and from (3.3) we have $$h_k\in \H_{A_k}(\p\t\R_+)\quad\hbox{for all}\;k.$$

The condition (3.4) means that we exclude any twisting of $h_k$.

In order to show the above statements we apply Theorem 2.5 for $M^o$. This enables us to define recurrently $b_k>\bar b_k>\bar a_k>a_k$ and $h|_{\p\t[a_k,b_k]}$ for $k=1,2,\ldots$. In fact, let $f_t$ be an isotopy in $\hr$ connecting $f$ with the identity. Suppose we have defined $1>b_1>\ldots> a_{k-1}$ and $g\in\hr$ such that $g=f\quad\hbox{on}\quad\p\t\bigcup_{i=1}^{k-1} [\bar a_i,\bar
b_i]$, $\operatorname{supp}(g)\s\bigcup_{i=1}^{k-1}A_i$, and $g_i\in\H_{A_i}(\p\t\R_+)$ for all $i\leq k-1$. Now it suffices to take $a_{k-1}>b_k>\bar b_k>\bar a_k>a_k$ in such a way that $\p\t(0,b_k]$ is disjoint with $\bigcup_{t\in[0,1]}f_t^{-1}(\p\t[a_{k-1},1])$. In view of Theorem 2.5 we get an isotopy $\bar h_t$ such that $\bar h_t=f_t$ on $\p\t[\bar a_k,\bar b_k]$ and $\operatorname{supp}(\bar h_t)\s\p\t( a_k, b_k)$. Next we define $h$ on $\p\t[a_k,1]$ by gluing together $g$ and $\bar h_1$. Continuing this procedure we define $h\in\H^{\p}(\p\t\R_+)$ fulfilling (3.2), (3.3) and (3.4). Here we put $h(x,0)=(x,0)$ for all $x\in\p$.

Next we set $h':=h^{-1}f$, that is $f=hh'$. It follows that $h'$ also enjoys the properties (3.2), (3.3) and (3.4) with a suitably chosen sequence similar to (3.1).

Let $U_k= (\tilde a_k,\tilde b_k)$, where $\tilde a_k,\tilde
b_k\in(0,1)$, $k=1,2,\ldots$, are such that $\tilde a_{k-1}>\tilde
b_k>b_k>a_k>\tilde a_k$ for all $k$ 