Observation 2.2. In the notation of Lemma 2.1, since the words W ₀ and $W_1^{\prime}$ are reduced and $W_0W_1^{\prime}$ is quasi-reduced, there exists an initial interval $I \subseteq \overline{W_0}$ and a terminal interval $I^{\prime} \subseteq \overline{W_1^{\prime}}$ such that either

1. (a) $|\overline{W_0} \setminus I| = 0 = |\overline{W_1^{\prime}} \setminus I^{\prime}|$ (in case $W_0W_1^{\prime}$ is reduced) or

2. (b) $|\overline{W_0} \setminus I| = 1 = |\overline{W_1^{\prime}} \setminus I^{\prime}|$ (in case $W_0W_1^{\prime}$ is not reduced).

In case (b) one obtains $\operatorname{Red}(W_0W_1^{\prime})$ by performing the multiplication

\begin{equation*}W_0(\max(\overline{W_0}))W_1^{\prime}(\min(\overline{W_1^{\prime}}))\end{equation*}

in the common group where they lie.

Proof. (1) ( $\Rightarrow$) Suppose that $W \in \mathcal{W}(\{G_n\}_{n \in \omega})$ is such that $W(i) \neq 1$ for all $i \in \overline{W}$ and that $W \sim E$. We will define a reduction scheme on W. For each $n \in \omega$ let $X_n = \{i \in \overline{W}: d(W(i)) = n\}$. As $W(i) \neq 1$ for all $i \in \overline{W}$, we have $\overline{W} = \bigsqcup_{n \in \omega} X_n$ and as W is a word, we know that each X_n is finite. We have $p_N(W) = W \upharpoonright \bigsqcup_{n = 0}^N X_n$ for each $N \in \omega$, and as $W \sim E$, we know $p_N(W)$ is equal in $*_{n = 0}^N G_n$ to E. From combinatorics in free products, for every $N \in \omega$ we have a finite reduction scheme $\mathcal{S}_N$ such that $\pi(W, C) \equiv E$ for all $C \in \mathcal{S}_N$ and $\bigcup_{C \in \mathcal{S_N}}\operatorname{set}(C) = \overline{p_N(W)} = \bigsqcup_{n = 0}^N X_n$. Let $Y_{-1} = \omega$. If $X_0 = \emptyset$ then we let $Y_0 = Y_{-1}$. Else we know, since X ₀ is finite, that there is an infinite $Y_0 \subseteq Y_{-1}$ such that for all $m, m^{\prime} \in Y_0$ we have $\{C \in \mathcal{S}_m: d(C) = 0\} = \{C \in \mathcal{S}_{m^{\prime}}: d(C) = 0\}$. Supposing that we have defined infinite $Y_N \subseteq \omega$, if $X_{N+1} = \emptyset$ then let $Y_{N+1} = Y_N$, else we select infinite $Y_{N+1} \subseteq Y_N$ such that $m, m^{\prime} \in Y_{N+1}$ we have $\{C \in \mathcal{S}_m: d(C) = N+1\} = \{C \in \mathcal{S}_{m^{\prime}}: d(C) = N+1\}$. Let

\begin{equation*} \mathcal{S} = \{C: (\exists m, N \in \omega)m \in Y_N \wedge d(C) = N \wedge C \in \mathcal{S}_{m}\}. \end{equation*}

Letting $C_0, C_1 \in \mathcal{S}$ be distinct, we let $N = \max\{d(C_0), d(C_1)\}$. Pick $m \in Y_N$ and we have $C_0, C_1 \in \mathcal{S}_m$, and so $\operatorname{set}(C_0) \cap \operatorname{set}(C_1) = \emptyset$. Next, we let $C = (i_0; \ldots; i_k) \in \mathcal{S}$ be given, $0 \leq j \lt k$ and i in the interval $(i_j, i_{j+1}) \subseteq \overline{W}$. Let $N = \max\{d(W(i)), d(C)\}$ and select $m \in Y_N$. There is a unique $C_0 \in \mathcal{S}_m$ such that $i \in \operatorname{set}(C_0) \subseteq (i_j, i_{j+1})$, and $C_0 \in \mathcal{S}$. Also, $\pi(W, C_0) \equiv E$, as indeed $\pi(W, C_1) \equiv E$ for every $C_1 \in \mathcal{S}$. Thus $\mathcal{S}$ is a reduction scheme on W and it is clear that $\bigcup_{C \in \mathcal{S}} \operatorname{set}(C) = \overline{W}$ and $\pi(W, C) = E$ for all $C \in \mathcal{S}$.

(1) ( $\Leftarrow$) Suppose that $W \in \mathcal{W}(\{G_n\}_{n \in \omega})$ is such that $W(i) \neq 1$ for all $i \in \overline{W}$. Suppose that there is a reduction scheme $\mathcal{S}$ on W such that $\bigcup_{C \in \mathcal{S}} \operatorname{set}(C) = \overline{W}$ and $\pi(W, C) \equiv E$ for all $C \in \mathcal{S}$. For a given $N \in \omega$ we let $\mathcal{S}_N = \{C \in \mathcal{S}: d(C) \leq N\}$, and notice that $\mathcal{S}_N$ witnesses that $p_N(W)$ is equal to identity in $*_{n = 0}^N G_n$, and therefore $W \sim E$.

(2) ( $\Rightarrow$) Suppose that $W \in \mathcal{W}(\{G_n\}_{n \in \omega})$ has $W(i) \neq 1$ for all $i \in \overline{W}$. Suppose that there exists a nonempty reduction scheme $\mathcal{S}$ on W. Let $C = (i_0; \ldots; i_k) \in \mathcal{S}$. If i ₀ and i ₁ are neighbouring then we know that W was not reduced, since $W(i_0)$ and $W(i_1)$ are in the group $G_{d(C)}$, and if i ₀ and i ₁ are not neighbouring then the reduction scheme $\mathcal{S}^{\prime} = \{C \in \mathcal{S}: \operatorname{set}(C) \cap (i_0, i_1) \neq \emptyset\}$ witnesses that the nonempty subword $W \upharpoonright (i_0, i_1)$ is $\sim E$, by part (1). Thus in either case, W is not reduced.

(2) ( $\Leftarrow$) Suppose that $W \in \mathcal{W}(\{G_n\}_{n \in \omega})$ has $W(i) \neq 1$ for all $i \in \overline{W}$. Suppose that the only reduction scheme on W is the empty scheme. If there are neighboring $i_0, i_1 \in \overline{W}$ such that $W(i_0)$ and $W(i_1)$ are in the same G_n, then $\mathcal{S} = \{C\}$ where $C = (i_0; i_1)$ is a nonempty reduction scheme, contradiction. If there is a nonempty interval $I \subseteq \overline{W}$ such that $W \upharpoonright I \sim E$ then by part (1) we have a reduction scheme $\mathcal{S}$ on $W \upharpoonright I$ such that $\bigcup_{C \in \mathcal{S}} \operatorname{set}(C) = I$, so in particular $\mathcal{S}$ is a nonempty reduction scheme on W itself, contradiction.

Proof. Certainly if W may be expressed as a finite concatenation as in (2) then $W \in \langle \operatorname{Sub}(\{W_x\}_{x \in X}) \rangle$ because each $W_j \in \operatorname{Sub}(\{W_x\}_{x \in X}) \rangle$, so that condition (2) implies condition (1). But by Lemma 2.1 and Observation 2.2 it is clear that the set of all words satisfying condition (2) forms a subgroup of $\operatorname{Red}(\{G_n\}_{n \in \omega})$ which includes the set $\operatorname{Sub}(\{W_x\}_{x \in X})$. Thus condition (1) implies condition (2).

Observation 2.12. For $W \in \operatorname{Red}(\{G_n\}_{n \in \omega})$ expressed as a finite concatenation $W \equiv W_0W_1 \cdots W_k$ and $J \subseteq \{0, \ldots, k\}$ such that $\overline{W_j}$ is finite for each $j \in J$, we may write $[[W]] = \prod_{0 \leq j \leq k, j \notin J}[[W_i]]$, since $\overline{W_j}$ finite implies that $W_j \in *_{n \in \omega}G_n$ and more particularly $[[W_j]]=[[E]]$.

