Orientation-preserving homeomorphisms of
Euclidean space are commutators

Let $A$ be the locally flat annulus co-bounded by $\Sigma_{1}$ and $\Sigma_{2}$. Choose an embedding $\iota_{A}\colon\thinspace\mathbb{S}^{n-1}\times[0,1]\to\mathbb{S}^{n}$ whose image is $A$ and such that $\iota_{A}|_{\mathbb{S}^{n-1}\times\{0\}}=\iota_{1}$. Let $\iota=\iota_{A}|_{\mathbb{S}^{n-1}\times\{1\}}$. Then $\tau=\iota^{-1}\circ\iota_{2}$ is an orientation-preserving homeomorphism of $\mathbb{S}^{n-1}$; hence, $\tau$ is isotopic to the identity. Choose an isotopy $H\colon\thinspace\mathbb{S}^{n-1}\times[0,1]\to\mathbb{S}^{n-1}$ such that $H(x,0)=x$ and $H(x,1)=\tau(x)$. Define $h\colon\thinspace\mathbb{S}^{n-1}\times[0,1]\to\mathbb{S}^{n-1}\times[0,1]$ by $h(x,t)=(H(x,t),t)$. It is readily checked that $h$ is a homeomorphism and that $\varphi:=\iota_{A}\circ h$ is the desired map. ∎

Let $\tau,\sigma\in\operatorname{Homeo}_{0}(\mathbb{S}^{n})$ be topologically loxodromic. Let $\{A_{n}\}_{n\in\mathbb{Z}}$ and $\{B_{n}\}_{n\in\mathbb{Z}}$ be sequences of locally flat annuli suited to $\tau$ and $\sigma$, respectively. Let $\Sigma_{A}$ be the component of $A_{0}$ such that $\partial A_{0}=\Sigma_{A}\cup\tau(\Sigma_{A})$, and similarly, define $\Sigma_{B}$ so that $\partial B_{0}=\Sigma_{B}\cup\sigma(\Sigma_{B})$.

Fix embeddings $\iota_{A},\iota_{B}\colon\thinspace\mathbb{S}^{n-1}\to\mathbb{S}^{n}$ such that the images of $\iota_{A}$ and $\iota_{B}$ are $\Sigma_{A}$ and $\Sigma_{B}$, respectively. Applying 2.4, we obtain embeddings $\varphi_{A},\varphi_{B}\colon\thinspace\mathbb{S}^{n-1}\times[0,1]\to\mathbb{S}^{n}$ whose images are $A_{0}$ and $B_{0}$, respectively, and such that $\varphi_{A}$ (resp., $\varphi_{B}$) is an isotopy between $\iota_{A}$ and $\tau\circ\iota_{A}$ (resp., $\iota_{B}$ and $\sigma\circ\iota_{B}$). Set $\varphi=\varphi_{A}\circ\varphi_{B}^{-1}$.

Define $h\colon\thinspace\mathbb{S}^{n}\to\mathbb{S}^{n}$ by $h(\sigma_{\pm})=\tau_{\pm}$ and $h(x)=(\tau^{n}\circ\varphi\circ\sigma^{-n})(x)$ for $x\in B_{n}$. It is now readily checked that $h$ is well-defined for the elements of $B_{n}\cap B_{n+1}$, establishing that $h$ is a well-defined orientation-preserving homeomorphism of $\mathbb{S}^{n}$. Now, for $x\in A_{n}$,

Hence, $\tau=h\circ\sigma\circ h^{-1}$. ∎

Viewing two given topologically loxodromic homeomorphisms in $G$ as homeomorphisms of $\mathbb{S}^{n}$ with a shared sink, a shared source, or both, the conjugating map constructed in 2.7 will preserve any shared source or sink, and hence the conjugation occurs in the appropriate group. ∎

