Proof. This is an easy corollary of Lemma 2.4(3).

Proof. By [Reference Ding3, Theorem 3.4(ii)], we have $E(G;0)\le _B E([0,1];0)$ .

Let $f:[0,1]\to H$ be an embedding. By the uniformly continuity of f and $f^{-1}:f([0,1])\to [0,1]$ , we see that $E([0,1];0)\sim _B E(f([0,1]);0)\le _B E(H;0)$ . So $E_*(G)\le _B E_*(H)\times E(H;0)$ .

Finally, by [Reference Ding and Zheng5, Lemma 6.3], we obtain $E_*(H)\times E(H;0)\le _B E_*(H)$ .

Proof. If not, then there exist an $\varepsilon _0>0$ , a strictly increasing sequence $(j_k)$ , and $g_k,g_k'\in F_{n_{j_k}-1}$ for each k, such that

$$ \begin{align*}d_G(g_k,g_k')<2^{-(k+1)},\quad d_H^\infty(T_{n_{j_k}}(u_{n_{j_k}}^{-1}g_k),T_{n_{j_k}}(u_{n_{j_k}}^{-1}g_k'))\ge\varepsilon_0.\end{align*} $$

Since $d_G$ is right-invariant, we have $\lim _kg_k'g_k^{-1}=1_G$ . We shall inductively find a strictly increasing sequence of natural numbers $k_0<k_1<\cdots $ so that

$$ \begin{align*}d_G(g_{k_0}'\cdots g_{k_{p-1}}'g_{k_{p-1}}^{-1}\cdots g_{k_0}^{-1},g_{k_0}'\cdots g_{k_p}'g_{k_p}^{-1}\cdots g_{k_0}^{-1})<2^{-p}\end{align*} $$

for each integer $p>0$ . Put $k_0=0$ first. Assume that $k_0,\ldots ,k_{p-1}$ have been found. By the continuity of group operations, there is some $\delta>0$ such that, for $g,g'\in G$ with $d_G(1_G,g'g^{-1})<\delta $ , we have

$$ \begin{align*}d_G(g_{k_0}'\cdots g_{k_{p-1}}'1_Gg_{k_{p-1}}^{-1}\cdots g_{k_0}^{-1},g_{k_0}'\cdots g_{k_{p-1}}'g'g^{-1}g_{k_{p-1}}^{-1}\cdots g_{k_0}^{-1})<2^{-p}.\end{align*} $$

Then we can find a $k_p>k_{p-1}$ large enough so that $d_G(1_G,g_{k_p}'g_{k_p}^{-1})<\delta $ . This completes the induction. It follows that, for $m>p$ , we have

$$ \begin{align*}d_G(g_{k_0}'\cdots g_{k_p}'g_{k_p}^{-1}\cdots g_{k_0}^{-1},g_{k_0}'\cdots g_{k_m}'g_{k_m}^{-1}\cdots g_{k_0}^{-1})<\sum_{i=p+1}^m 2^{-i}<2^{-p}.\end{align*} $$

Since $d_G$ is complete, we get that

$$ \begin{align*}\lim_pg_{k_0}'\cdots g_{k_p}'g_{k_p}^{-1}\cdots g_{k_0}^{-1}\text{ converges.}\end{align*} $$

For each $n\in I$ , put

$$ \begin{align*}x(n)=\left\{\begin{array}{ll}u_n^{-1}g_{k_p}', & n=n_{j_{k_p}}, \\ u_n^{-1}, & \text{otherwise,}\end{array}\right.\quad y(n)=\left\{\begin{array}{ll}u_n^{-1}g_{k_p}, & n=n_{j_{k_p}}, \\ u_n^{-1}, & \text{otherwise.}\end{array}\right.\end{align*} $$

For any $n\ge n_{j_{k_0}}$ , let $p_n$ be the largest p such that $n_{j_{k_p}}\le n$ . Note that

$$ \begin{align*}(x\oplus w,y\oplus w)|_0^n=g_{k_0}'\cdots g_{k_{p_n}}'g_{k_{p_n}}^{-1}\cdots g_{k_0}^{-1},\end{align*} $$

so $(x\oplus w)E_*(G)(y\oplus w)$ . But for each $p\in \omega $ , if $k=k_p$ , then

$$ \begin{align*}\max_{l_{j_k}\le i\le m<l_{j_k+1}}d_H(\psi(x),\psi(y))|_{[i,m]}= d_H^\infty(T_{n_{j_k}}(u_{n_{j_k}}^{-1}g_k'),T_{n_{j_k}}(u_{n_{j_k}}^{-1}g_k))\ge\varepsilon_0. \end{align*} $$

So $\psi (x)E_*(H)\psi (y)$ fails, contradicting that $\psi $ is a reduction.

Proof. If $\psi (x') E_*(H)\psi (y')$ , then $\lim _p(\psi (x'),\psi (y'))|_0^{p}\text { converges}.$ In particular, $\lim _j(\psi ( x'),\psi (y'))|_0^{l_j-1}\text {converges}.$

Conversely, suppose $\lim _j(\psi (x'),\psi (y'))|_0^{l_j-1}$ converges. For any $q\in \omega $ , by Lemma 3.1, there is a $\delta _q\in \omega $ such that

$$ \begin{align*}\forall^\infty n\in I\,\forall g,g'\in F_{n-1}\,(d_G(g,g')<\delta_q\Rightarrow d_H^\infty(T_n(u_n^{-1}g),T_n(u_n^{-1}g'))<2^{-q}).\end{align*} $$

Since $\lim _j d_G(x(n_j),y(n_j))=0$ and $x(n_j),y(n_j)\in F_{n_j-1}$ for any integer $j>0$ , there exists a $j_0\in \omega $ such that

$$ \begin{align*}\forall j>j_0\, (d_H^\infty(T_{n_j}(u_{n_j}^{-1}x(n_j)),T_{n_j}(u_{n_j}^{-1}y(n_j)))<2^{-q}).\end{align*} $$

Then for any $j,k\in \omega $ with $j>j_0$ and $l_j\leq k<l_{j+1}$ , since $d_H$ is two-sided invariant, we have

$$ \begin{align*}\begin{array}{r@{\;}l} d_H((\psi(x'),\psi(y'))|_0^{l_j-1},(\psi(x'),\psi(y'))|_0^{k})&=d_H(\psi(x'),\psi(y'))|_{[l_j,k]} \\ &\leq d_H^\infty(T_{n_j}(x'(n_j)),T_{n_j}(y'(n_j))) \\ &<2^{-q}.\end{array}\end{align*} $$

So by the convergency of $\lim _j(\psi (x'),\psi (y'))|_0^{l_j-1}$ , we obtain that

$$ \begin{align*}\lim_k(\psi(x'),\psi(y'))|_0^{k}\text{ converges},\end{align*} $$

and hence $\psi (x') E_*(H)\psi (y')$ .

Proof. Suppose G is TSI, and let $d_G$ be a compatible complete two-sided invariant metric on G. Then we have $d_G(1_G,g_pg_p')=d_G(g_p^{-1},g_p')=d_G(1_G,g_p'g_p)$ , so $\lim _pg_pg_p'=1_G\iff \lim _pg_p'g_p=1_G$ .

On the other hand, fix an open neighborhood basis $(V_n)$ of $1_G$ . We claim that

$$ \begin{align*}\forall n\,\exists m_n\,\forall g\in G\,(gV_{m_n}g^{-1}\subseteq V_n).\end{align*} $$

If not, there exist an $n_0$ and two sequences $(g_p),(h_p)$ in G such that $\lim _ph_p=1_G$ , but $g_ph_pg_p^{-1}\notin V_{n_0}$ for each $p\in \omega $ . Put $g_p'=h_pg_p^{-1}$ . It is clear that $\lim _pg_p'g_p=1_G$ , but $g_pg_p'\nrightarrow 1_G$ , which is a contradiction.

Now we put $U_n=\bigcup _{g\in G}gV_{m_n}g^{-1}$ . Note that $U_n\subseteq V_n$ for each $n\in \omega $ . Therefore, $(U_n)$ is also an open neighborhood basis of $1_G$ with $gU_ng^{-1}=U_n$ for all $g\in G$ and $n\in \omega $ . By [Reference Gao8, Exercise 2.1.4], G admits a compatible two-sided invariant metric. Hence G is TSI (see [Reference Becker and Kechris2, Corollary 1.2.2]).

Proof. Assume toward a contradiction that $E(G)\leq _B E(H)$ . By [Reference Ding and Zheng5, Theorem 4.3], it suffices to consider the case that G is CLI. Let $d_G$ be a compatible complete right-invariant metric on G and $d_H$ a compatible complete two-sided invariant metric on H.

We use the notation defined earlier in this section.