Proof. We must show that the described function is well-defined; in other words the function is independent of all the choices made in the description. Therefore, suppose that we can write

• • $\overline{W} \equiv J_0J_1 \cdots J_k$, where $W \equiv (W \upharpoonright J_0) \cdots (W \upharpoonright J_k)$ is a decomposition as in Lemma 2.8;

• • $0 \leq r_0 \lt r_1 \lt \cdots \lt r_s \leq k$ are such that $\{r_0, \ldots, r_s\} = \{r \in \{0, 1, \ldots, k\}\mid |J_r| \geq 2\}$; and

• • $W \upharpoonright J_{r_j} \equiv (W_{x_{r_j}} \upharpoonright I_j)^{\delta_j}$ with $I_j \subseteq \overline{W_{x_{r_j}}}$ an interval and $\delta_j \in \{-1, 1\}$ for each $0 \leq j \leq s$;

and also we may write

• • $\overline{W} \equiv L_0L_1 \cdots L_p$, where $W \equiv (W \upharpoonright L_0) \cdots (W \upharpoonright L_p)$ is a decomposition as in Lemma 2.8;

• • $0 \leq q_0 \lt q_1 \lt \cdots \lt q_t \leq p$ are such that $\{q_0, \ldots, q_t\} = \{q \in \{0, 1, \ldots, p\}\mid |L_q| \geq 2\}$; and

• • $W \upharpoonright L_{q_l} \equiv (W_{x_{q_l}^{\prime}} \upharpoonright E_l)^{\epsilon_l}$, with $E_l being \subseteq \overline{W_{x_{q_l}^{\prime}}}$ an interval and $\epsilon_l \in \{-1, 1\}$ for each $0 \leq l \leq t$.

We must show that the elements

\begin{equation*} [[(U_{x_{r_0}} \upharpoonright \alpha(I_0, \iota_{x_{r_0}}))^{\delta_0}]]\cdots[[(U_{x_{r_s}}, \alpha(I_s, \iota_{x_{r_s}}))^{\delta_s}]] \end{equation*}

and

\begin{equation*} [[(U_{x_{q_0}^{\prime}} \upharpoonright \alpha(E_0, \iota_{x_{q_0}^{\prime}}))^{\epsilon_0}]]\cdots[[(U_{x_{q_t}^{\prime}} \upharpoonright \alpha(E_t, \iota_{x_{q_t}^{\prime}}))^{\epsilon_t}]] \end{equation*}

are equal in $\mathcal{A}(\{K_n\}_{n \in \omega})$.

First, we will let $f_{j}: J_{r_j} \rightarrow I_j^{\delta_j}$ be the order isomorphism witnessing $W \upharpoonright J_{r_j} \equiv (W_{x_{r_j}} \upharpoonright I_j)^{\delta_j}$ for each $0 \leq j \leq s$. Let $g_l: L_{q_l} \rightarrow E_l^{\epsilon_l}$ be the order isomorphism witnessing $W \upharpoonright L_{q_l} \equiv (W_{x_{q_l}} \upharpoonright E_l)^{\epsilon_l}$ for each $0 \leq l \leq t$.

We will take $\mathbb{J}$ to be the set of intervals consisting of nonempty intersections of one of the J_r with one of the L_q, where $0 \leq r \leq k$ and $0 \leq q \leq p$. The elements of $\mathbb{J}$ are pairwise disjoint intervals in $\overline{W}$, ordered in the natural way, such that $\overline{W} \equiv \prod_{M \in \mathbb{J}} M$. For each $0 \leq r \leq k$ we let $\mathbb{J}_{r}$ be the set of those elements of $\mathbb{J}$ which are subsets of J_r, and for each $0 \leq q \leq p$ we let $\mathbb{J}_{(q)}$ be the set of those elements of $\mathbb{J}$ which are subsets of L_q. Let $\overline{\mathbb{J}} \subseteq \mathbb{J}$ be the subset in $\mathbb{J}$ consisting of those intervals which are of cardinality greater than 1, and $\overline{\mathbb{J}_{r}} = \mathbb{J}_{r} \cap \overline{\mathbb{J}}$ for each $0 \leq r \leq k$ and $\overline{\mathbb{J}_{(q)}} = \mathbb{J}_{(q)} \cap \overline{\mathbb{J}}$ and $0 \leq q \leq p$.

Define $f_{-}: \overline{\mathbb{J}} \rightarrow \{0, \ldots, s\}$ by $f_{-}(M) = j$, where $M \subseteq J_{r_j}$ and $f^{-}: \overline{\mathbb{J}} \rightarrow \{0, \ldots, t\}$ by $f^{-}(M) = l$, where $M \subseteq L_{q_l}$. Notice that for each $M \in \overline{\mathbb{J}}$, we have $f_{f_{-}(M)}(M) \subseteq (\overline{W_{x_{r_{f_{-}(M)}}}})^{\delta_{f_{-}(M)}}$, and so $(f_{f_{-}(M)}(M))^{\delta_{f_{-}(M)}} \subseteq \overline{W_{x_{r_{f_{-}(M)}}}}$. Similarly $(f_{f^{-}(M)}(M))^{\epsilon_{f^{-}(M)}} \subseteq \overline{W_{x_{q_{f^{-}(M)}}^{\prime}}}$ and also

\begin{equation*} \begin{aligned} (W_{x_{r_{f_{-}(M)}}} \upharpoonright (f_{f_{-}(M)}(M))^{\delta_{f_{-}(M)}})^{\delta_{f_{-}(M)}} & \equiv W \upharpoonright M\\ & \equiv (W_{x_{q_{f^{-}(M)}}^{\prime}} \upharpoonright (f_{f^{-}(M)}(M))^{\epsilon_{f^{-}(M)}})^{\epsilon_{f^{-}(M)}}. \end{aligned} \end{equation*}

Therefore, by the coherence of the collection of coi triples, we see for each $M \in \overline{\mathbb{J}}$ that

\begin{equation*}[[(U_{x_{r_{f_{-}(M)}}} \upharpoonright \alpha((f_{f_{-}(M)}(M))^{\delta_{f_{-}(M)}} , \iota_{x_{r_{f_{-}(M)}}}))^{\delta_{f_{-}(M)}}]]\end{equation*}

(*)\begin{equation}= [[(U_{x_{q_{f^{-}(M)}}^{\prime}} \upharpoonright \alpha((g_{f^{-}(M)}(M))^{\epsilon_{f^{-}(M)}} , \iota_{x_{q_{f^{-}(M)}}^{\prime}}))^{\epsilon_{f^{-}(M)}}]]. \end{equation}

Next, we claim that for each $0 \leq j \leq s$, we have

(**)\begin{equation} \begin{array}{ll} [[(U_{x_{r_j}}\upharpoonright \alpha(I_j, \iota_{x_{r_j}}))^{\delta_j}]] & = \prod_{M \in \overline{\mathbb{J}_{r_j}}} [[(U_{x_{r_j}}\upharpoonright \alpha((f_j(M))^{\delta_j}, \iota_{x_{r_j}}))^{\delta_j}]]. \end{array} \end{equation}

To see why this is true, we recall that for each $M \in \mathbb{J}_{r_j}$ the set $f_j(M)$ is an interval in $I_j^{\delta_j}$, and so $(f_j(M))^{\delta_j}$ is an interval in I_j. Then $I_j \equiv \prod_{M \in \mathbb{J}_{r_j}^{\delta_j}} (f_j(M))^{\delta_j}$. Then we may write

\begin{equation*} \alpha(I_j, \iota_{x_{r_j}}) \equiv (\prod_{M \in \mathbb{J}_{r_j}^{\delta_j}} Q_M\alpha((f_j(M))^{\delta_j}, \iota_{x_{r_j}}))Q_f, \end{equation*}

where each element of $\{Q_M\}_{M \in \mathbb{J}_{r_j}^{\delta_j}} \cup \{Q_f\}$ is a finite interval, by Lemma 3.8. Also, whenever $|M| = 1$ (that is, when $M \in \mathbb{J}_{r_j} \setminus \overline{\mathbb{J}_{r_j}}$), we have by Lemma 3.9 that $\alpha(f_j(M), \iota_{x_{r_j}})$ is finite. Now

\begin{equation*} \begin{array}{ll} U_{x_{r_k}}\upharpoonright \alpha(I_k, \iota_{x_{r_j}}) & \equiv (\prod_{M \in \mathbb{J}_{r_j}^{\delta_j}} (U_{x_{r_j}}\upharpoonright Q_M)(U_{x_{r_j}}\upharpoonright\alpha((f_j(M))^{\delta_j}, \iota_{x_{r_j}})))\\ & \cdot (U_{x_{r_j}}\upharpoonright Q_f) \end{array} \end{equation*}

and by taking the $[[\cdot]]$ class of both sides and deleting $[[U_{x_{r_j}}\upharpoonright Q_f]]$, and all $[[U_{x_{r_j}}\upharpoonright Q_M]]$, and all $U_{x_{r_j}}\upharpoonright\alpha((f_j(M))^{\delta_j}, \iota_{x_{r_j}})$ for $M \in \mathbb{J}_{r_j} \setminus \overline{\mathbb{J}_{r_j}}$, we see that

\begin{equation*} \begin{array}{ll} [[U_{x_{r_j}}\upharpoonright \alpha(I_j, \iota_{x_{r_j}})]] & = \prod_{M \in \overline{\mathbb{J}_{r_j}}^{\delta_j}} [[U_{x_{r_j}}\upharpoonright\alpha((f_j(M))^{\delta_j}, \iota_{x_{r_j}})]], \end{array} \end{equation*}

and by taking the δ_j power of both sides, we obtain the desired equality. By the same reasoning, we have for each $0 \leq l \leq t$ that

\begin{equation*} \begin{array}{ll} [[(U_{x_{q_l}^{\prime}} \upharpoonright \alpha(E_l, \iota_{x_{q_l}^{\prime}}))^{\epsilon_l}]] & = \prod_{M \in \overline{\mathbb{J}_{(q_l)}}}[[(U_{x_{q_l}^{\prime}} \upharpoonright \alpha((g_l(M))^{\epsilon_l} , \iota_{x_{q_l}^{\prime}}))^{\epsilon_l}]]. \end{array} \end{equation*}

