the above equalities yield that $h$ cannot be in the commutator subgroup.

Finally, let us indicate further perfectness result concerning homeomorphism groups. Let $M$ be a compact manifold with boundary $\p$. Let $\p=\p_i$, $i=1\ld k$, be the family of all connected components of the boundary $\p$ of $M$, that is $\p=\p_1\cup\ldots\cup\p_k$. Let $K=\{1\ld k\}$. For any $J\s K$ let $\H(M^o,J)$ denote all the elements of $\H(M^o)$ that can be joined with the identity by an isotopy which stabilizes near $\p_J$, where $\p_J:=\bigcup_{i\in J}\p_i$. In particular, $\H(M^o)=\H(M^o,\emptyset)$ and $\H_c(M^o)=\H(M^o,K)$. Then we have

[@MD] The groups $\H(M^o,J)$, where $J\s K$, are perfect.

For the proof, see also [@ry6]. The proof is no longer valid if we drop the assumption that $M^o$ is the interior of a manifold with boundary, e.g. if $M$ is the cylinder $\mathbb S^1\t\R$ with attached infinitely many handles.

On the simplicity of $\H_c(M)$
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The following result is related to Ling’s paper [@li].

Under the hypothesis of 1.1, there does not exist any fixed point free normal subgroup of $\H_c(M)$.

Suppose that $G$ is a fixed point free normal subgroup of $H_c(M)$. It follows that if $M$ has boundary then $\dim M\geq 2$. Choose a cover $\U\s\B$ such 