Since G is not TSI, by Proposition 3.4, there are two sequences $(g_p),(g_p')$ of elements of G such that $\lim _pg^{\prime }_pg_p=1_G$ , but $g_pg_p'\nrightarrow 1_G$ . Since $d_G$ is right-invariant, $\lim _p d_G(g^{\prime }_p,g_p^{-1})=0$ . By transferring to a subsequence, we may assume that, there is a $\delta>0$ such that $\inf _p d_G(1_G,g_pg^{\prime }_p)>\delta $ .

Since $\bigcup _n F_n$ is dense, by perturbation, we may assume that $\{g_p,g_p':p\in \omega \}\subseteq \bigcup _n F_n$ . By Lemma 3.1 and Definition 3.2, we can find two sequence $(p(i))$ and $(q(i))$ of natural numbers so that

1. (i) $g_{p(i)},g_{p(i)}'\in F_{n_{q(i)}-1}$ ;

2. (ii) $d_H(S_{n_{q(i)+1}}(g^{\prime }_{p(i)}),S_{n_{q(i)+1}}(g_{p(i)}^{-1}))<2^{-i}$ ; and

3. (iii) for each $i\in \omega $ , $0<p(i)<q(i)<q(i)+1<p(i+1)$ .

For each $n\in I$ , define

$$ \begin{align*}x(n)=\left\{\begin{array}{ll} u_n^{-1} g_{p(i)}, & n=n_{q(i)}, \\ u_n^{-1}g^{\prime}_{p(i)}, & n=n_{q(i)+1}, \\ u_n^{-1},& \text{otherwise}, \end{array}\right. \quad y(n)=\left\{\begin{array}{ll} u_n^{-1}g_{p(i)}, & n=n_{q(i)}, \\ u_n^{-1}g_{p(i)}^{-1}, & n=n_{q(i)+1}, \\ u_n^{-1},& \text{otherwise}. \end{array}\right. \end{align*} $$

For $j\in \omega $ , by letting $h^{\prime }_j=S_{n_j}(u_{n_j}x(n_j))$ and $h_j=S_{n_j}(u_{n_j}y(n_j))$ , we have

$$ \begin{align*}(\psi(x),\psi(y))|_0^{l_{j+1}-1}=h^{\prime}_0\cdots h^{\prime}_jh_j^{-1}\cdots h_0^{-1}.\end{align*} $$

For $i\in \omega $ and $m>j>q(i)+1$ , it follows from (ii) and Lemma 2.4(1) that

$$ \begin{align*}d_H(h^{\prime}_j\cdots h^{\prime}_m,h_j\cdots h_m)<\sum_{k>i} 2^{-k}<2^{-i}.\end{align*} $$

Now by Lemma 2.4(2),

$$ \begin{align*}\lim_j(\psi(x),\psi(y))|_0^{l_{j+1}-1}=\lim_jh^{\prime}_0\cdots h^{\prime}_jh_j^{-1}\cdots h_0^{-1}\text{ converges}.\end{align*} $$

Note that $\lim _i d_G(g^{\prime }_{p(i)},g_{p(i)}^{-1})=0$ . So by Lemma 3.3, we have

$$ \begin{align*}\psi(x) E_*(H)\psi(y),\end{align*} $$

and hence $(x\oplus w) E_*(G)(y\oplus w)$ . In particular,

$$ \begin{align*}\lim_i(x\oplus w,y\oplus w)|_0^{n_{q(i)+1}}=\lim_i g_{p(0)}g^{\prime}_{p(0)}\cdots g_{p(i)}g^{\prime}_{p(i)}\text{ converges}.\end{align*} $$

Therefore, it follows that

$$ \begin{align*}\lim_i g_{p(i)}g^{\prime}_{p(i)}=\lim_i(g_{p(0)}g^{\prime}_{p(0)}\cdots g_{p(i-1)}g^{\prime}_{p(i-1)})^{-1}g_{p(0)}g^{\prime}_{p(0)}\cdots g_{p(i)}g^{\prime}_{p(i)}=1_G.\end{align*} $$

We obtain a contradiction.

Proof. Let $e_n=(0,\ldots ,0,\overset {n}1,0,\ldots )$ for each integer $n>0$ . Then $(e_n)$ is the canonical Schauder basis of $c_0$ . Note that $E(c_0,(e_n))={\mathbb {R}}^\omega /c_0$ . By applying [Reference Ding4, Lemma 4.2] to $E(c_0,(e_n))$ , we conclude the proof.

Proof. Suppose $E(G)\leq _B {\mathbb {R}}^\omega /c_0$ . Note that ${\mathbb {R}}^\omega /c_0<_B{\mathbb {R}}^\omega /\mathrm {cs}=E_*({\mathbb {R}})$ (see [Reference Ding4, Theorem 5.9(i)]). By Theorem 3.5, G is also TSI. Let $d_G$ be a compatible complete two-sided invariant metric on G. Put $V_k=\{g\in G:d_G(1_G,g)<2^{-k}\}$ .

Assume for contradiction that G is not non-archimedean. Then there is a $K\in \omega $ such that $V_K$ contains no open subgroups of G. It is clear that $V_k=V_k^{-1}$ and $\bigcup _mV_k^m$ is an open subgroup which is not contained in $V_K$ . So there exists $m_k>0$ with $V_k^{m_k}\not \subseteq V_K$ for each k. We can find $g_{k,0},\ldots ,g_{k,m_k-1}\in V_k$ such that $g_{k,0}\cdots g_{k,m_k-1}\notin V_K$ . Put $h_j=g_{k,i}$ for $j=\sum _{l<k}m_l+i$ with $i<m_k$ . Then we have $\lim _j h_j=1_G$ . By Cauchy Criterion,

$$ \begin{align*}\lim_jh_0\cdots h_j\text{ diverges}.\end{align*} $$

Assume that $E_*(G)\le _B{\mathbb {R}}^\omega /c_0$ . Let $(F_n)$ be a sequence of finite subsets of G such that

1. (i) $1_G,h_0\in F_n=F_n^{-1}$ ;

2. (ii) $F_{n-1}^{n}\subseteq F_n$ ; and

3. (iii) for $n>0,h_n\in F_{n-1}$ .

Denote by E the restriction of $E_*(G)$ on $\prod _nF_n$ . By Lemma 4.1, there exist an infinite set $I\subseteq \omega $ and a $w\in \prod _{n\notin I}F_n$ such that $E|_I^w\le _A{\mathbb {R}}^\omega /c_0$ . So there are natural numbers $0=n_0<n_1<n_2<\cdots $ with $I=\{n_j:j\in \omega \}$ , $0=l_0<l_1<l_2<\cdots $ , maps $T_{n_j}:F_{n_j}\to {\mathbb {R}}^{l_{j+1}-l_j}$ , and $\psi :\prod _{n\in I}F_n\to {\mathbb {R}}^\omega $ with

$$ \begin{align*}\psi(x)=T_{n_0}(x(n_0))^\smallfrown T_{n_1}(x(n_1))^\smallfrown T_{n_2}(x(n_2))^\smallfrown\cdots,\end{align*} $$

such that $\psi $ is an additive reduction of $E|_I^w$ to ${\mathbb {R}}^\omega /c_0$ .

Put $u_{n_0}=1_G$ , and for $j>0$ , let

$$ \begin{align*}u_{n_j}=w(n_{j-1}+1)\cdots w(n_j-1).\end{align*} $$

Assume again for contradiction that there are $\varepsilon _0>0$ and natural numbers $0<j(0)<j(1)<\cdots <j(p)<\cdots $ such that

$$ \begin{align*}\max_{0\leq k <l_{j(p)+1}-l_{j(p)}}|T_{n_{j(p)}}(u^{-1}_{n_{j(p)}}h_{j(p)})(k)-T_{n_{j(p)}}(u^{-1}_{n_{j(p)}})(k)|\geq\varepsilon_0.\end{align*} $$

By $\lim _p h_{j(p)}=1_G$ , we can find natural numbers $p_0<p_1<\cdots <p_i<\cdots $ so that $d_G(1_G,h_{j(p_i)})<2^{-i}$ for each $i\in \omega $ . Thus $(h_{j(p_0)}\cdots h_{j(p_i)})$ is $d_G$ -Cauchy, so $\lim _ih_{j(p_0)}\cdots h_{j(p_i)}$ converges. For each $n\in I$ , put $x_0(n)=u^{-1}_{n}$ and

$$ \begin{align*}y_0'(n)=\left\{\begin{array}{ll} u^{-1}_{n}h_{j(p_i)}, & n=n_{j(p_i)}, \\ u^{-1}_n, & \text{otherwise.}\end{array}\right.\end{align*} $$

For any $n> n_{j(p_0)}$ , let $i_n$ be the largest i with $n_{j(p_{i})}\leq n$ . Then

$$ \begin{align*}(y_0'\oplus w,x_0\oplus w)|_0^n=h_{j(p_0)}\cdots h_{j(p_{i_n})},\end{align*} $$

and thus $(y_0'\oplus w) E_*(G)(x_0\oplus w)$ . Note that for all $i\in \omega $ ,

$$ \begin{align*}\max_{l_{j(p_i)}\leq k<l_{j(p_i)+1}}|\psi(y_0')(k)-\psi(x_0)(k)|\geq\varepsilon_0.\end{align*} $$

So $\psi (y_0'){\mathbb {R}}^\omega /c_0 \psi (x_0)$ fails, which is a contradiction.