Thus

\begin{equation*} \begin{array}{ll} \prod_{j = 0}^s[[(U_{x_{r_j}}\upharpoonright \alpha(I_j, \iota_{x_{r_j}}))^{\delta_j}]]\\ = \prod_{j = 0}^s \prod_{M \in \overline{\mathbb{J}_{r_j}}} [[(U_{x_{r_j}}\upharpoonright \alpha((f_j(M))^{\delta_j}, \iota_{x_{r_j}}))^{\delta_j}]]\\ = \prod_{j = 0}^s\prod_{M \in \overline{\mathbb{J}_{r_j}}}[[(U_{x_{r_{f_{-}(M)}}}\upharpoonright \alpha((f_{f_{-}(M)}(M))^{\delta_{f_{-}(M)}} , \iota_{x_{r_{f_{-}(M)}}}))^{\delta_{f_{-}(M)}}]]\\ = \prod_{j = 0}^s\prod_{M \in \overline{\mathbb{J}_{r_j}}}[[(U_{x_{q_{f^{-}(M)}}^{\prime}} \upharpoonright \alpha((g_{f^{-}(M)}(M))^{\epsilon_{f^{-}(M)}} , \iota_{x_{q_{f^{-}(M)}}^{\prime}}))^{\epsilon_{f^{-}(M)}}]]\\ = \prod_{M \in \overline{\mathbb{J}}}[[(U_{x_{q_{f^{-}(M)}}^{\prime}} \upharpoonright \alpha((g_{f^{-}(M)}(M))^{\epsilon_{f^{-}(M)}} , \iota_{x_{q_{f^{-}(M)}}^{\prime}}))^{\epsilon_{f^{-}(M)}}]]\\ = \prod_{l = 0}^t \prod_{M \in \overline{\mathbb{J}_{(q_l)}}}[[(U_{x_{q_{f^{-}(M)}}^{\prime}} \upharpoonright \alpha((g_{f^{-}(M)}(M))^{\epsilon_{f^{-}(M)}} , \iota_{x_{q_{f^{-}(M)}}^{\prime}}))^{\epsilon_{f^{-}(M)}}]]\\ = \prod_{l = 0}^t \prod_{M \in \overline{\mathbb{J}_{(q_l)}}}[[(U_{x_{q_l}^{\prime}} \upharpoonright \alpha((g_l(M))^{\epsilon_l} , \iota_{x_{q_l}^{\prime}}))^{\epsilon_l}]]\\ = \prod_{l = 0}^t[[(U_{x_{q_l}^{\prime}} \upharpoonright \alpha(E_l, \iota_{x_{q_l}^{\prime}}))^{\epsilon_l}]], \end{array} \end{equation*}

where the first equality was established in (**), the second equality is an obvious relabelling, the third equality is a finite term-by-term replacement using (*), the fourth and fifth equalities are each a re-indexing, the sixth equality is an obvious relabelling and the seventh equality was established in the analogue of (**). The proof for the well-definedness of the comparably defined ϕ ₁ is symmetric.

Proof. We first point out that for $W \in \operatorname{Fine}(\{W_x\}_{x \in X})$ we have $\phi_0(W^{-1}) = (\phi_0(W))^{-1}$. To see this, we let

• • $\overline{W} \equiv J_0J_1 \cdots J_k$, where $W \equiv (W \upharpoonright J_0) \cdots (W \upharpoonright J_k)$ is a decomposition as in Lemma 2.8;

• • $0 \leq r_0 \lt r_1 \lt \cdots \lt r_s \leq k$ are such that $\{r_0, \ldots, r_s\} = \{r \in \{0, 1, \ldots, k\}\mid |J_r| \geq 2\}$; and

• • $W \upharpoonright J_{r_j} \equiv (W_{x_{r_j}} \upharpoonright I_j)^{\delta_j}$ with $I_j \subseteq \overline{W_{x_{r_j}}}$ an interval;

and as

• • $\overline{W^{-1}} \equiv J_k^{-1} \cdots J_1^{-1}J_0^{-1}$, where $W^{-1} \equiv (W^{-1} \upharpoonright J_k^{-1}) \cdots (W^{-1} \upharpoonright J_0^{-1})$ is a decomposition as in Lemma 2.8;

• • $0 \leq r_0 \lt r_1 \lt \cdots \lt r_s \leq k$ are such that $\{r_0, \ldots, r_s\} = \{r \in \{0, 1, \ldots, k\}\mid |J_r^{-1}| \geq 2\}$; and

• • $W^{-1} \upharpoonright J_{r_j}^{-1} \equiv (W_{x_{r_j}} \upharpoonright I_j)^{-\delta_j}$ with $I_j \subseteq \overline{W_{x_{r_j}}}$ an interval;

we get

\begin{equation*} \begin{array}{ll} (\phi_0(W))^{-1} & = (\prod_{j = 0}^s[[(U_{x_{r_j}}\upharpoonright \alpha(I_j, \iota_{x_{r_j}}))^{\delta_j}]])^{-1}\\ & = \prod_{j = s}^0[[(U_{x_{r_j}}\upharpoonright \alpha(I_j, \iota_{x_{r_j}}))^{-\delta_j}]]\\ & = \phi_0(W^{-1}). \end{array} \end{equation*}

Also, if $W \in \operatorname{Fine}(\{W_x\}_{x \in X})$ is finite (i.e. $|\overline{W}| \lt \infty$), then $\phi_0(W) = [[E]]$. This is seen by writing $\overline{W} \equiv J_0J_1 \cdots J_k$ as in Lemma 2.8, and letting $0 \leq r_0 \lt r_1 \lt \cdots \lt r_s \leq k$ be such that $\{r_0, \ldots, r_s\} = \{r \in \{0, 1, \ldots, j\}\mid |J_r| \geq 2\}$, and $W \upharpoonright J_{r_j} \equiv (W_{x_{r_j}} \upharpoonright I_j)^{\delta_j}$. Since each $J_{r_j}$ is finite, we see that $(W_{x_{r_j}} \upharpoonright I_j)^{\delta_j}$ is a finite word, and so $(U_{x_{r_j}}\upharpoonright \alpha(I_j, \iota_{x_{r_j}}))^{\delta_j}$ will be finite by Lemma 3.9. Then $\phi_0(W) = \prod_{j = 0}^s[[(U_{x_{r_j}}\upharpoonright \alpha(I_j, \iota_{x_{r_j}}))^{\delta_j}]] = [[E]]$.

Next we notice that when $W \in \operatorname{Fine}(\{W_x\}_{x \in X})$ and $\overline{W} \equiv E_aE_b$ we get $\phi_0(W) = \phi_0(W \upharpoonright E_a)\phi_0(W \upharpoonright E_b)$. To see this, we let $J_0, J_1, \ldots, J_k \subseteq \overline{W}$, $0 \leq r_0 \lt \cdots \lt r_s \leq k$, etc. be as before. The claim obviously holds if either E_a or E_b is empty, so without loss of generality, we assume that $E_a \neq \emptyset \neq E_b$. Let $0 \leq r \leq k$ be maximal such that $J_r \cap E_a \neq \emptyset$. If $E_a \equiv J_0 \cdots J_r$ and $E_b \equiv J_{r + 1} \cdots J_k$, then these decompositions are as in Lemma 2.8 and it is easily seen that

\begin{equation*} \begin{array}{ll} \phi_0(W) & = \prod_{j = 0}^s[[(U_{x_{r_j}}\upharpoonright \alpha(I_j, \iota_{x_{r_j}}))^{\delta_j}]]\\ & = (\prod_{0\leq j \leq s, r_j \leq r}[[(U_{x_{r_j}}\upharpoonright \alpha(I_j, \iota_{x_{r_j}}))^{\delta_j}]])\\ & \cdot (\prod_{0\leq j \leq s, r_j \gt r}[[(U_{x_{r_j}}\upharpoonright \alpha(I_j, \iota_{x_{r_j}}))^{i\delta_j}]]))\\ & = \phi_0(W \upharpoonright E_a)\phi_0(W \upharpoonright E_b). \end{array} \end{equation*}

