Let $M$ be a compact manifold and let $f : M \rightarrow M$ be a homeomorphism which is isotopic to the identity. We will say that $f$ can be *fragmented* if it satisfies the following property. Let $\mathcal{U}$ be any open cover of $M$. There then exists homeomorphisms $f_1,\ldots,f_n$ from $M$ to itself which are isotopic to the identity and open sets $U_1,\ldots,U_n \in \mathcal{U}$ such that the support of $f_i$ is in $U_i$ and $f = f_1 \cdots f_n$.

It is known that if $M$ is a smooth manifold and $f$ is a diffeomorphism that is smoothly isotopic to the identity, then $f$ may be fragmented (and the resulting $f_i$ are also diffeomorphisms smoothly isotopic to the identity). Indeed, this is is one of the ingredients of Thurston's proof that the identity component of the group of diffeomorphisms of a compact manifold is simple. A proof can be found in chapter 2 of Banyaga's book "The structure of classical diffeomorphism groups".

However, the proof in the smooth case appears to use smoothness in a strong way.

Question : Can an arbitrary homeomorphism $f : M \rightarrow M$ which is isotopic to the identity be fragmented?