Therefore, for all $\varepsilon>0$ , we have

$$ \begin{align*}\forall^\infty j\,\forall k\in[0,l_{j+1}-l_j)\,(|T_{n_j}(u^{-1}_{n_j}h_j)(k)-T_{n_j}(u^{-1}_{n_j})(k)|<\varepsilon).\end{align*} $$

For each $n\in I$ , put $y_0(n)=u^{-1}_nh_j$ , where $n=n_j$ . Then we have

$$ \begin{align*}\psi(y_0){\mathbb{R}}^\omega/c_0\psi(x_0),\end{align*} $$

and hence $(y_0\oplus w) E_*(G)(x_0\oplus w)$ . In particular,

$$ \begin{align*}\lim_j(y_0\oplus w,x_0\oplus w)|_0^{n_j}=\lim_jh_0\cdots h_j\text{ converges}.\end{align*} $$

This leads to a contradiction.

Proof. $(1)\Rightarrow (2)$ follows from [Reference Ding and Zheng5, Theorem 3.5]. $(2)\Rightarrow (3)$ follows from [Reference Gao8, Lemma 8.5.3]. $(3)\Rightarrow (1)$ follows from Theorem 4.2.

Again by [Reference Ding and Zheng5, Theorem 3.5], we see that $E(G)\sim _B E_0^\omega $ iff G is TSI uncountable non-archimedean.

Question 4.4. For any Polish group G, does it hold that

$$ \begin{align*}E(G)<_BE({\mathbb{R}})\iff E(G)\le_BE_0^\omega?\end{align*} $$

Proof. (1) Let V be an unenclosed set of G. We have

$$ \begin{align*}\forall g\in G\,(g\neq 1_G\Rightarrow \exists m\,(g^m\notin V)).\end{align*} $$

Thus V contains no non-trivial subgroups of G, so G is NSS.

(2) By (1), we only need to prove another direction. Suppose that G is NSS and locally compact. Let $V\ni 1_G$ be an open subset of G such that $\overline {V}$ is compact and contains no non-trivial subgroups of G. Then

$$ \begin{align*}\forall g\in \overline{V}\,(g\neq 1_G\Rightarrow \exists m\,(g^m\notin \overline{V})).\end{align*} $$

We claim that V is an unenclosed set of G. Fix a $(g_n)\in G^\omega $ with $ g_n\nrightarrow 1_G$ . By the definition of unenclosed set, it suffices to consider the case that $(g_n)\in V^\omega $ . Since $\overline {V}$ is compact, there exist a subsequence $(g_{n_i})$ of $(g_n)$ and an element $1_G\neq h\in \overline {V}$ such that $\lim _i g_{n_i}=h$ . Then we can find an $m\in \omega $ with $h^m\notin \overline {V}$ . By the continuity of group operations, there are $i_0<i_1<\cdots <i_{m-1}$ such that $g_{n_{i_0}}g_{n_{i_1}}\cdots g_{n_{i_{m-1}}}\notin \overline {V} $ . Therefore G is strongly NSS.

Proof. Let $d_G,d_{H_m}\leq 1$ be compatible complete two-sided invariant metrics on G and $H_m$ respectively. A compatible complete two-sided invariant metric $d_H$ on H is defined as

$$ \begin{align*}d_H(h,h')=\sum_{m=0}^\infty 2^{-m}d_{H_m}(h(m),h'(m))\end{align*} $$

for $h,h'\in H$ . Assume that $E_*(G)\le _BE_*(H)$ .

We use the notation defined in the arguments before Lemma 3.1 and before Lemma 3.3.

For $m\in \omega $ , let $\pi _m:\prod _n H_n\to H_m$ be the canonical projection map, so $\pi _m(h)=h(m)$ for each h. It is clear that every $\pi _m$ is a continuous homomorphism.

It is worth noting that

$$ \begin{align*}\forall h,h'\in H\, (d_{H_m}(\pi_m(h),\pi_m(h'))\leq 2^md_H(h,h')).\end{align*} $$

For $m\in \omega $ , let $X_m$ be the set of all $g\in \bigcup _n F_n$ such that:

$$ \begin{align*}\forall\delta>0\, \exists j_\delta\,(g\in F_{n_{j_\delta}-1}\wedge \forall j'> j\geq j_\delta\,(d_{H_m}(1_{H_m},\pi_m(S_{n_j}(g)S_{n_{j'}}(g^{-1})))<\delta)).\end{align*} $$

Note that $S_{n_j}(1_G)=1_H$ for all $j\in \omega $ , so $1_G\in X_m$ .

Claim 1. For any $m\in \omega $ , there is an $r_m>0$ such that

$$ \begin{align*}\forall g,g'\in\bigcup_n F_n\,((d_G(g,g')<r_m\wedge g\in X_m)\Rightarrow g'\in X_m).\end{align*} $$

Proof of Claim 1.

If not, there are an $m_0\in \omega $ and two sequences $(g_i),(g^{\prime }_i)\in \bigcup _n F_n$ such that

1. (i) $d_G(g_i,g^{\prime }_i)<2^{-i}$ ; and

2. (ii) $g_i\in X_{m_0}$ and $g^{\prime }_i\notin X_{m_0}$ .

Note that $H_{m_0}$ is strongly NSS. Let open set $V\ni 1_{H_{m_0}}$ be an unenclosed set of $H_{m_0}$ . Put $p(-1,2M_{-1}-1)=0$ . For any $i\in \omega $ , we will find an $M_i\in \omega $ and natural numbers $0<p(i,0)<p(i,1)<\cdots <p(i,2M_i-1)$ so that

1. (1) $p(i-1,2M_{i-1}-1)<p(i,0)$ ;

2. (2) $g_i,g_i'\in F_{n_{p(i,0)-1}}$ ;

3. (3) for $i\in \omega $ ,

   $$ \begin{align*}\pi_{m_0}(S_{n_{p(i,0)}}(g^{\prime}_i)S_{n_{p(i,1)}}((g^{\prime}_i)^{-1})\cdots S_{n_{p(i,2M_i-2)}}(g^{\prime}_i)S_{n_{p(i,2M_i-1)}}((g^{\prime}_i)^{-1}))\notin V;\end{align*} $$

4. (4) for $i\in \omega $ ,

   $$ \begin{align*}\sum_{l=0}^{M_i-1}d_{H_{m_0}}(1_{H_{m_0}},\pi_{m_0}(S_{n_{p(i,2l)}}(g_i)S_{n_{p(i,2l+1)}}({g_i^{-1}})))<2^{-i}.\end{align*} $$

Let us begin with $i=0$ . Since $g^{\prime }_0\notin X_{m_0}$ , there are a $\delta _0>0$ and natural numbers $p(0)<p(1)<\cdots <p(j)<\cdots $ such that $g^{\prime }_0\in F_{n_{p(0)}-1}$ and for each $j\in \omega $ ,

$$ \begin{align*}d_{H_{m_0}}(1_{H_{m_0}},\pi_{m_0}(S_{n_{p(2j)}}(g^{\prime}_0)S_{n_{p(2j+1)}}((g_0')^{-1})))\geq\delta_0.\end{align*} $$

By the definition of $X_{m_0}$ and $g_0\in X_{m_0}$ , there exists a strictly increasing sequence $(j_k)$ of natural numbers such that $g_0\in F_{n_{p(j_0)}-1}$ and for each $k\in \omega $ ,

$$ \begin{align*}d_{H_{m_0}}(1_{H_{m_0}},\pi_{m_0}(S_{n_{p(2j_k)}}(g_0)S_{n_{p(2j_k+1)}}({g_0^{-1}})))<2^{-(k+1)}.\end{align*} $$