Otherwise we have $J_r \cap E_a \neq \emptyset \neq J_r \cap E_b$, which implies in particular that $|J_r| \geq 2$ and therefore $r = r_{k^{\prime}}$ for some $0 \leq k^{\prime} \leq s$. Let $J_r^{\prime} = J_r \cap E_a$ and $J_r^{\prime\prime} = J_r \cap E_b$. If, for example, $|J_r^{\prime}| = 1$ and $|J_r^{\prime\prime}| \geq 2$, then we let $f: J_{r_{k^{\prime}}} \rightarrow I_{k^{\prime}}^{\delta_{k^{\prime}}}$ be an order isomorphism witnessing $W \upharpoonright J_{r_{k^{\prime}}} \equiv (W_{x_{r_{k^{\prime}}}} \upharpoonright I_{k^{\prime}})^{\delta_{k^{\prime}}}$. We get

\begin{equation*}[[(U_{x_{r_{k^{\prime}}}}\upharpoonright \alpha(I_{k^{\prime}}, \iota_{x_{r_{k^{\prime}}}}))^{\delta_{k^{\prime}}}]] = [[(U_{x_{r_{k^{\prime}}}}\upharpoonright \alpha((f(J_r^{\prime\prime}))^{\delta_{k^{\prime}}}, \iota_{x_{r_{k^{\prime}}}}))^{\delta_{k^{\prime}}}]]\end{equation*}

because $|I_{k^{\prime}} \setminus f(J_r^{\prime\prime})^{\delta_{k^{\prime}}}| = 1$ (by applying Lemmas 3.8 and 3.9). Thus

\begin{equation*} \begin{array}{ll} \phi_0(W) & = \prod_{j = 0}^s[[(U_{x_{r_j}}\upharpoonright \alpha(I_j, \iota_{x_{r_j}}))^{\delta_j}]]\\ & = (\prod_{0\leq j \leq k^{\prime}}[[(U_{x_{r_j}}\upharpoonright \alpha(I_k, \iota_{x_{r_j}}))^{\delta_j}]])\\ & \cdot (\prod_{k^{\prime} \lt j \leq s}[[(U_{x_{r_j}}\upharpoonright \alpha(I_j, \iota_{x_{r_j}}))^{\delta_j}]]))\\ & = (\prod_{0\leq j \lt k^{\prime}}[[(U_{x_{r_j}}\upharpoonright \alpha(I_j, \iota_{x_{r_j}}))^{\delta_j}]])\\ & \cdot [[(U_{x_{r_{k^{\prime}}}}\upharpoonright \alpha(I_{k^{\prime}}, \iota_{x_{r_{k^{\prime}}}}))^{\delta_{k^{\prime}}}]])\\ & \cdot (\prod_{k^{\prime} \lt j \leq s}[[(U_{x_{r_j}}\upharpoonright \alpha(I_j, \iota_{x_{r_j}}))^{\delta_j}]]))\\ & = \phi_0(W \upharpoonright E_a)\phi_0(W \upharpoonright E_b). \end{array} \end{equation*}

If $|J_r^{\prime}| \geq 2$ and $|J_r^{\prime\prime}| = 1$, or $|J_r^{\prime}| = 1 = |J_r^{\prime\prime}|$, the proof is similar. If $|J_r^{\prime}|, |J_r^{\prime\prime}| \geq 2$, then

\begin{equation*} \begin{array}{ll} [[(U_{x_{r_{k^{\prime}}}}\upharpoonright \alpha(I_{k^{\prime}}, \iota_{x_{r_{k^{\prime}}}}))^{\delta_{k^{\prime}}}]] & = [[(U_{x_{r_{k^{\prime}}}}\upharpoonright \alpha((f(J_r^{\prime}))^{\delta_{k^{\prime}}}, \iota_{x_{r_{k^{\prime}}}}))^{\delta_{k^{\prime}}}]]\\ & \cdot [[(U_{x_{r_{k^{\prime}}}}\upharpoonright \alpha((f(J_r^{\prime\prime}))^{\delta_{k^{\prime}}}, \iota_{x_{r_{k^{\prime}}}}))^{\delta_{k^{\prime}}}]] \end{array} \end{equation*}

by using Lemmas 3.8 and 3.9 as was done in establishing (**) in Lemma 3.13 (if, say, $\delta_{k^{\prime}} = -1$ then one has $I_{k^{\prime}} \equiv (f(J_r^{\prime\prime}))^{-1}(f(J_r^{\prime}))^{-1}$, write

\begin{equation*}\alpha(I_{k^{\prime}}, \iota_{x_{r_{k^{\prime}}}}) \equiv Q_0\alpha(f(J_r^{\prime}), \iota_{x_{r_{k^{\prime}}}})Q_1\alpha(f(J_r^{\prime\prime}), \iota_{x_{r_{k^{\prime}}}})Q_2\end{equation*}

where each of Q ₀, Q ₁, Q ₂ is a finite interval, etc.) Thus $\phi_0(W) = \phi_0(W \upharpoonright E_a)\phi_0(W \upharpoonright E_b)$, as required.

Now we let $W, W^{\prime} \in \operatorname{Fine}(\{W_x\}_{x \in X})$ be given. Suppose we can write, as in Lemma 2.1, $W \equiv W_0W_aW_1$ and $W^{\prime} \equiv W_0^{\prime}W_bW_1^{\prime}$ so that $|\overline{W_a}| = 1 = |\overline{W_b}|$, $\operatorname{Let}(W_a)= \{g\}$, $\operatorname{Let}(W_b) = \{g^{\prime}\}$, $\{g, g^{\prime}\} \subseteq G_n$ and $gg^{\prime} = g^{\prime\prime} \neq 1$ in G_n and further that

1. (1) $W \equiv W_0W_aW_1$;

2. (2) $W^{\prime} \equiv W_0^{\prime}W_bW_1^{\prime}$;

3. (3) $W_0^{\prime} \equiv (W_1)^{-1}$; and

4. (4) $W_0g^{\prime\prime}W_1^{\prime}$ is reduced.

Then we have

\begin{equation*} \begin{array}{ll} \phi_0(W \circledcirc W^{\prime}) & = \phi_0(W_0g^{\prime\prime}W_1^{\prime})\\ & = \phi_0(W_0)\phi_0(g^{\prime\prime})\phi_0(W_1^{\prime})\\ & = \phi_0(W_0)\phi_0(W_1^{\prime})\\ & = \phi_0(W_0)\phi_0(W_1)(\phi_0(W_1))^{-1}\phi_0(W_1^{\prime})\\ & = \phi_0(W_0)\phi_0(W_a)\phi_0(W_1)\phi_0(W_0^{\prime})\phi_0(W_b)\phi_0(W_1^{\prime})\\ & = \phi_0(W)\phi_0(W^{\prime}). \end{array} \end{equation*}

In the case where $W \circledcirc W^{\prime} \equiv W_0W_1^{\prime}$, W_a and W_b are not necessary and the proof is even easier.

Thus, we have established that ϕ ₀ is a homomorphism, and the symmetric argument shows that ϕ ₁ is also a homomorphism. Moreover, all finite words are in the kernel of ϕ ₀, and also that of ϕ ₁, and so the homomorphisms descend to homomorphisms

\begin{equation*} \Phi_0: \beth(\operatorname{Fine}(\{W_x\}_{x \in X})) \rightarrow \beth(\operatorname{Fine}(\{U_x\}_{x \in X})) \end{equation*}

and

\begin{equation*} \Phi_1: \beth(\operatorname{Fine}(\{U_x\}_{x \in X})) \rightarrow \beth(\operatorname{Fine}(\{W_x\}_{x \in X})). \end{equation*}

It remains to see that Φ₀ and Φ₁ are isomorphisms such that $\Phi_0^{-1} = \Phi_1$. By applying Lemma 2.8, we can write any element of $\beth(\operatorname{Fine}(\{W_x\}_{x \in X}))$ as a product $[[W_0]]\cdots[[W_s]]$, where $W_j \in \operatorname{Sub}(\{W_x\}_{x\in X} \cup \{W_x^{-1}\}_{x \in X})$ for each $0 \leq j \leq s$. For each $0 \leq j \leq s$ we therefore select $x_j \in X$, $\delta_j \in \{1, -1\}$, and interval $I_j \subseteq \overline{W_{x_j}}$ so that $W_j \equiv (W_{x_j} \upharpoonright I_j)^{\delta_j}$. Then

\begin{equation*} \begin{array}{ll} \Phi_1 \circ \Phi_0(\prod_{j = 0}^s[[W_j]]) & = \prod_{j = 0}^s\Phi_1([[(U_{x_j}\upharpoonright \alpha (I_j, \iota_{x_j}))^{\delta_j}]])\\ & = \prod_{j = 0}^s (\Phi_1([[U_{x_j}\upharpoonright \alpha (I_j, \iota_{x_j}))]]))^{\delta_j}\\ & = \prod_{j = 0}^s[[W_{x_j}\upharpoonright \alpha(\alpha (I_j, \iota_{x_j})), \iota_{x_j}^{-1})]]^{\delta_j}\\ & = \prod_{j = 0}^s[[W_{x_j}\upharpoonright I_j]]^{i_j}\\ & = \prod_{j = 0}^s[[W_j]], \end{array} \end{equation*}

where the fourth equality uses the fact that $I_j \equiv I\alpha(\alpha (I_j, \iota_{x_j})), \iota_{x_j}^{-1})I^{\prime}$, where I and I′ are finite intervals (Lemma 3.4). Thus, $\Phi_1 \circ \Phi_0$ is the identity map, and by the same argument, we get $\Phi_0 \circ \Phi_1$ as the identity map. The theorem is proved.

