Next, a topological group $G$ is *continuously factorizable* if there exist $d\in\N$, $B_1\ld
    B_d\in\B$, and continuous mappings $S_i:G\r G_{B_i}$, $i=1\ld r$, such that for all $g\in G$ $$g=S_1(g)\ldots S_{d}(g).$$

Under the assumption of Theorem 1.1 on $M$, the groups $\hm$ and $\hdm$ are factorizable with respect to any cover $\U\s\B$. The same is true for the isotopy groups $\P\hm$ and $\P\hdm$.

If $M$ is compact, it follows from Theorem 2.3. If $M$ admits a compact exhaustion, the reasoning is similar to that in the proof of 1.1.

For any $g\in\hm, g\neq\operatorname{id}$, denote by $\operatorname{frag}_M(g)$ the smallest $d$ such that (5.2) holds. By definition $\operatorname{frag}_M(\operatorname{id})=0$. Clearly $\operatorname{frag}_M$ is a norm on $\hm$. Likewise we define $\operatorname{frag}_M^{iso}$ on the isotopy group $\mathcal P\hm$. Clearly $\operatorname{frag}_M(f)\leq\operatorname{frag}_M^{iso}(f_t)$ if $f_t$ is an isotopy connecting $f$ with the identity.

The significance of $\operatorname{frag}_M$ is illustrated by Theorem 1.4.

1.  A topological group $G$ is *continuously perfect* if there exist $r\in\N$ and continuous mappings $S_i:G\r G$, $\bar
    S_i:G\r G$, $i=1\ld r$, satisfying the equality $$g=[S_1(g),\bar S_1(g)]\ldots[S_{r}(g),\bar
    S_{r}(g)]$$ for all $g\in G$.

2.  Let $H$ be a subgroup of $G$. $H$ is said to be *continuously perfect in G* if there exist $r\in\N$ and continuous mappings $S_i:H\r G$, 