Let $h\in H_{m_0}^\omega $ be such that

$$ \begin{align*}h(k)=\pi_{m_0}(S_{n_{p(2j_{k})}}(g^{\prime}_0)S_{n_{p(2j_{k}+1)}}((g^{\prime}_0)^{-1}))\end{align*} $$

for each $k\in \omega $ . Then $ h(k)\nrightarrow 1_{H_{m_0}}$ . Since V is an unenclosed set of $H_{m_0}$ , there are finitely many natural numbers $k_0<k_1<\cdots <k_{q}$ such that $h({k_0})\cdots h({k_{q}})\notin V$ , i.e.,

$$ \begin{align*}\pi_{m_0}(S_{n_{p(2j_{k_0})}}(g^{\prime}_0)S_{n_{p(2j_{k_0}+1)}}((g^{\prime}_0)^{-1})\cdots S_{n_{p(2j_{k_{q}})}}(g^{\prime}_0)S_{n_{p(2j_{k_{q}}+1)}}((g^{\prime}_0)^{-1}))\notin V.\end{align*} $$

Now we put $M_0=q+1$ , $p(0,2l)=p(2j_{k_l})$ , and $p(0,2l+1)=p(2j_{k_l}+1)$ for all $l<q+1$ . Then we have $g_0,g_0'\in F_{n_{p(j_0)}-1}\subseteq F_{n_{p(0,0)}-1}$ , and

$$ \begin{align*}\sum_{l=0}^{q}d_{H_{m_0}}(1_{H_{m_0}},\pi_{m_0}(S_{n_{p(0,2l)}}(g_0)S_{n_{p(0,2l+1)}}({g_0^{-1}})))<\sum_{l=0}^{q}2^{-(k_l+1)}<1.\end{align*} $$

We can see that clauses (1)–(4) hold for $i=0$ .

Now assume that $M_0,\ldots ,M_{i}$ and $p(0,0)<\cdots <p(0,2M_0-1)<\cdots <p(i,0)<\cdots <p(i,2M_{i}-1)$ have been defined. Using similar arguments as those presented in the preceding paragraph, pick a $p'>p(i,2M_{i}-1)$ with $g_{i+1},g_{i+1}'\in F_{n_{p'}-1}$ , then we can find an $M_{i+1}\in \omega $ and natural numbers $p'\le p(i+1,0)<\cdots <p(i+1,2M_{i+1}-1)$ so that clauses (1)–(4) hold.

For each $n\in I$ , define

$$ \begin{align*}x(n)=\left\{\begin{array}{ll} u_n^{-1} g^{\prime}_{i}, & n=n_{p(i,2k)},0\leq k<M_i , \\ u_n^{-1}{(g_i')}^{-1}, & n=n_{p(i,2k+1)},0\leq k<M_i , \\ u_n^{-1},& \text{otherwise}, \end{array}\right.\end{align*} $$

$$ \begin{align*}y(n)=\left\{\begin{array}{ll} u_n^{-1}g_i, & n=n_{p(i,2k)},0\leq k<M_i, \\ u_n^{-1}{g_i^{-1}}, & n=n_{p(i,2k+1)},0\leq k<M_i, \\ u_n^{-1},& \text{otherwise}. \end{array}\right. \end{align*} $$

Then for $q\in \omega $ , the following equality holds:

$$ \begin{align*}(x\oplus w,y\oplus w)|_0^q=\left\{\begin{array}{ll} g^{\prime}_ig_i^{-1}, & n_{p(i,2k)}\leq q<n_{p(i,2k+1)},0\leq k<M_i, \\ 1_G,& \text{otherwise}. \end{array}\right.\end{align*} $$

By (i), we have

$$ \begin{align*}\lim_q(x\oplus w,y\oplus w)|_0^q=1_G.\end{align*} $$

So $(x\oplus w) E_*(G)(y\oplus w)$ , and thus $\psi (x)E_*(H)\psi (y)$ holds. It follows from Lemma 2.4(2) that

$$ \begin{align*}\lim_id_H(\psi(x),\psi(y))|_{[l_{p(i,0)},\, l_{p(i,2M_i-1)+1})}=0.\end{align*} $$

For each $j\in \omega $ , let $h_j'=S_{n_j}(u_{n_j}x(n_j))$ and $h_j=S_{n_j}(u_{n_j}y(n_j))$ . Note that $S_{n_j}(1_G)=1_H$ for all $j>0$ , so

$$ \begin{align*} (\psi(x),\psi(y))|_{l_{p(i,0)}}^{l_{p(i,2M_i-1)+1}-1} &=h^{\prime}_{p(i,0)}h^{\prime}_{p(i,1)}\cdots h^{\prime}_{p(i,2M_i-1)}(h_{p(i,0)}h_{p(i,1)}\cdots h_{p(i,2M_i-1)})^{-1}.\end{align*} $$

Now we have

$$ \begin{align*}\lim_id_H(h^{\prime}_{p(i,0)}h^{\prime}_{p(i,1)}\cdots h^{\prime}_{p(i,2M_i-1)},h_{p(i,0)}h_{p(i,1)}\cdots h_{p(i,2M_i-1)})=0,\end{align*} $$

and hence

$$ \begin{align*}\lim_i h^{\prime}_{p(i,0)}h^{\prime}_{p(i,1)}\cdots h^{\prime}_{p(i,2M_i-1)}(h_{p(i,0)}h_{p(i,1)}\cdots h_{p(i,2M_i-1)})^{-1}=1_H.\end{align*} $$

Let $V_1\ni 1_{H_{m_0}}$ be an open subset of $H_{m_0}$ with $V_1^2\subseteq V$ . Since $\pi _{m_0}$ is a continuous homomorphism, there exists an $i_0\in \omega $ such that

$$ \begin{align*}\pi_{m_0}(h^{\prime}_{p(i,0)}h^{\prime}_{p(i,1)}\cdots h^{\prime}_{p(i,2M_i-1)})\pi_{m_0}(h_{p(i,0)}h_{p(i,1)}\cdots h_{p(i,2M_i-1)})^{-1}\in V_1\end{align*} $$

holds for any $i>i_0$ . By $(4)$ , we have

$$ \begin{align*}\lim_i d_{H_{m_0}}(1_{H_{m_0}},\pi_{m_0}(h_{p(i,0)}h_{p(i,1)}\cdots h_{p(i,2M_i-1)}))\leq\lim_i 2^{-i}=0.\end{align*} $$

Therefore, for i large enough,

$$ \begin{align*}\pi_{m_0}(h_{p(i,0)}h_{p(i,1)}\cdots h_{p(i,2M_i-1)})\in V_1.\end{align*} $$

It follows that

$$ \begin{align*}\pi_{m_0}(h^{\prime}_{p(i,0)}h^{\prime}_{p(i,1)}\cdots h^{\prime}_{p(i,2M_i-1)})\in V,\end{align*} $$

i.e.,

$$ \begin{align*}\pi_{m_0}(S_{n_{p(i,0)}}(g^{\prime}_i)S_{n_{p(i,1)}}((g^{\prime}_i)^{-1})\cdots S_{n_{p(i,2M_i-2)}}(g^{\prime}_i)S_{n_{p(i,2M_i-1)}}((g^{\prime}_i)^{-1}))\in V,\end{align*} $$

contradicting (3).

For any $m\in \omega $ , let $V_m=\{g\in G:d_G(1_G,g)<r_m\}$ and $W_m=\bigcup _iV_m^i$ . Note that $V_m^{-1}=V_m$ . It is clear that $W_m$ is an open, and thus a clopen subgroup of G. We claim that $W_m\cap \bigcup _n F_n\subseteq X_m$ . For $i=0$ , note that $1_G\in X_m$ , so Claim 1 gives that $V_m\cap \bigcup _n F_n\subseteq X_m$ . Assume that $V_m^i\cap \bigcup _n F_n\subseteq X_m$ . Let $h\in V_m^{i+1}\cap \bigcup _n F_n$ with $h=gg'$ , where $g\in V_m^i$ and $g'\in V_m$ . Note that $\bigcup _n F_n$ is a dense subgroup of G. We can find $\hat g,\hat g'\in \bigcup _n F_n$ such that $d_G(h,\hat g\hat g')<r_m$ with $\hat g\in V_m^i$ and $\hat g'\in V_m$ . Since $d_G(\hat g,\hat g\hat g')=d_G(1_G,\hat g')<r_m$ and $\hat g\in X_m$ , we have $\hat g\hat g'\in X_m$ , and thus $h\in X_m$ . So $W_m\cap \bigcup _n F_n\subseteq X_m$ .