Proof. We prove claim (1), and claim (2) follows in precisely the same way as the first claim, given the symmetric nature of the definition of coi triples.

If W is empty then we let U and ι be empty, the check that the larger coi collection is coherent is easy. Else we write

\begin{equation*} W \equiv W_0 \cdots W_k \end{equation*}

so that for each $0 \leq r \leq k$ we have W_r a subword of an element in $\{W_x\}_{x\in X} \cup \{W_x^{-1}\}_{x \in X}$, or $|\overline{W_r}| = 1$, as in Lemma 2.8. Let $J = \{r_0, \ldots, r_s\}$ be the set of those elements in $0 \leq r \leq k$ such that $W_{r_j}$ is infinite, with $r_0 \lt r_1 \lt \cdots \lt r_s$. Select $x_j \in X$ and $\delta_j \in \{-1, 1\}$ and intervals $I_j \subseteq \overline{W_{x_j}}$ such that $W_{r_j} \equiv (W_{x_j} \upharpoonright I_j)^{\delta_j}$. For each $r_j \in J$ we let $U_j^{\prime} \equiv (U_{x_j} \upharpoonright \alpha(I_j, \iota_{x_j}))^{\delta_j}$. By Lemma 3.9 we know that for each $0 \leq j \leq s$ the word $U_j^{\prime}$ is infinite.

If $J = \emptyset$ then the word W is finite, and we select $h \in K_0\setminus \{1\}$, let $U \equiv h$, select $\lambda \in \overline{W}$ and let $\iota: \{\lambda\} \rightarrow \overline{U}$ be the unique bijection. Again, the check that the extended coi collection is coherent is straightforward.

Now we may assume that $J \neq \emptyset$. For each $0 \leq j \lt s$, if both $\max\overline{U_j^{\prime}}$ and $\min\overline{U_{j+1}^{\prime}}$ exist and $d(U_j^{\prime}(\max\overline{U_j^{\prime}})) = d(U_{j+1}(\min\overline{U_{j+1}^{\prime}}))$ then we select

\begin{equation*}h_j \in K_{d(U_j^{\prime}\max\overline{U_j^{\prime}}) + 1} \setminus \{1\}\end{equation*}

and let $U_j \equiv U_j^{\prime}h_j$, otherwise we let $U_j \equiv U_j^{\prime}$. By Lemma 2.1 it is clear that the word U ₀ is reduced, that $U_0U_1$ is reduced, that $U_0U_1U_2$ is reduced, etc., and that $U_0\cdots U_s$ is reduced. We let $U \equiv U_0 \cdots U_s$.

We define a coi ι from W to U. For each $0 \leq j \leq s$ we let $f_j: \overline{W_{r_j}} \rightarrow I_j^{\delta_j}$ be the order isomorphism witnessing $W_{r_j} \equiv (W_{x_j} \upharpoonright I_j)^{\delta_j}$ and $k_j: \overline{U_j^{\prime}} \rightarrow \alpha(I_j, \iota_{x_j})^{\delta_j}$ be the order isomorphism witnessing $U_j^{\prime} \equiv (U_{x_j} \upharpoonright \alpha(I_j, \iota_{x_j}))^{\delta_j}$. Let the domain of ι be given by $\operatorname{dom}(\iota) = \bigcup_{j=0}^s f_j^{-1}(\operatorname{dom}(\iota_{x_j}) \cap I_j)$, the range of ι be given by $\bigcup_{j=0}^s k_j^{-1} \circ \iota_{x_j}(\operatorname{dom}(\iota_{x_j}) \cap I_j)$ and ι be the unique function which extends $ k_j^{-1} \circ \iota_{x_j} \circ f_j$ for each $0 \leq j \leq s$. By Lemma 3.2 (iii) we have that the domain (respectively range) of ι is close in $\overline{W}$ (respectively, $\overline{U}$).

We check that $\{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x\in X} \cup \{\operatorname{coi}(W, \iota, U)\}$ is coherent. Suppose that $y\in X$ and intervals $I \subseteq \overline{W}$ and $I^{\prime} \subseteq \overline{W_y}$ and $\delta \in \{-1, 1\}$ are such that $W \upharpoonright I \equiv (W_y \upharpoonright I^{\prime})^{\delta}$. Let $L \subseteq J$ denote the set of those r_j such that $\overline{W_{r_j}} \cap I \neq \emptyset$. For each $r_j \in L \cap J$ we have $W\upharpoonright (\overline{W_{r_j}} \cap I) \equiv (W_{x_j}\upharpoonright \Lambda_j^*)^{\delta_j}$ for the obvious choice of interval $\Lambda_j^* \subseteq I_j \subseteq \overline{W_{x_j}}$. Thus $(W_{x_j}\upharpoonright \Lambda_j^*)^{\delta \cdot \delta_j} \equiv W_y \upharpoonright I_j^{\prime}$ for the obvious choice of interval $I_j^{\prime} \subseteq I^{\prime}$. By the coherence of $\{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x\in X}$ we therefore have

\begin{equation*} \begin{array}{ll} [[U \upharpoonright \alpha(I, \iota)]] & = \prod_{r_j\in L} [[U \upharpoonright \alpha(\overline{W_{r_j}} \cap I, \iota)]]\\ & = \prod_{r_j \in L} [[U_{x_j} \upharpoonright \alpha(\Lambda_j^*, \iota_{x_j})]]^{\delta_j}\\ & = \prod_{r_j \in L^{\epsilon}} [[U_{y}\upharpoonright \alpha(I_j^{\prime}, \iota_y)]]^{\delta}\\ & = [[(U_y \upharpoonright \alpha(I^{\prime}, \iota_y))^{\delta}]], \end{array} \end{equation*}

where the first and last equalities hold by Lemma 3.9.

If we select intervals $I, I^{\prime} \subseteq \overline{W}$ and $\delta \in \{-1, 1\}$ such that $W \upharpoonright I \equiv (W \upharpoonright I^{\prime})^{\delta}$, then a similar strategy of finitely decomposing I and I′ is employed to show $[[U\upharpoonright \alpha(I, \iota)]] = [[(U\upharpoonright \alpha(I^{\prime}, \iota))^{\delta}]]$.

The check that if $U \upharpoonright Q \equiv (U_z \upharpoonright Q^{\prime})^{\epsilon}$, where $z\in X$, then the appropriate elements of $\mathcal{A}(\{G_n\}_{n \in \omega})$ are equal is similar to that above, with slight modifications (although $U \notin \operatorname{Fine}(\{U_x\}_{x\in X})$ is possible, the word U is a finite concatenation of words in $\operatorname{Fine}(\{U_x\}_{x \in X})$ and words of length 1). Similarly modifications apply for intervals $Q, Q^{\prime} \subseteq \overline{U}$, and the proof is complete.