Let $f\in\operatorname{Homeo}_{0}(\mathbb{S}^{n}\times\mathbb{R})$. Let $\Sigma_{0}=\mathbb{S}^{n}\times\{0\}$, and set $t_{0}=0$. Let $n_{0}\in\mathbb{N}$ such that $A_{0}:=\mathbb{S}^{n}\times[-n_{0},n_{0}]$ contains $\Sigma_{0}\cup f(\Sigma_{0})$ in its interior. Now, choose $t_{1},n_{1}\in\mathbb{N}$ such that $\Sigma_{1}=\mathbb{S}^{n}\times\{t_{1}\}$ satisfies $\Sigma_{1}\cup f(\Sigma_{1})$ is contained in the interior of $A_{1}:=\mathbb{S}^{n}\times[n_{0},n_{1}]$. Continuing in this fashion in both directions, we construct sequences $\{n_{k}\}_{k\in\mathbb{Z}}$ and $\{t_{k}\}_{k\in\mathbb{Z}}$ of integers such that, by setting $\Sigma_{k}=\mathbb{S}^{n}\times\{t_{k}\}$ and $A_{k}=\mathbb{S}^{n}\times[n_{k},n_{k+1}]$, we have $\Sigma_{k}\cup f(\Sigma_{k})$ is contained in the interior of $A_{k}$.

Now, let $g^{\prime}\in\operatorname{Homeo}_{0}(\mathbb{S}^{n}\times\mathbb{R})$ such that $g^{\prime}(A_{k})=A_{k+1}$. Note, by construction, $\bigcup_{k\in\mathbb{Z}}A_{k}$ is all of $\mathbb{S}^{n}\times\mathbb{R}$, and so $g^{\prime}$ is topologically loxodromic. As both $(g^{\prime}\circ f)(\Sigma_{k})$ and $\Sigma_{k+1}$ are locally flat annuli contained in the interior of $A_{k+1}$, we can choose $h_{k}\in\operatorname{Homeo}_{0}(\mathbb{S}^{n}\times\mathbb{R})$ that is supported in the interior of $A_{k+1}$ and satisfies $h_{k}(g^{\prime}(f(\Sigma_{k})))=\Sigma_{k+1}$. The sequence $\{h_{k}\}_{k\in\mathbb{Z}}$ consists of homeomorphisms with pairwise-disjoint supports, and hence, we can define $h=\prod_{k\in\mathbb{Z}}h_{k}$. Set $g=h\circ g^{\prime}$.

As the support of each $h_{k}$ is contained in the interior of $A_{k+1}$, we have that

in particular, $g$ is topologically loxodromic. Let $B_{k}$ be the locally flat annulus co-bounded by $\Sigma_{k}$ and $\Sigma_{k+1}$. Then, as $(g\circ f)(\Sigma_{k})=\Sigma_{k+1}$, we have $(g\circ f)(B_{k})=B_{k+1}$. Therefore, $g\circ f$ is topologically loxodromic. Moreover, $g$ and $g\circ f$ share the same sink, and hence they are conjugate by 2.8. This establishes that $f$ can be expressed as a commutator. ∎

Letting $\mathbb{S}^{0}$ be a singleton, the proof in the annulus case with $n=0$ shows that every element of $\operatorname{Homeo}_{0}(\mathbb{R})$ can be expressed as a commutator. We may now assume that $n>1$.

Fix $z\in\mathbb{S}^{n}$, and identify $\operatorname{Homeo}_{0}(\mathbb{R}^{n})$ with the stabilizer of $z$ in $\operatorname{Homeo}_{0}(\mathbb{S}^{n})$. Let $f$ be an element of $\operatorname{Homeo}_{0}(\mathbb{S}^{n})$ such that $f(z)=z$. We will show that $f$ can be expressed as a commutator of a pair of elements in $\operatorname{Homeo}_{0}(\mathbb{S}^{n})$ that each fix $z$. The identity homeomorphism is clearly a commutator, so we may assume that $f$ is not the identity, and therefore there exists $y\in\mathbb{S}^{n}$ such that $f(y)\neq y$. By continuity, there exists a locally flat ball $D_{0}$ centered at $y$ such that $f(D_{0})\cap D_{0}=\varnothing$. Choose a locally flat ball $D_{1}$ centered at $z$ that is disjoint from $D_{0}\cup f(D_{0})$. Again by continuity, by shrinking $D_{1}$ if necessary, we may assume that $f(D_{1})$ is also disjoint from $D_{0}\cup f(D_{0})$.