Let $g\in W_m\cap \bigcup _n F_n\subseteq X_m$ . Pick a $j_g\in \omega $ with $g\in F_{n_{j_g}-1}$ . For any $\delta>0$ , by the definition of $X_m$ , there exists a $j_\delta \geq j_g$ such that

$$ \begin{align*}\forall j'> j\geq j_{\delta}\,(d_{H_m}(1_{H_m},\pi_m(S_{n_j}(g)S_{n_{j'}}(g^{-1})))<\delta).\end{align*} $$

Let $j,k\in \omega $ with $j_\delta <j<k$ . Fix a $k'>k$ . Since $\pi _{m}$ is a homomorphism, we have

$$ \begin{align*}d_{H_m}(1_{H_m},\pi_m(S_{n_j}(g))\pi_m(S_{n_{k'}}(g^{-1})))<\delta,\end{align*} $$

$$ \begin{align*}d_{H_m}(1_{H_m},\pi_m(S_{n_k}(g))\pi_m(S_{n_{k'}}(g^{-1})))<\delta.\end{align*} $$

So $d_{H_m}(\pi _m(S_{n_j}(g)),\pi _m(S_{n_k}(g)))<2\delta .$ Thus $(\pi _m(S_{n_{j}}(g)))$ is a Cauchy sequence in $H_m$ . By the completeness of $d_{H_m}$ , we can define

$$ \begin{align*}\pi^S_m(g)=\lim_{j}\pi_m(S_{n_{j}}(g))\in H_{m}.\end{align*} $$

Note that for $g,g'\in F_n$ with $n<n_j$ ,

$$ \begin{align*}d_H(S_{n_j}(g),S_{n_j}(g'))\le d_H^\infty(T_{n_j}(u_{n_j}^{-1}g),T_{n_j}(u_{n_j}^{-1}g')),\end{align*} $$

and

$$ \begin{align*}\forall h,h'\in H\, (d_{H_m}(\pi_m(h),\pi_m(h'))\leq 2^md_H(h,h')).\end{align*} $$

By Lemma 3.1, $\pi ^S_m$ is uniformly continuous on $W_m\cap \bigcup _n F_n$ , which can be uniquely extended to a uniformly continuous map from $W_m$ to $H_m$ , still denoted by $\pi ^S_m$ . Put $W=\bigcap _m W_m$ , for any $g\in W$ , we let

$$ \begin{align*}S(g)=(\pi^S_0(g),\pi^S_1(g),\ldots,\pi^S_m(g),\ldots).\end{align*} $$

We claim that $S(g)\in H$ . Indeed, fix a $g\in W$ and an arbitrary neighborhood O of $S(g)$ , then there are $\varepsilon>0$ and natural number L such that

$$ \begin{align*}\left\{h\in \prod_m H_m: \forall m\leq L\, (d_{H_m}(\pi_m(h),\pi^S_m(g))<\varepsilon)\right\}\subseteq O.\end{align*} $$

Clearly $g\in \bigcap _{m\leq L}W_m$ . Since $\bigcup _n F_n$ is dense in G, by the continuity of $\pi ^S_m$ , we can find some $g'\in \bigcap _{m\leq L}W_m\cap \bigcup _n F_n$ so that $d_{H_m}(\pi ^S_m(g'),\pi ^S_m(g))<\varepsilon $ for $m\leq L$ . Note that $\pi ^S_m(g')=\lim _{j}\pi _m(S_{n_{j}}(g'))\in H_{m}$ holds for any $m\leq L$ . So there is a $j'\in \omega $ such that $S_{n_{j'}}(g')\in H\cap O$ . Thus $S(g)\in H$ by the fact that H is closed in $\prod _m H_m$ . The map S is a continuous map from W to H.

It is worth noting that $S(1_G)=1_H$ and W is a closed subgroup of G.

Now we prove clause (ii) first:

Proof of Claim 2.

Let $x,y\in W^\omega $ . Note that $xE_*(G)y\iff xE_*(W)y$ as W is a closed subgroup of G. By Proposition 2.8, we only need to show that

$$ \begin{align*}x E_*(G)y\iff S^\omega(x)E_*(H)S^\omega(y)\end{align*} $$

holds whenever $\lim _p d_G(x(p),y(p))=0$ . Since $\bigcup _n F_n$ is dense, by the definitions of $d_H$ and S, for each $p\in \omega $ , we can find a sufficiently large $j(p)\in \omega $ and $g_p,g^{\prime }_p\in \bigcup _n F_n $ so that

1. (1) $0<j(0)<j(1)<\cdots j(p)<\cdots $ ;

2. (2) $g_p,g^{\prime }_p\in F_{n_{j(p)}-1}\cap \bigcap _{m\le p} W_m$ ;

3. (3) $d_G(x(p),g^{\prime }_p)<2^{-p}$ and $d_G(y(p),g_p)<2^{-p}$ ; and

4. (4) for all $m\le p$ ,

   $$ \begin{align*}d_{H_m}(\pi_m(S(x(p))),\pi_m(S_{n_{j(p)}}(g^{\prime}_p)))<2^{-(p+1)},\end{align*} $$
   $$ \begin{align*}d_{H_m}(\pi_m(S(y(p))),\pi_m(S_{n_{j(p)}}(g_p)))<2^{-(p+1)}.\end{align*} $$

Now we define, for each $n\in I$ ,

$$ \begin{align*}\hat x(n)=\left\{\begin{array}{ll} u_n^{-1}g^{\prime}_p, & n=n_{j(p)}, \\ u_n^{-1}, & \text{otherwise,} \end{array}\right. \quad \hat y(n)=\left\{\begin{array}{ll} u_n^{-1}g_p, & n=n_{j(p)}, \\ u_n^{-1}, & \text{otherwise.} \end{array}\right.\end{align*} $$

For any $i<k$ , by (3) and Lemma 2.4(1),

$$ \begin{align*}d_G((x,y)|_i^k,g^{\prime}_{i}\cdots g^{\prime}_{k}{g_k^{-1}}\cdots {g_i^{-1}})<\sum_{q=i}^{k}2^{-q+1}<2^{-i+2}.\end{align*} $$

Then by Lemma 2.4(2),

$$ \begin{align*}xE_*(G)y\iff\lim_pg^{\prime}_{0}\cdots g^{\prime}_p{g_p^{-1}}\cdots {g_0^{-1}}\text{ converges}.\end{align*} $$

For any $k>n_{j(0)}$ , let $p_k$ be the largest p with $n_{j(p)}\leq k$ . Note that

$$ \begin{align*}(\hat x\oplus w,\hat y\oplus w)|_0^k=g^{\prime}_{0}\cdots g^{\prime}_{p_k}{g_{p_k}^{-1}}\cdots {g_0^{-1}}.\end{align*} $$

It follow that

$$ \begin{align*}x E_*(G)y\iff(\hat x\oplus w) E_*(G)(\hat y\oplus w)\iff\psi(\hat x)E_*(H)\psi(\hat y).\end{align*} $$

By (4), we have

$$ \begin{align*}\begin{array}{l@{\;}l}d_H(S(x(p)),S_{n_{j(p)}}(g^{\prime}_p))&=\sum_{m=0}^{\infty}2^{-m}d_{H_m}(\pi_m(S(x(p))),\pi_m(S_{n_{j(p)}}(g^{\prime}_p))) \\ &<\sum_{m=0}^{p}2^{-(m+p+1)}+\sum_{m=p+1}^{\infty}2^{-m} \\ &<2^{-(p-1)},\end{array}\end{align*} $$

$$ \begin{align*}\begin{array}{l@{\;}l}d_H(S(y(p)),S_{n_{j(p)}}(g_p))&=\sum_{m=0}^{\infty}2^{-m}d_{H_m}(\pi_m(S(y(p))),\pi_m(S_{n_{j(p)}}(g_p))) \\ &<\sum_{m=0}^{p}2^{-(m+p+1)}+\sum_{m=p+1}^{\infty}2^{-m} \\ &<2^{-(p-1)}.\end{array}\end{align*} $$

Note that $S_{n_j}(1_G)=1_H$ for all $j\in \omega $ , by similar arguments as above, we see that

$$ \begin{align*}\begin{array}{ll}& S^\omega(x)E_*(H)S^\omega(y) \\ \iff & \lim_p S_{n_{j(0)}}(g^{\prime}_0)\cdots S_{n_{j(p)}}(g^{\prime}_p)S_{n_{j(p)}}(g_p)^{-1}\cdots S_{n_{j(0)}}(g_0)^{-1}\text{ converges.}\end{array}\end{align*} $$

For any $i> j(0)$ , let $p_{i}$ be the largest p with $j(p)\leq i$ . Then we have

$$ \begin{align*}(\psi(\hat x),\psi(\hat y))|_0^{l_{i+1}-1}=S_{n_{j(0)}}(g^{\prime}_0)\cdots S_{n_{j(p_{i})}}(g^{\prime}_{p_{i}})S_{n_{j(p_{i})}}(g_{p_{i}})^{-1}\cdots S_{n_{j(0)}}(g_0)^{-1}.\end{align*} $$

This implies that

$$ \begin{align*}\lim_i(\psi(\hat x),\psi(\hat y))|_0^{l_{i+1}-1}\text{ converges}\iff S^\omega(x)E_*(H)S^\omega(y).\end{align*} $$

Note that $\lim _pd_G(g^{\prime }_p,g_p)=0$ , it follows from Lemma 3.3 that

$$ \begin{align*}\lim_i(\psi(\hat x),\psi(\hat y))|_0^{l_{i+1}-1}\text{ converges}\iff \psi(\hat x) E_*(H)\psi(\hat y).\end{align*} $$

So $S^\omega (x)E_*(H)S^\omega (y)\iff \psi (\hat x) E_*(H)\psi (\hat y)\iff xE_*(G)y.$

Subsequently, we prove clause (i) as follows:

Claim 3. For $m\in \omega $ , if $x,y\in W_m^\omega $ , then

$$ \begin{align*}x E_*(G)y\Longrightarrow(\pi^S_m)^\omega(x)E_*(H_m)(\pi^S_m)^\omega(y).\end{align*} $$

Proof of Claim 3.