Proof. Once again we will prove (1), and the proof of (2) follows the symmetric approach. Assume the hypotheses. Select $h \in K_{N+1} \setminus \{1\}$. If U_y is finite, then so is W_x by Lemma 3.9. In this case we let $U \equiv h$, and if W_y is nonempty we select $\lambda \in \overline{W_y}$ and let $\iota: \{\lambda\} \rightarrow \overline{U}$ be the unique function. That $\{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x\in X} \cup \{\operatorname{coi}(W_y, \iota, U)\}$ is coherent is straightforward. Now we suppose that U_y (and also W_y) is infinite. If $i \in \overline{U_y}$ with $d(U_y(i)) \leq N + 2$, then there exists a maximal interval $i \in I \subseteq \overline{U_y}$ such that $d(U_y(i^{\prime})) \leq N+2$ for all $i^{\prime} \in I$, and as U_y is a word we know that I is finite. If $i \in \overline{U_y}$ with $d(U_y(i)) \gt N + 2$, then there exists a maximal interval $i \in I \subseteq \overline{U_y}$ such that $d(U_y(i^{\prime})) \gt N + 2$ for all $i^{\prime} \in I$ (one can argue this by Zorn’s lemma). Thus we may write U_y as a finite concatenation

\begin{equation*} U_y \equiv U_0^{\prime} \cdots U_k^{\prime} \end{equation*}

where for each $0 \leq r \leq k$, we have either all letters in $U_r^{\prime}$ having $d(\cdot) \leq N + 2$ or all letters having $d(\cdot) \gt N + 2$, and moreover if all letters of $U_r^{\prime}$ have $d(\cdot) \leq N + 2$, then all letters in $U_{r + 1}^{\prime}$ have $d(\cdot) \gt N + 2$ and vice versa. Now we let

\begin{equation*} U_r = \left\{ \begin{array}{ll} h & \text{if } d(U_r^{\prime}) \leq N+2, \\ U_r^{\prime} & \text{if }d(U_r^{\prime}) \gt N + 2. \end{array} \right. \end{equation*}

and $U \equiv U_0U_1\cdots U_k$. By Lemma 2.1 it is clear that U is reduced, for if $U_1, U_3, \ldots, U_{2p+1}$ are all $\equiv h$ then we have U ₀ reduced, $U_0U_1 \equiv U_0h$ is reduced, $U_0U_1U_2 \equiv U_0hU_2$ is reduced, etc., and in case $U_0, U_2, \ldots$ are $\equiv h$ then the reasoning is similar.

Let J be the set of those $0 \leq r \leq k$ such that $U_r \not\equiv h$, so that J consists either of all even or all odd natural numbers which are at least 0 and at most k. For each $r \in J$ we have U_r as a subword of the original word U_y, so $\overline{U_r}$ is an interval in $\overline{U_y}$. Let $\mathbb{J} \subseteq J$ be those $0 \leq r \leq k$ such that U_r is infinite. The word U is obtained from U_y by deleting finitely many finite words and inserting finitely many finite words, and as U_y was infinite we know $\mathbb{J} \neq \emptyset$. As U_r is an infinite word for each $r \in \mathbb{J}$, by Lemma 3.2 (i) we know $\operatorname{Close}(\operatorname{im}(\iota_y) \cap \overline{U_r}, \overline{U_r})$ for such an r, and by Lemma 3.2 (iii), we have $\operatorname{Close}(\bigcup_{r \in J} (\operatorname{im}(\iota_y) \cap \overline{U_r}), \overline{U})$. Now we define a coi ι from W_y to U by letting $\operatorname{im}(\iota) = \bigcup_{r \in J} \operatorname{im}(\iota_y) \cap \overline{U_r}$, $\operatorname{dom}(\iota) = \iota_y^{-1}(\bigcup_{r \in J} \operatorname{im}(\iota_y) \cap \overline{U_r})$ and $\iota(i) = \iota_y(i)$ for all $i \in \operatorname{dom}(\iota)$. We have that $\operatorname{dom}(\iota_y) \setminus \operatorname{dom}(\iota)$ is finite by Lemma 3.9, so by Lemma 3.2 (i) we have $\operatorname{Close}(\operatorname{dom}(\iota), \overline{W_y})$. Also, it is clear that ι is a bijection.

So, ι is a coi from W_y to U, and $d(U) \gt N$. If I is an interval in $\overline{W_y}$, then one can obtain $U_y \upharpoonright \alpha(I, \iota_y)$ from $U \upharpoonright \alpha(I, \iota)$ by deleting finitely many finite subwords and inserting finitely many finite subwords, so $[[U_y \upharpoonright \alpha(I, \iota_y)]] = [[U \upharpoonright \alpha(I, \iota)]]$. Thus, if we let $z \in X$ and intervals $I \subseteq \overline{W_y}$ and $I^{\prime} \subseteq \overline{W_z}$ and $\delta \in \{-1, 1\}$ are such that $W_y \upharpoonright I \equiv (W_z \upharpoonright I^{\prime})^{\delta}$, we have

\begin{equation*}\begin{array}{c} [[U \upharpoonright \alpha(I, \iota)]] = [[U_y \upharpoonright \alpha(I, \iota_y)]]\\ = [[(U_z \upharpoonright \alpha(I^{\prime}, \iota_z))^{\delta}]], \end{array} \end{equation*}

where the second equality comes from the fact that the original coi collection $\{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x\in X}$ is coherent. The comparable proof shows that if $I, I^{\prime}$ are intervals in $\overline{W_y}$ and $\delta \in \{-1, 1\}$ with $W_y \upharpoonright I \equiv (W_y \upharpoonright I^{\prime})^{\delta}$ then we have

\begin{equation*} [[U \upharpoonright \alpha(I, \iota)]] = [[(U \upharpoonright \alpha(I^{\prime}, \iota))^{\delta}]]. \end{equation*}

Now we suppose that $z\in X$ and I is an interval in $\overline{U}$, I′ is an interval in $\overline{U_z}$, and $\epsilon \in \{-1, 1\}$ are such that $U \upharpoonright I \equiv (U_z \upharpoonright I^{\prime})^{\epsilon}$. Let $L = \{0 \leq r \leq k: \overline{U_r} \cap I \neq \emptyset\}$ and notice that

\begin{equation*} \begin{array}{ll} [[W_y \upharpoonright \alpha(I, \iota^{-1})]] & = \prod_{r \in L} [[W_y \upharpoonright \alpha(\overline{U_r} \cap I, \iota^{-1})]]\\ & = \prod_{r \in L \cap J} [[W_y \upharpoonright \alpha(\overline{U_r} \cap I, \iota^{-1})]]\\ & = \prod_{r \in L \cap J} [[W_y \upharpoonright \alpha(\overline{U_r} \cap I, \iota_y^{-1})]]\\ & = \prod_{r \in (L \cap J)^{\epsilon}} [[(W_z \upharpoonright \alpha(K_r,\iota_z^{-1}))^{\epsilon}]]\\ & = [[(W_z \upharpoonright \alpha(I^{\prime}, \iota_z^{-1}))^{\epsilon}]], \end{array} \end{equation*}

where K_r is the subinterval in $I^{\prime} \subseteq \overline{W_z}$ obtained from $\overline{U_r} \cap I$ via the order isomorphism witnessing $U \upharpoonright I \equiv (U_z \upharpoonright I^{\prime})^{\epsilon}$. The first equality is clear, the second follows from Lemma 3.9 and the third is because $\iota_y^{-1}$ and ι ⁻¹ coincide on $\overline{U_r} \cap I$ for each $r \in L \cap J$, the fourth is from the fact that $\{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x\in X}$ is coherent and the fifth is by Lemma 3.9. The similar claim when $I, I^{\prime}$ are intervals in $\overline{U}$ and $\epsilon \in \{-1, 1\}$ are such that $U\upharpoonright I \equiv (U \upharpoonright I^{\prime})^{\epsilon}$ follows via similar reasoning. Thus $\{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x\in X} \cup \{\operatorname{coi}(W_y, \iota, U)\}$ is coherent, and also in this case $\operatorname{dom}(\iota)$ is nonempty (as $\overline{W_y}$ is infinite).

Proof. If Λ is empty, then the claim is obvious. Thus, we may assume $\Lambda \neq \emptyset$. If there exists such an order embedding p, then we know that for each $n \in \omega$ the preimage $f^{-1}(\{n\})$ in Λ is finite. Thus suppose that $\Lambda \neq \emptyset$ and that such an order embedding exists. Select $\lambda^{\prime} \in \Lambda$. We claim that if $p: \Lambda \rightarrow \overline{V}$ is an order embedding with $p(\Lambda)$ an interval in $\overline{V}$ and $f(\lambda) = d(V(p(\lambda)))$ for all $\lambda \in \Lambda$, then p is totally determined by $p(\lambda^{\prime})$. To see this, we let $\lambda \in \Lambda$ be given. Letting $n = d(V(p(\lambda))) = f(\lambda)$ we have that $f^{-1}(\{n\})$ is finite. If $\lambda^{\prime} \leq \lambda$ in Λ, then we let $\lambda_0 \lt \lambda_1 \lt \cdots \lt \lambda_k$ be the set of all those elements in $f^{-1}(\{n\})$, which are $\geq \lambda^{\prime}$. As $p(\Lambda)$ is an interval in $\overline{V}$, we know that $p(\lambda_0) \in \overline{V}$ is the minimal element greater than or equal to $p(\lambda^{\prime})$ with $d(p(\lambda_0)) = n$, and $p(\lambda_1) \in \overline{V}$ is the minimal element above $p(\lambda_0)$ with $d(p(\lambda_1)) = n$, etc. Thus if $\lambda = \lambda_i$, then we see that $p(\lambda)$ is uniquely determined by $p(\lambda^{\prime})$, and the argument in case $\lambda \lt \lambda^{\prime}$ is entirely analogous.

