($\tilde b_0=1$). Now in view of Corollaries 3.2 and 3.3 with $M=\p\t\R_+$ $h$ belongs to the commutator subgroup of the group $\H_{[U]}^{\p}(\p\t\R_+)$, where $U=\bigcup_k U_k$. More precisely $h=[\tilde h,\bar h]$ for $\tilde h,\bar h\in \H^{\p}_{[U]}(\p\t\R_+)$. It is easily seen that $\tilde h,\bar h\in \H^{\p}_c(\p\t\R_+)$. The same is true for $h'$. Thus $\H^{\p}_c(\p\t\R_+)$ is a perfect group. $\square$

\(1) Tsuboi gave another proof of the perfectness of $\H(\R_+)$ in [@tsu2]. He did not use [@ed-ki] in it.

\(2) Given a smooth manifold with boundary $M$ of dimension $\geq
2$, it is known that the group $\D(M)$ is perfect (see Rybicki [@Ryb98]; also Abe and Fukui [@AF01] by using a different method). For $n=1$ $\D(M)$ is not perfect. In particular, Fukui in [@Fuk80] calculated that $H_1(\D(\R_+))=\R$.

\(3) Let $\R^n_+=[0,\infty)\t\R^{n-1}$ be the half-space and $0\leq
s\leq\infty$. Let $\D_s(\R^n_+)$ be the compactly supported identity component of the subgroup of all elements of $\D(\R^n_+)$ which are $s$-tangent to the identity on $\p_{\R^n_+}$. Here $f$ is 0-tangent to id means that $f=\operatorname{id}$ on $\p_{\R^n_+}$. If $0\leq
s<\infty$ then $\D_s(\R^n_+)$ is not perfect. In fact, for any diffeomorphisms $f,g\in \D_s(\R^n_+)$ we have $$D^{s+1}(fg)(0)= D^{s+1}f(0)+D^{s+1}g(0),\quad    D^{s+1}f^{-1}(0) = -D^{s+1}f(0).$$ Therefore if we choose $h\in \D_s(\R^n_+)$ such that $D^{s+1}h(0)\not =0$, 