Let $x,y\in W_m^\omega $ with $x E_*(G)y$ . Then for each $p\in \omega $ , we can find a $j(p)\in \omega $ and $g_p,g_p'\in W_m\cap \bigcup _n F_n$ so that

1. (1) $0<j(0)<j(1)<\cdots j(p)<\cdots $ ;

2. (2) $g_p,g^{\prime }_p\in F_{n_{j(p)}-1}\cap W_m$ ;

3. (3) $d_G(x(p),g^{\prime }_p)<2^{-p}$ and $d_G(y(p),g_p)<2^{-p}$ ; and

4. (4) we have

   $$ \begin{align*}d_{H_m}(\pi^S_m(x(p)),\pi_m(S_{n_{j(p)}}(g^{\prime}_p)))<2^{-p},\end{align*} $$
   $$ \begin{align*}d_{H_m}(\pi^S_m(y(p)),\pi_m(S_{n_{j(p)}}(g_p)))<2^{-p}.\end{align*} $$

Now define, for each $n\in I$ ,

$$ \begin{align*}\hat x(n)=\left\{\begin{array}{ll} u_n^{-1}g^{\prime}_p, & n=n_{j(p)}, \\ u_n^{-1}, & \text{otherwise,} \end{array}\right. \quad \hat y(n)=\left\{\begin{array}{ll} u_n^{-1}g_p, & n=n_{j(p)}, \\ u_n^{-1}, & \text{otherwise.} \end{array}\right.\end{align*} $$

Following a similar approach to the proof of Claim 2, we obtain

$$ \begin{align*}x E_*(G)y\iff(\hat x\oplus w) E_*(G)(\hat y\oplus w)\iff\psi(\hat x)E_*(H)\psi(\hat y).\end{align*} $$

So we have

$$ \begin{align*}\lim_p(\psi(\hat x),\psi(\hat y))|_0^{l_{j(p)+1}-1}\text{ converges},\end{align*} $$

and thus

$$ \begin{align*}\lim_p S_{n_{j(0)}}(g^{\prime}_0)\cdots S_{n_{j(p)}}(g^{\prime}_{p})S_{n_{j(p)}}(g_{p})^{-1}\cdots S_{n_{j(0)}}(g_0)^{-1}\text{ converges}.\end{align*} $$

Since $\pi _m$ is a continuous homomorphism, we obtain

$$ \begin{align*}\lim_p \pi_m(S_{n_{j(0)}}(g^{\prime}_0)\cdots S_{n_{j(p)}}(g^{\prime}_{p})S_{n_{j(p)}}(g_{p})^{-1}\cdots S_{n_{j(0)}}(g_0)^{-1})\text{ converges}.\end{align*} $$

Following again the similar arguments as in the proof of Claim 2, we get $(\pi ^S_m)^\omega (x)E_*(H_m)(\pi ^S_m)^\omega (y)$ .

It follows from Claims 2 and 3 that $W_m,\pi ^S_m,W$ , and S are as desired. Finally, by Lemma 2.9, we can conclude that the converse is also true if $G=W$ and if the interval $[0,1]$ can be embedded into H.

Proof. (1) $\Rightarrow $ (2). Let W be an open subgroup of G such that $S\upharpoonright W:W\rightarrow H$ is a continuous homomorphism. Then

$$ \begin{align*}\forall x,y\in W^\omega\,\forall g\in W\,(\lim_n(x,y)|_0^n=g\Rightarrow \lim_n(S^\omega(x),S^\omega(y))|_0^n=S(g)).\end{align*} $$

Since W is also closed, we see that $S^\omega $ is an $(E_*(W),E_*(H))$ -homomorphism. Hence $W'=W$ is as required.

(2) $\Rightarrow $ (1). Let $W'$ be an open subgroup of G such that the map $S^\omega $ is an $(E_*(W'),E_*(H))$ -homomorphism, and let $d_G,d_H$ be compatible left-invariant metrics on G and H respectively.

Define $X=\{g\in W':S(g^{-1})=S(g)^{-1}\}$ . Then $X\ne \emptyset $ , since $1_G\in X$ .

Claim 1. There exists an $r_0>0$ such that

$$ \begin{align*}\forall g,g'\in W'\, ((d_G(g,g')<r_0\wedge g\in X)\Rightarrow g'\in X).\end{align*} $$

Proof of Claim 1.

If not, then we can find two sequences $(g_q)$ and $(g^{\prime }_q)$ in $W'$ so that

1. (i) $g_q\in X$ and $\,g^{\prime }_q\notin X$ ;

2. (ii) $\lim _q (g^{\prime }_q)^{-1}g_q=1_G.$

Let $h_q=S((g^{\prime }_q)^{-1})S(g^{\prime }_q)$ , then $h_q\neq 1_H$ for $q\in \omega $ . Since H has no small subgroups, there exists some $D>0$ such that, for each $q\in \omega $ , we can find an $m_q>0$ with $ d_{H}(h_q^{m_q},1_{H})\geq D$ .

Put $M_{-1}=0$ and $M_q=m_0+\cdots +m_q$ for $q\in \omega $ . For each $p\in \omega $ , define

$$ \begin{align*}x(p)=\left\{\begin{array}{ll}(g_q')^{-1}, & p=2(M_{q-1}+i),0\leq i<m_q, \\ g^{\prime}_q, & p=2(M_{q-1}+i)+1,0\leq i<m_q,\end{array}\right.\end{align*} $$

$$ \begin{align*}y(p)=\left\{\begin{array}{ll}g_q^{-1}, & p=2(M_{q-1}+i),0\leq i<m_q, \\ g_q, & p=2(M_{q-1}+i)+1,0\leq i<m_q.\end{array}\right.\end{align*} $$

For any $k\in \omega $ , let $q_k$ be the largest q such that $M_{q-1}\leq k$ . Clearly,

$$ \begin{align*}(x,y)|_0^{2k+1}=1_G,\quad (x,y)|_0^{2k}=(g^{\prime}_{q_k})^{-1}g_{q_k}.\end{align*} $$

By (ii), we have $x E_*(G)y$ . Since $S^\omega $ is an $(E_*(G),E_*(H))$ -homomorphism, $S^\omega (x)E_*(H)S^\omega (y)$ holds, i.e.,

$$ \begin{align*}\lim_k(S^\omega(x),S^\omega(y))|_0^k\text{ converges}.\end{align*} $$

Then by Lemma 2.4(2) and $g_q\in X$ , we have

$$ \begin{align*}\lim_qh_q^{m_q}=\lim_q(S^\omega(x),S^\omega(y))|_{2M_{q-1}}^{2M_q-1}=1_H.\end{align*} $$

A contradiction!

Put $V=\{g\in W':d_G(1_G, g)<r_0\}$ . Note that $V=V^{-1}$ . Define $W_1=\bigcup _iV^i$ , which is an open subgroup of $W'$ . We claim that $W_1\subseteq X$ . Note that $1_G\in X$ , so by Claim 1, we have $V\subseteq X$ . Assume that $V^i\subseteq X$ . For any $g=g_1g_2\in V^{i+1}$ with $g_1\in V^i$ and $g_2\in V$ , we note that $g_1\in X$ and $d_G(g_1,g)=d_G(g_1,g_1g_2)=d_G(1_G,g_2)<r_0$ . Again by Claim 1, we have $g\in X$ . This shows that $W_1\subseteq X$ .

Claim 2. There exists $0<r_1<r_0$ such that

$$ \begin{align*}\forall g,g'\in W_1\,(d_G(1_G,g)<r_1\Rightarrow S(gg')=S(g)S(g')).\end{align*} $$

Proof of Claim 2.

If not, there are two sequences $(g_q),(g^{\prime }_q)$ in $W_1$ such that $\lim _qg_q=1_G$ and $S(g_qg^{\prime }_q)\ne S(g_q)S(g_q')$ for $q\in \omega $ .