Now, since $\lambda^{\prime}$ must be assigned to one of the finitely many elements $i \in \overline{V}$ such that $f(\lambda^{\prime}) = d(V(i))$, there are only finitely many order embeddings p of Λ, as an interval, into $\overline{V}$ with $f(\lambda) = d(V(p(\lambda)))$ for all $\lambda \in \Lambda$.

Proof. We let $\{\iota_y\}_{y \in Y}$ be the collection of all order embeddings with domain Λ and codomain an element $\overline{V_y} \in \{\overline{V_x}\}_{x \in X} \cup \{\overline{V_x}^{-1}\}_{x \in X}$ such that $f_0 = d \circ V_y \circ \iota_y$ and also $\iota_y(\Lambda)$ is an interval in $\overline{V_y}$. We also assume that the indexing Y does not include any duplicates, so $y_0 \neq y_1$ implies that $\iota_{y_0} \neq \iota_{y_1}$. By Lemma 4.3, we know that for a fixed $x\in X$, there can be only finitely many ι_y with codomain $\overline{V_x}$ or with codomain $\overline{V_x^{-1}}$. Thus, in particular, we know that $|Y| \lt 2^{\aleph_0}$. As the set of functions $q: f_1(\omega) \rightarrow \omega \setminus \{0\}$ is of cardinality $2^{\aleph_0}$, we select a q such that $(\forall y \in Y)(\exists \lambda_y \in f_1(\omega))$ $V_y(\iota_y(\lambda_y)) \neq f_3(\lambda_y)^{f_2(\lambda_y)q(\lambda_y)}$. It is clear that this q satisfies the conclusion.

Proof. Assume the hypotheses. For each $n \in \omega$, select $h_n \in H_n$ of infinite order. Write $\omega = \bigsqcup_{m \in \omega} Z_m$ where each Z_m is infinite and define $z: \omega \rightarrow \omega$ by $n \in Z_{z(n)}$. Let $f_{0, m}: \omega \cap [\min Z_m, \infty) \rightarrow \omega$ be the identity map. For each $m \in \omega$, we let $f_{1, m}: \omega \rightarrow Z_m$ be a bijection. For each $m \in \omega$, we let $f_{2, m}$ be the constant function with domain $Z_m = f_{1, m}(\omega)$ and output 1. For each $m \in \omega$, we let $f_{3, m}$ be the function with domain $Z_m = f_{1, m}(\omega)$ given by $f_{3, m}(n) = h_n$.

By Lemma 4.4 (letting $\Lambda = \omega \cap [\min Z_m, \infty)$ for each m), we select for each $m \in \omega$ a function $q_m: f_{1, m}(\omega) \rightarrow \omega \setminus \{0\}$ such that there exist no $V \in \operatorname{Sub}(\{V_x\}_{x \in X} \cup \{V_x^{-1}\}_{x\in X})$ and order isomorphism $\iota: \omega \cap [\min Z_m, \infty) \rightarrow \overline{V}$ such that

1. (a) $d \circ V \circ \iota = f_{0, m}$ on $\omega \cap [\min Z_m, \infty)$ and

2. (b) $V(\iota(n)) = f_{3, m}(n)^{f_{2, m}(n)q_m(n)}$ for all $n \in f_{1, m}(\omega)$.

Now we let

\begin{equation*}V^{\prime} \equiv h_0^{q_{z(0)}(0)}h_1^{q_{z(1)}(1)} \cdots .\end{equation*}

It is easy to see that V′ is reduced (each finite initial word $h_0^{q_{z(0)}(0)}h_1^{q_{z(1)}(1)} \cdots h_k^{q_{z(k)}(k)}$ is reduced, and if V′ were not reduced there would exist a nonempty reduction scheme on V′ by Proposition 2.6, but then a finite initial word would have a nonempty reduction scheme, contradiction).

Were it the case that $V^{\prime} \in \operatorname{Fine}(\{V_x\}_{x \in X})$, we know by Lemma 2.8 that we can decompose V′ as a finite concatenation

\begin{equation*} V^{\prime} \equiv V_0^{\prime}V_1^{\prime} \cdots V_k^{\prime}, \end{equation*}

and for each $0 \leq j \leq k$ at least one of the following holds:

• • $V_j^{\prime} \in \operatorname{Sub}(\{V_x\}_{x \in X} \cup \{V_x^{-1}\}_{x \in X})$ is nonempty;

• • $|\overline{V_j^{\prime}}| = 1$ with $\operatorname{Let}(V_j^{\prime}) = \{h\}$ such that $g \in H_{n_j}\setminus \{1\}$ and h is a product of elements in $H_{n_j} \cap \operatorname{Let}(\{V_x\}_{x \in X} \cup \{(V_x^{\prime})^{-1}\}_{x \in X})$.

As $\overline{V^{\prime}} = \omega$, we know $\overline{V_k^{\prime}}$ is a nonempty terminal interval in ω (and therefore of order type ω) and $V_0^{\prime}\cdots V_{k-1}^{\prime}$ is a finite word. Therefore, $V_k^{\prime} \in \operatorname{Sub}(\{V_x\}_{x \in X} \cup \{V_x^{-1}\}_{x \in X})$, say $V_k^{\prime} \in \operatorname{Sub}(V_{\underline{x}}^{\epsilon})$, with $\epsilon \in \{-1,1\}$. As $\overline{V_k^{\prime}}$ is a nonempty terminal interval in ω, we can select some $m \in \omega$ so that $[\min Z_m, \infty)\cap \omega \subseteq \overline{V_k^{\prime}}$. As $V_k^{\prime} \in \operatorname{Sub}(V_{\underline{x}}^{\epsilon})$, we know that $V = V_k^{\prime} \upharpoonright ([\min Z_m, \infty) \cap \omega) \in \operatorname{Sub}(V_{\underline{x}}^{\epsilon})$, which contradicts the conditions for our selection of $h_m \in H_m$.

Proof. As usual, we will prove only (1) since (2) is analogous. Select, by Lemmas 4.1 and 4.2, a nonempty word $U_0^{\prime} \in \operatorname{Red}(\{K_n\}_{n \in \omega})$ with $d(U_0^{\prime}) \gt 1$ and a coi ι ₀ from W ₀ to $U_0^{\prime}$ such that $\{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x \in X} \cup \{\operatorname{coi}(W_0, \iota_0, U_0^{\prime})\}$ is coherent and ι ₀ has nonempty domain. Suppose that we have already selected nonempty words $U_0^{\prime}, \ldots, U_k^{\prime} \in \operatorname{Red}(\{K_n\}_{n \in \omega})$ with $d(U_j) \gt j + 1$ and $\iota_0, \ldots, \iota_k$, ι_j a coi from W_j to $U_j^{\prime}$, ι_j having nonempty domain and $\{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x \in X} \cup \{\operatorname{coi}(W_0, \iota_0, U_0^{\prime}), \ldots, \operatorname{coi}(W_k, \iota_k, U_k^{\prime})\}$ coherent. Then by Lemmas 4.1 and 4.2 select a nonempty word $U_{k+1}^{\prime} \in \operatorname{Red}(\{K_n\}_{n \in \omega})$ with $d(U_{k+1}^{\prime}) \gt k + 1$ and a coi $\iota_{k+1}$ from $W_{k+1}$ to $U_{k+1}^{\prime}$ with $\iota_{k+1}$ having nonempty domain and $\{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x \in X} \cup \{\operatorname{coi}(W_0, \iota_0, U_0^{\prime}), \ldots, \operatorname{coi}(W_{k+1}, \iota_{k+1}, U_{k+1}^{\prime})\}$ coherent. By Lemma 3.12 we have that $\{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x \in X} \cup \{\operatorname{coi}(W_n, \iota_n, U_n^{\prime})\}_{n \in \omega}$ is coherent. For each $n \in \omega$, select an element $k_n \in K_n$ of infinite order.