First, in $D_{1}$, we proceed identically as we did in the annulus case. Choose a locally flat annulus $A_{1}$ such that $\partial D_{1}\cup f(\partial D_{1})$ is contained in the interior of $A_{1}$ and such that $A_{1}$ is disjoint from $D_{0}\cup f(D_{0})$. Now choose a locally flat ball $D_{2}$ centered at $z$ such that $D_{2}\cup f(D_{2})$ is disjoint from $A_{1}$. Choose a locally flat annulus $A_{2}$ such that $A_{1}$ and $A_{2}$ share a boundary component and $\partial D_{2}\cup f(\partial D_{2})$ is contained in the interior of $A_{2}$. Continuing in this fashion, we build a sequence of locally flat annuli $\{A_{k}\}_{k\in\mathbb{N}}$ and locally flat balls $\{D_{k}\}_{k\in\mathbb{N}}$ such that $A_{k}$ and $A_{k+1}$ share a boundary, $\partial D_{k}\cup f(\partial D_{k})$ is contained in the interior of $A_{k}$, and $\bigcap_{k\in\mathbb{N}}D_{k}=\{z\}$. Note that the last condition is not a priori guaranteed, but at each stage we are free to choose $D_{k}$ to have radius less than $1/k$, forcing the intersection of the $D_{k}$ to be $\{z\}$.

Let $g_{0}^{\prime}$ be a topologically loxodromic homeomorphism of $\mathbb{S}^{n}$ that fixes $z$, that maps $A_{k+1}$ to $A_{k}$ for $k\in\mathbb{N}$, and that maps $f(\partial D_{1})$ to $\partial D_{0}$. For $k\leq 0$, let $A_{k}=(g_{0}^{\prime})^{\circ(n+1)}(A_{1})$. Then, $(g_{0}^{\prime}\circ f)(\partial D_{k+1})$ and $\partial D_{k}$ are locally flat spheres in the annulus $A_{k}$, and so we may choose $h_{k}\in\operatorname{Homeo}(\mathbb{S}^{n})$ supported in the interior of $A_{k}$ such that $h_{k}(g_{0}^{\prime}(f(\partial D_{k+1})))=\partial D_{k}$. As the $h_{k}$ have pairwise disjoint support, we can define $h=\prod_{k\in\mathbb{N}}h_{k}$. Let $g_{0}=h\circ g_{0}^{\prime}$. Then, for $k\in\mathbb{N}\cup\{0\}$, setting $T_{k}$ to be the locally flat annulus bounded by $\partial D_{k}$ and $\partial D_{k+1}$, we have $(g_{0}\circ f)(T_{k+1})=T_{k}$.

At this point, $g_{0}\circ f$ behaves like a topologically loxodromic homeomorphism when restricted to $D_{1}$, and so now we have to edit $g_{0}$ in the complement of $D_{1}$. This portion of the argument is a version of Tsuboi’s argument for spheres. Let $B_{0}=g_{0}(f(D_{0}))$, and note that $f(B_{0})\subset f(D_{0})$ and $g_{0}(f(B_{0}))\subset B_{0}$. Moreover, observe that $B_{0}\cup f(D_{0})\subset A_{0}$; hence, modifying $g_{0}$ in either $f(D_{0})$ or $B_{0}$ will not change the fact that $g_{0}$ is topologically loxodromic.

Fix $x\in B_{0}$. We may assume that $g_{0}(f(x))=x$; indeed, if not, then we may post compose $g_{0}$ with a homeomorphism of $\mathbb{S}^{n}$ supported in $B_{0}$ that maps $g_{0}(f(x))$ to $x$. And, as just noted above, this edited version of $g_{0}$ remains topologically loxodromic. We will now recursively edit $g_{0}$ in $f(D_{0})$ so that $g_{0}\circ f$ will be topologically loxodromic with $z$ its sink and $x$ its source.