For each $q\in \omega $ , let $h_q=S(g_qg^{\prime }_q)S((g^{\prime }_q)^{-1})S(g_q^{-1})$ . It follows from $W_1\subseteq X$ that

$$ \begin{align*}h_q=S(g_qg_q')S(g_q')^{-1}S(g_q)^{-1}\ne 1_H.\end{align*} $$

Since H has no small subgroups, there exists some $D>0$ such that, for each $q\in \omega $ , we can find an $m_q\in \omega $ with $ d_{H}(h_q^{m_q},1_{H})\geq D$ .

Put $M_{-1}=0$ and $M_q=m_0+\cdots +m_q$ for $q\in \omega $ . For each $p\in \omega $ , define

$$ \begin{align*}x(p)=\left\{\begin{array}{ll}g_qg_q', & p=3(M_{q-1}+i),0\leq i<m_q, \\ (g^{\prime}_q)^{-1}, & p=3(M_{q-1}+i)+1,0\leq i<m_q, \\ g_q^{-1}, &p=3(M_{q-1}+i)+2,0\leq i<m_q,\end{array}\right.\end{align*} $$

$$ \begin{align*}y(p)=\left\{\begin{array}{ll}g^{\prime}_q, & p=3(M_{q-1}+i), 0\leq i<m_q, \\ (g^{\prime}_q)^{-1}, & p=3(M_{q-1}+i)+1,0\leq i<m_q, \\ 1_G, &p=3(M_{q-1}+i)+2,0\leq i<m_q.\end{array}\right.\end{align*} $$

For any $k\in \omega $ , let $q_k$ be the largest q such that $M_{q-1}\leq k$ . Clearly,

$$ \begin{align*}(x,y)|_0^{3k}=(x,y)|_0^{3k+1}= g_{q_k},\quad (x,y)|_0^{3k+2}=1_G.\end{align*} $$

Note that $\lim _qg_q=1_G$ , so $x E_*(G)y$ holds. Since $S^\omega $ is an $(E_*(G),E_*(H))$ -homomorphism, $S^\omega (x)E_*(H)S^\omega (y)$ holds, i.e.,

$$ \begin{align*}\lim_k(S^\omega(x),S^\omega(y))|_0^k\text{ converges}.\end{align*} $$

By Lemma 2.4(2) and $g_q'\in W_1\subseteq X$ , we have $S(g_q')S((g_q')^{-1})=1_H$ . So

$$ \begin{align*}\lim_qh_q^{m_q}=\lim_q(S^\omega(x),S^\omega(y))|_{3M_{q-1}}^{3M_q-1}=1_H.\end{align*} $$

A contradiction!

Finally, let $V_1=\{g\in W': d_G(1_G,g)<r_1\}$ and $W=\bigcup _iV^i_1$ . Note that $V_1=V_1^{-1}$ and $V_1\subseteq V$ , so W is an open subgroup of $W_1$ . For any $g,g'\in W$ , we shall check that $S(gg')=S(g)S(g')$ . There are $g_0,g_1,\ldots ,g_m\in V_1$ such that $g=g_0 g_1\cdots g_m$ . Note that $W\subseteq W_1$ , so by Claim 2, we have

$$ \begin{align*}S(gg')=S(g_0)S(g_1\cdots g_mg')=S(g_0)\cdots S(g_m)S(g')=S(g)S(g').\end{align*} $$

Then W is as required.

Proof. Let $d_G,d_H$ be compatible left-invariant metrics on G and H respectively. Define $V_k=\{g\in \ker (\varphi ):d_G(1_G,g)<2^{-k}\}$ .

First, suppose $x\in \mathrm {IPC(\varphi )}$ for any $x\in G^\omega $ with $\lim _px(p)=1_G$ .

Assume toward a contradiction that $\ker (\varphi )$ is not non-archimedean. Then there exists a $K\in \omega $ such that $V_{K}$ contains no open subgroups of $\ker (\varphi )$ . Since $\bigcup _m V_k^m$ is an open subgroup of $\ker (\varphi )$ , we have $\bigcup _m V_k^m\nsubseteq V_{K}$ . Thus for each $k\in \omega $ , we can find an $m_k\in \omega $ and $g_{k,0},\dots ,g_{k,m_k-1}\in V_k$ such that $g_{k,0}g_{k,1}\cdots g_{k,m_k-1}\notin V_{K}$ .

Put $M_{-1}=0$ and $M_k=m_0+\dots +m_k$ for $k\in \omega $ . Let $x\in G^\omega $ be defined as $x(p)=g_{k,i}$ for $p=M_{k-1}+i$ with $0\leq i<m_k$ . Note that $\lim _px(p)=1_G$ , so $x\in \mathrm {IPC(\varphi )}$ . Since $d_G$ is left-invariant, we have

$$ \begin{align*}\forall k\, (d_G(x(0)\cdots x(M_{k-1}-1),x(0)\cdots x(M_k-1))\geq 2^{-K}).\end{align*} $$

Thus $(x(0)\cdots x(p))$ is not $\iota $ -Cauchy. But $(\varphi (x(0))\cdots \varphi (x(p)))$ is $\iota $ -Cauchy since $\varphi (x(p))=1_H$ for all $p\in \omega $ , contradicting that $x\in \mathrm {IPC}(\varphi )$ .

Now assume that $\varphi (G)$ is closed in H and $\ker (\varphi )$ is non-archimedean. Let $x\in G^\omega $ with $\lim _px(p)=1_G$ . We prove $x\in \mathrm {IPC(\varphi )}$ as follows:

On the one hand, suppose that $(x(0)\cdots x(p))$ is $\iota $ -Cauchy. For any $\varepsilon>0$ , since $\varphi $ is continuous, there is a $\delta>0$ such that

$$ \begin{align*}\forall g\in G\,(d_G(1_G,g)<\delta\Rightarrow d_H(1_H,\varphi(g))<\varepsilon).\end{align*} $$

Since $(x(0)\cdots x(p))$ is $\iota $ -Cauchy, we can find an $N\in \omega $ such that

$$ \begin{align*}\forall m\geq n>N\, (d_G(1_G,x(n)\cdots x(m))<\delta).\end{align*} $$

Thus we have that, for any $ m\geq n>N$ ,

$$ \begin{align*}d_H(1_H,\varphi(x(n))\cdots\varphi(x(m)))=d_H(1_H,\varphi(x(n)\cdots x(m)))<\varepsilon.\end{align*} $$

This shows that $(\varphi (x(0))\cdots \varphi (x(p)))$ is $\iota $ -Cauchy.

On the other hand, suppose that $(\varphi (x(0))\cdots \varphi (x(p)))$ is $\iota $ -Cauchy. Let $(W_p)$ be a decreasing neighborhood basis of $1_{\ker (\varphi )}$ such that each $W_p$ is an open subgroup of $\ker (\varphi )$ . Let $\widetilde {\varphi }:G/{\ker (\varphi )}\rightarrow H$ be defined as

$$ \begin{align*}\widetilde{\varphi}(\ker(\varphi)g)=\varphi(g).\end{align*} $$

Note that $\varphi (G)$ is closed, so it is also a Polish group under the topology inherited from H. Thus $\widetilde {\varphi }$ is a topological isomorphism of $G/{\ker (\varphi )}$ onto $\varphi (G)$ (see [Reference Gao8, Corollary 2.3.4]). Note that $\mu =\widetilde {\varphi }^{-1}\circ \varphi $ is the canonical projection map, where $\mu (g)=\ker (\varphi )g$ for $g\in G$ . Since $\widetilde {\varphi }$ is a topological isomorphism, we have $(\mu (x(0))\cdots \mu (x(p)))$ is $\iota $ -Cauchy.

Let $d_\mu (\ker (\varphi )g,\ker (\varphi )g')=\inf \{d_G(hg,h'g'):h,h'\in \ker (\varphi )\}$ , then $d_\mu $ is a compatible left-invariant metric on $G/\ker (\varphi )$ (see [Reference Gao8, Lemma 2.2.8]).

