Let $B^{n}$ denote the euclidean closed ball of dimension $n$. Alexander's trick shows that every homeomorphism of $B^{n}$ fixing $\partial B^n$ is $C^{0}$-isotopic to the identity via a boundary fixing isotopy.
On the other hand, there are values of $n$ for which the group $\mathit{Diff}_{0}(B^n,\partial B^n)$ of all diffeomorphisms of $B^n$ that are $C^\infty$-isotopic to the identity through a boundary-fixing isotopy does not coincide with the full group of boundary fixing diffeos, $\mathit{Diff}(B^n,\partial B^n)$. If I understood/remember correctly, equality is known to hold for $n=2,3,5$ and a few other values and fails for $n=6$ and other higher dimensions, while the question remains open in dimension $4$. Are there any surveys you could recommend where one could read on the current status of the problem?
What I really wanted to ask, though, is the following. Is it known whether $\mathit{Diff}_0(B^{n},\partial B_{n})$ is dense in $\mathit{Homeo}(B^n,\partial B_{n})$ with respect to the uniform convergence topology? The discussion here would seem to answer this in the affirmative for all values of $n$ other than $n=4,5$, while case $n=5$ follows from the equality between $\mathit{Diff}(B^5,\partial B^5)$ and $\mathit{Diff}_{0}(B^5,\partial B^5)$.
If all the above checks out that leaves us only with $n=4$. Is the question above known to be true for $n=4$? Or is there any reason to hope for a relatively accesible proof that wouldn't involve proving $\textit{Diff}(B^4,\partial B^4)=\textit{Diff}_{0}(B^4,\partial B^4)$?
