Homeomorphism of the ball isotopic to the identity on the sphere Let $h:B^{n}\to B^{n}$ be a homeomorphism of the $n$-dimensional unit ball in $\mathbb{R}^n$. If $h(x)=x$ for $x\in S^{n-1}$, then it is relatively simple to show that $h$ is isotopic to the identity on the full $B^{n}$: I can construct the isotopy as follows:
$$
h_{t}(x):=
\left\{
\begin{array}
xx & \mbox{if } t\leq |x| \leq 1 \\
t\cdot h\left(\dfrac{x}{t} \right) & \mbox{if }0\leq |x| \leq t
\end{array}
\right.
$$
My question is: if I just assume that $h$ is isotopic to the identity on $S^{n-1}$, does it still follow that $h$ is isotopic to the identity on the full ball $B^{n}$?
I want to use this result to prove that if I have two homeomorphisms $h,k: B^{n}\to B^{n}$ such that their restrictions to $S^{n-1}$ are isotopic, then they are isotopic on $B^{n}$. Any hints or suggestions would be very much appreciated!
 A: Here's the rough idea, coming from what Mike Miller said in the comments.  We will view the ball $B^n$ as a union of concentric spheres.  Let $B_+$ denote the union of the spheres of radius at least 1/2 and $B_-$ denote the union of the spheres of radius at most $1/2$ (including the origin as well.)  Note that $B_+\cap B_-$ is the sphere of radius $1/2$.
We will first isotope $h$ to $h_1$, where $h_1$ copies $h$ on all spheres in $B_+$, and then "speeds" through $h$ on $B_-$.
In formulas, we use $H_1(x,t) = \begin{cases}|x| h\left(\frac{x}{|x|}\right) & \text{if } t\leq |x| \\ th\left(\frac{x}{t}\right) & \text{if } |x|\leq t \end{cases}$ where we are thinking of $t\in [1/2,1]$.  Then $H_1(x,1) = h(x)$ and $H_1(x,1/2) = h_1$ is described above.  In particular, $h_1$ on the sphere of radius 1/2 is just a scaled version of $h$.
$ \ $
Next, we isotope $h_1$ to $h_2$ where $h_2|_{B_-} = h_1|_{B_-}$ but where $h_2|_{S^{n-1}} = Id_{S^{n-1}}$.
To do this, first pick an isotopy $F:S^{n-1}\times [0,1]\rightarrow S^{n-1}$ between $h|_{S^{n-1}}:S^{n-1}\rightarrow S^{n-1}$ and the identity with $F(x,0) = h$ and $F(x,1) = Id$.  Then, to isotope $h_1$ to $h_2$, we use the formula $H_2(x,t) = \begin{cases} |x|F(x/|x|, (2|x|-1)t)  & \text{if }  |x|\geq 1/2\\ h_1(x) & \text{if } |x| \leq 1/2\end{cases}.$
$ \ $
At this stage, $h_2$ is the identity on $S^{n-1}$, so to finish isotoping $h$ to $Id|_{B^n}$, we apply the homotopy you gave in your post.