For any $\varepsilon>0$ , we can find $0<\varepsilon '<\varepsilon $ and $p_0\in \omega $ satisfying

$$ \begin{align*}\{g\in\ker(\varphi):d_G(1_G,g)<3\varepsilon'\}\subseteq W_{p_0}\subseteq\{g\in\ker(\varphi):d_G(1_G,g)<\varepsilon\}.\end{align*} $$

Note that $\lim _px(p)=1_G$ and $(\mu (x(0))\cdots \mu (x(p)))$ is $\iota $ -Cauchy. Thus there is an $N\in \omega $ such that

$$ \begin{align*}d_G(1_G,x(n))<\varepsilon',\end{align*} $$

$$ \begin{align*}d_\mu(\mu(x(N)\cdots x(n)),\ker(\varphi)1_H)<\varepsilon'\end{align*} $$

for each $n>N$ . By the definition of $d_\mu $ , there exists some $g_n\in \ker (\varphi )$ for each $n>N$ such that

$$ \begin{align*}d_G(x(N)\cdots x(n),g_n)<\varepsilon'.\end{align*} $$

For any $n>N$ , we have

$$ \begin{align*}d_G(x(N)\cdots x(n),x(N)\cdots x(n+1))=d_G(1_G,x(n+1))<\varepsilon'.\end{align*} $$

All these together imply that $d_G(1_G,g_n^{-1}g_{n+1})=d_G(g_n,g_{n+1})<3\varepsilon '$ . So $g_n^{-1}g_{n+1}\in W_{p_0}$ . Since $W_{p_0}$ is a subgroup, it follows that $g_n^{-1}g_m\in W_{p_0}$ for any $m\geq n\geq N$ . This gives that

$$ \begin{align*}\begin{aligned} d_G(x(0)\cdots x(n),x(0)\cdots x(m))&=d_G(x(N)\cdots x(n),x(N)\cdots x(m)) \\ &<2\varepsilon'+d_G(g_n,g_m) \\ &<2\varepsilon'+\varepsilon<3\varepsilon. \end{aligned} \end{align*} $$

Therefore, $(x(0)\cdots x(p))$ is $\iota $ -Cauchy.

Proof. Since G and H are both TSI, for $x\in {G}^\omega $ , we have

$$ \begin{align*}(x(0)\cdots x(p))\text{ is } \iota\text{-Cauchy}\iff xE_*(G)1_{G^\omega},\end{align*} $$

$$ \begin{align*}(\varphi(x(0))\cdots\varphi(x(p)))\text{ is } \iota\text{-Cauchy}\iff \lim_p\varphi^\omega(x)E_*(H)1_{H^\omega}.\end{align*} $$

So (2) $\Rightarrow $ (1) follows from Proposition 2.8. We prove (1) $\Rightarrow $ (2) as follows:

Let $d_G$ be a compatible compete two-sided invariant metric on G.

Given $x,y\in G^\omega $ . Suppose $xE_*(G)y$ . Since $\varphi $ is a continuous homomorphism, we have $\varphi ^\omega (x)E_*(H)\varphi ^\omega (y)$ holds and $\lim _pd_G(x(p),y(p))=0$ .

On the other hand, suppose $\varphi ^\omega (x)E_*(H)\varphi ^\omega (y)$ holds and $\lim _pd_G(x(p), y(p))=0$ . Define $z\in G^\omega $ as:

$$ \begin{align*}z(p)=y(0)\cdots y(p-1)x(p)y(p)^{-1}y(p-1)^{-1}\cdots y(0)^{-1}\end{align*} $$

for $p\in \omega $ . Since $d_G$ is two-sided invariant,

$$ \begin{align*}\lim_pd_G(z(p),1_G)=\lim_pd_G(x(p),y(p))=0.\end{align*} $$

So $z\in \mathrm {IPC}(\varphi )$ . Since $\varphi $ is homomorphism, it holds that

$$ \begin{align*}\varphi(z(p))=\varphi(y(0))\cdots\varphi(y(p-1))\varphi(x(p))\varphi(y(p))^{-1}\varphi(y(p-1))^{-1}\cdots\varphi(y(0))^{-1}.\end{align*} $$

Therefore,

$$ \begin{align*}\qquad\qquad xE_*(G)y\iff(z(0)\cdots z(p))\text{ is } \iota\text{-Cauchy},\end{align*} $$

$$ \begin{align*}\varphi^\omega(x)E_*(H)\varphi^\omega(y)\iff(\varphi(z(0))\cdots\varphi(z(p)))\text{ is } \iota\text{-Cauchy}.\end{align*} $$

So (1) $\Rightarrow $ (2) follows from Proposition 2.8 again.

Proof. Suppose $E(G)\le _BE(H)$ . It follows from Theorem 5.3 that, for each $m\in \omega $ , there exists an open subgroup $W_m'$ of G and a continuous map $\pi ^S_m:W_m'\to H_m$ with $\pi ^S_m(1_G)=1_{H_m}$ such that (1) $(\pi ^S_m)^\omega :(W_m')^\omega \rightarrow H_m^\omega $ is an $(E_*(W_m'),E_*(H_m))$ -homomorphism; and (2) let $W'=\bigcap _mW_m'$ and define $S:W'\rightarrow H$ as $S(g)=(\pi ^S_0(g),\pi ^S_1(g),\ldots )$ , then $S^\#$ is a reduction of $E_*(W')$ to $E_*(H)\times E(W';0)$ .

Note that $H_m$ is NSS. By Lemma 6.1, there is an open subgroup $W_m$ of $W_m'$ such that $\pi ^S_m\upharpoonright W_m$ is a homomorphism. We put $W=\bigcap _m W_m$ , and still denote $S\upharpoonright W$ by S for brevity. Then S is a continuous homomorphism. It is clear that $G_0\subseteq W\subseteq W'$ .

As both G and H are TSI, according to Lemma 6.2 and Proposition 6.3, $\ker (S)$ is non-archimedean. Thus, W and S fulfill the requirements.

On the other hand, let S be a continuous homomorphism from G to H such that $\ker (S)$ is non-archimedean. Suppose that $S(G)$ is closed in H and the interval $[0,1]$ can be embedded into H. Again by Lemma 6.2 and Proposition 6.3, the map $S^\#$ is a reduction of $E_*(G)$ to $E_*(H)\times E(G;0)$ . Finally, it follows from Lemma 2.9 that $E(G)\leq _B E(H)$ .

Proof. Assume for contradiction that $E(G)\leq _B E(H)$ . By Theorem 6.4, there is a continuous homomorphism $S:G_0\rightarrow H$ such that $\ker (S)$ is non-archimedean. Note that $S(G_0)$ is also connected, so $S(G_0)\subseteq H_0$ . Since H is totally disconnected, we have $S(G_0)=\{1_H\}$ . So $G_0=\ker (S)$ , which is connected and non-archimedean. This implies that $G_0=\{1_G\}$ , contradicting that G is not totally disconnected.

Proof. Let $d_G,d_H$ be compatible complete two-sided invariant metrics on G and H respectively. By Proposition 2.8, we only need to show that

$$ \begin{align*}xE_*(G)y\iff S^\omega(x)E_*(H)S^\omega(y)\end{align*} $$

holds whenever $\lim _pd_G(x(p),y(p))=0$ .

Suppose $xE_*(G)y$ holds. Since $S:G\to H$ is a continuous homomorphism, it follows trivially that $S^\omega (x)E_*(H)S^\omega (y)$ .

On the other hand, suppose $\lim _pd_G(x(p),y(p))=0$ and $S^\omega (x)E_*(H)S^\omega (y)$ . We prove $xE_*(G)y$ as follows:

If not, by Lemma 2.4(2), $\sup _{q>p}{d_G(x,y)|_{[p,q]}}\nrightarrow 0$ . So there exist $r>0$ and two strictly increasing sequences of natural numbers $(p_k),(q_k)$ such that $p_k<q_k$ and $d_G(1_G,(x,y)|_{p_k}^{q_k})=d_G(x,y)|_{[p_k,q_k]}>r$ for each $k\in \omega $ . Let $0<r_0<r$ satisfying that $\{g\in G:d_G(1_G,g)<r_0\}\subseteq U$ . There exists a $K\in \omega $ such that $d_G(x(p),y(p))<r_0/2$ for $p>p_K$ . For $p>p_k$ and $k>K$ , since $d_G$ is two-sided invariant, we have

$$ \begin{align*}d_G((x,y)|_{p_k}^p,(x,y)|_{p_k}^{p-1}))=d_G(x(p)y(p)^{-1},1_G)<r_0/2,\end{align*} $$

and hence

$$ \begin{align*}d_G(1_G,(x,y)|_{p_k}^p)<d_G(1_G,(x,y)|_{p_k}^{p-1})+r_0/2.\end{align*} $$

For each $k>K$ , we can find $p_k<p_k'<q_k$ such that

$$ \begin{align*}r_0/2\le d_G(1_G,(x,y)|_{p_k}^{p_k'})<r_0,\end{align*} $$

so $(x,y)|_{p_k}^{p_k'}\in U$ . By $S^\omega (x)E_*(H)S^\omega (y)$ , it is clear that

$$ \begin{align*}S((x,y)|_{p_k}^{p_k'})=(S^\omega(x),S^\omega(y))|_{p_k}^{p_k'}\to 1_H.\end{align*} $$

Since $S\upharpoonright U$ is a homeomorphism from U to $S(U)$ , we have $(x,y)|_{p_k}^{p_k'}\to 1_G$ . This contradicts that $d_G(1_G,(x,y)|_{p_k}^{p_k'})\geq r_0/2$ .