < s < 1$): $$\lim_{x \rightarrow 0}\frac{|F_s(x) - 0|}{|x -0|} = \lim_{x \rightarrow 0}\frac{|x|^{s-1}|x| - 0}{|x|} = \lim_{x \rightarrow 0} |x|^{s-1} = \infty.$$ And hence the function is still not differentiable at the origin.