Below, given a subset $X$, we let $X^{\mathrm{o}}$ denote its interior. Let $\Sigma_{0}=\partial B_{0}$, let $B_{0}^{\prime}=f(B_{0})$, let $B_{1}=g_{0}(B_{0}^{\prime})$, and let $\Sigma_{1}=\partial B_{1}$. Note $\Sigma_{0}\cap\Sigma_{1}=\varnothing$, and therefore $\Sigma_{0}$ and $\Sigma_{1}$ co-bound a locally flat annulus, which we label $T_{-1}$. Then, by continuity, we can choose a locally flat ball $B_{1}^{\prime}$ in the interior of $B_{0}^{\prime}$ centered at $f(x)$ such that $g_{0}(B_{1}^{\prime})\subset B_{1}^{\mathrm{o}}$. We can then choose $\sigma_{1}\in\operatorname{Homeo}(\mathbb{S}^{n})$ such that $\sigma_{1}$ is supported in $f(B_{0})$ and such that $\sigma_{1}(f(\Sigma_{1}))=\partial B_{1}^{\prime}$. Set $B_{2}=g_{0}(B_{1}^{\prime})$, and set $\Sigma_{2}=\partial B_{2}$. Now, set $T_{-2}$ to be the locally flat annulus co-bounded by $\Sigma_{1}$ and $\Sigma_{2}$, and set $g_{1}=g_{0}\circ\sigma_{1}$. Then, $(g_{1}\circ f)(T_{k})=T_{k-1}$ for all $k\in\mathbb{N}\cup\{0,-1\}$. Note that $B_{1}^{\prime}$ can be chosen arbitrarily small.

Continuing in this fashion, for each $m\in\mathbb{N}$, we obtain a locally flat annulus $T_{-m}$ and a homeomorphism $\sigma_{m}$ of $\mathbb{S}^{n}$ supported in a ball of radius $1/m$ centered at $f(x)$ such that

1. (1)

   $\mathbb{S}^{n}\smallsetminus\{x,z\}=\bigcup_{k\in\mathbb{Z}}T_{k}$,

2. (2)

   $(g_{m}\circ f)(T_{k})=T_{k-1}$ for every integer $k\geq-m$, and

3. (3)

   $g_{m}=g_{m-1}\circ\sigma_{m}=g_{0}\circ(\sigma_{1}\circ\sigma_{2}\circ\cdots\circ\sigma_{m})$.

These properties guarantee that $\lim_{m\to\infty}g_{m}$ exists, call it $g$, and that $(g\circ f)(T_{k})=T_{k-1}$ for every $k\in\mathbb{Z}$; hence, $g\circ f$ is topologically loxodromic. Moreover, $g$ is topologically loxodromic as $g$ agrees with $g_{0}$ outside of $B_{0}\cup f(D_{0})$; in particular, $g(A_{k})=g_{0}(A_{k})=A_{k+1}$ for all $k\in\mathbb{Z}$. And, as $g$ and $g\circ f$ share the same sink, they are conjugate in $\operatorname{Homeo}_{0}(\mathbb{R}^{n})$ by 2.8. Thus, $f$ can be expressed as a commutator in $\operatorname{Homeo}_{0}(\mathbb{R}^{n})$. ∎

Without loss of generality, it is enough to prove the statement with $p_{i}\in\mathbb{N}$ for each $i\in\{1,\ldots,r\}$, as we can always replace an element with its inverse in the decomposition to switch the sign of the exponent.

Fix $g\in G$. We have shown in the above proofs that there exist $f,h\in G$ such that $g=f\circ h$ and both $f$ and $h$ are topologically loxodromic. Applying the same techniques to $f$, we can write $f=f_{1}\circ f_{2}$ with both $f_{1}$ and $f_{2}$ topologically loxodromic. Therefore, continuing this splitting as many times as necessary, we can write $g=\prod_{i=1}^{r}f_{i}$ with $f_{i}$ topologically loxodromic. Now, every power of a topologically loxodromic homeomorphism is itself topologically loxodromic. Therefore, by 2.7 and 2.8, $f_{i}^{\circ p_{i}}$ is conjugate to $f_{i}$. Choose $h_{i}$ such that $f_{i}=h_{i}\circ f_{i}^{\circ p_{i}}\circ h_{i}^{-1}$. Then, setting $g_{i}=h_{i}\circ f_{i}\circ h_{i}^{-1}$, we have $f_{i}=g_{i}^{\circ p_{i}}$, and hence, $g=\prod_{i=1}^{r}g_{i}^{\circ p_{i}}$. ∎