As in Lemma 4.5, we write $\omega = \bigsqcup_{m \in \omega} Z_m$, where each Z_m is infinite and define $z: \omega \rightarrow \omega$ by $n \in Z_{z(n)}$. Consider a word of form

\begin{equation*} U \equiv U_0^{\prime}k_0^{q_{z(0)}(0)} U_1^{\prime}k_1^{q_{z(1)}(1)}U_2^{\prime}k_2^{q_{z(2)}(2)} \cdots, \end{equation*}

where $\{q_m\}_{m\in \omega}$ is a sequence of functions with $q_m: Z_m \rightarrow \omega \setminus\{0\}$. It is easy to see that $U_0^{\prime}k_0^{q_{z(0)}(0)}$ is reduced, since $d(U_0^{\prime}) \gt 0$, and similarly that $U_0^{\prime}k_0^{q_{z(0)}(0)} U_1^{\prime}$, $U_0^{\prime}k_0^{q_{z(0)}(0)} U_1^{\prime}k_1^{q_{z(1)}(1)}$, etc. are all reduced. Therefore, by applying Proposition 2.6, it is clear that such a U is reduced (if U is not reduced, there is a nonempty reduction scheme $\mathcal{S}$, from which we can find a nonempty reduction scheme $\mathcal{S}^{\prime} \subseteq \mathcal{S}$ on an initial $U_0^{\prime}k_0^{q_{z(0)}(0)}\cdots U_m^{\prime}k_m^{q_{z(m)}(0)})$. Let $\mathcal{F}$ denote the group of reduced words generated by $\{k_0, k_1, \ldots\}$. Now by Lemma 4.4 for each $m\in \omega$ we select the function $q_m: Z_m \rightarrow \omega \setminus \{0\}$ so that a word of form

\begin{align*}& k_{\min Z_m}^{q_m(\min Z_m)}U_{(\min Z_m) + 1}^{\prime}k_{(\min Z_m) + 1}^{q_{z((\min Z_m) +1)}((\min Z_m) + 1)}U_{(\min Z_m) + 2}^{\prime}\\ &\quad\times k_{(\min Z_m) + 2}^{q_{z((\min Z_m) + 2)}((\min Z_m) + 2)}\cdots\end{align*}

does not occur in $\operatorname{Sub}(\{U_x\}_{x \in X} \cup \{U_x^{-1}\}_{x \in X} \cup \{U_n^{\prime}\}_{n \in \omega} \cup \{(U_n^{\prime})^{-1}\}_{n \in \omega} \cup \mathcal{F})$.

Now we have produced a concrete word U and to simplify notation we set $q_{z(n)}(n) = r_n$, so that $U \equiv U_0^{\prime}k_0^{r_0}U_1^{\prime}k_1^{r_1}\cdots$. By arguing as in Lemma 4.5, we get that no nonempty terminal subword of U is an element in $\operatorname{Fine}(\{U_x\}_{x\in X} \cup \{U_n^{\prime}\}_{n \in \omega} \cup \mathcal{F})$. Define ι to be the union $\iota = \bigcup_{m \in \omega}\iota_m$, so $\operatorname{dom}(\iota) = \bigcup_{m \in \omega}\operatorname{dom}(\iota_m)$ and $\operatorname{im}(\iota) = \bigcup_{m \in \omega}\operatorname{im}(\iota_m)$. It is easy to see that ι is an order isomorphism. We get by Lemma 3.2 (iii) that $\operatorname{Close}(\operatorname{dom}(\iota), \overline{W})$ and $\operatorname{Close}(\operatorname{im}(\iota), \overline{U})$, and so ι is a coi from W to U. We will prove that the collection $\{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x \in X} \cup \{\operatorname{coi}(W_n, \iota_n, U_n^{\prime})\}_{n \in \omega} \cup \{\operatorname{coi}(W, \iota, U)\}$ is coherent, from which the coherence of $\{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x \in X} \cup \{\operatorname{coi}(W, \iota, U)\}$ is immediate.

We point out that every proper initial subword of W is an element in the subgroup $\operatorname{Fine}(\{W_x\}_{x \in X})$, for if $W \equiv V_0V_1$ with V ₁ nonempty, we can select $\lambda \in \overline{V_1}$, and $\lambda \in \overline{W_N}$ for some $N \in \omega$, so $\overline{V_0} \subseteq \bigcup_{n = 0}^{N-1}\overline{W_n}$, so V ₀ is a subword of $\prod_{n = 0}^{N-1}W_n$, so V ₀ is in $\operatorname{Fine}(\{W_x\}_{x \in X})$, since $W_0, \ldots, W_{N-1} \in \operatorname{Fine}(\{W_x\}_{x \in X})$. Next, every nonempty terminal subword of W is not in $\operatorname{Fine}(\{W_x\}_{x \in X})$, for if $W \equiv V_0V_1$ with $V_1 \not\equiv E$, we have $V_0 \in \operatorname{Fine}(\{W_x\}_{x \in X})$, so that if $V_1 \in \operatorname{Fine}(\{W_x\}_{x \in X})$ we get $W \in \operatorname{Fine}(\{W_x\}_{x \in X})$, contradiction. Thus, any nonempty subword V of W which is an element in $\operatorname{Fine}(\{W_x\}_{x \in X})$ has $\overline{V} \subseteq \bigcup_{n = 0}^{N-1}\overline{W_n}$, for otherwise we get that V is a nonempty terminal subword.

Let $y \in X \cup \omega$ and intervals $I \subseteq \overline{W}$ and $I^{\prime} \subseteq \overline{W_y}$ and $\delta \in \{-1, 1\}$ such that $W\upharpoonright I \equiv (W_y \upharpoonright I^{\prime})^{\delta}$. Then in particular $W \upharpoonright I \in \operatorname{Fine}(\{W_x\}_{x \in X} \cup \{W_n\}_{n \in \omega}) = \operatorname{Fine}(\{W_x\}_{x \in X})$ (this latter equality holds because each $W_n \in \operatorname{Fine}(\{W_x\}_{x \in X})$). Then I is not a terminal interval in $\overline{W}$, and so $I \subseteq \bigcup_{n = 0}^{N-1}\overline{W_n}$. Then $[[U \upharpoonright \alpha(I, \iota)]] = [[(U_y \upharpoonright \alpha(I^{\prime}, \iota_y))^{\delta}]]$, as by the proof of Lemma 4.1, we know that the collection

\begin{equation*}\begin{array}{c} \{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x \in X} \cup \{\operatorname{coi}(W_n, \iota_n, U_n^{\prime})\}_{n \in \omega}\\ \cup \{\operatorname{coi}(\prod_{n = 0}^NW_n, \bigcup_{n}^N \iota_n, U_0^{\prime}k_0^{r_0}\cdots U_N^{\prime}k_N^{r_N})\} \end{array}\end{equation*}

is coherent and

\begin{equation*} \begin{array}{ll} [[U \upharpoonright \alpha(I, \iota)]] & = [[U_0^{\prime}k_0^{r_0}\cdots U_N^{\prime}k_N^{r_N} \upharpoonright \alpha(I, \bigcup_{n}^N \iota_n)]]\\ & = [[(U_y \upharpoonright \alpha(I^{\prime}, \iota_y))^{\delta}]] \end{array} \end{equation*}

where the first equality holds because $U \upharpoonright \alpha(I, \iota) \equiv U_0^{\prime}k_0^{r_0}\cdots U_N^{\prime}k_N^{r_N} \upharpoonright \alpha(I, \bigcup_{n}^N \iota_n)$.

Suppose now that I and I′ are intervals in $\overline{W}$ and $\delta \in \{-1, 1\}$ such that $W \upharpoonright I \equiv (W \upharpoonright I^{\prime})^{\delta}$. Then $W \upharpoonright I \in \operatorname{Fine}(\{W_x\}_{x \in X})$ if and only if I is not a nonempty terminal interval. If $W \upharpoonright I \in \operatorname{Fine}(\{W_x\}_{x \in X})$ then of course $(W \upharpoonright I)^{\delta} \equiv W \upharpoonright I^{\prime} \in \operatorname{Fine}(\{W_x\}_{x \in X})$ as well, so that I′ is also not a nonempty terminal interval. Then we may select $N \in \omega$ which is large enough that $I, I^{\prime} \subseteq \bigcup_{n = 0}^N\overline{W_n}$ and argue as in Lemma 4.1 that $[[U \upharpoonright \alpha(I, \iota)]] = [[(U \upharpoonright \alpha(I^{\prime}, \iota))^{\delta}]]$ since the collection

\begin{equation*} \{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x \in X} \cup \{\operatorname{coi}(W_n, \iota_n, U_n^{\prime})\}_{n \in \omega}\\ \cup \{\operatorname{coi}(\prod_{n = 0}^NW_n, \bigcup_{n = 0}^N \iota_n, U_0^{\prime}k_0^{r_0}\cdots U_N^{\prime}k_N^{r_N})\} \end{equation*}

is coherent. Thus, we suppose that I and I′ are nonempty terminal intervals. If it were the case that $\delta = -1$, then select $\lambda_0 \in I$ and we have on one hand that $W \upharpoonright \{\lambda \in I: \lambda \leq \lambda_0\} \in \operatorname{Fine}(\{W_x\}_{x \in X})$ since $\{\lambda \in I: \lambda \leq \lambda_0\}$ is not a terminal interval in $\overline{W}$, but on the other hand $W \upharpoonright \{\lambda \in I: \lambda \leq \lambda_0\} \notin \operatorname{Fine}(\{W_x\}_{x \in X})$ since $(W \upharpoonright \{\lambda \in I: \lambda \leq \lambda_0\})^{-1}$ is a nonempty terminal subword of the terminal subword $W \upharpoonright I^{\prime}$ of $\overline{W}$. That is a contradiction. Thus we know δ = 1. So I and I′ are each nonempty terminal intervals in $\overline{W}$, and hence either $I \subseteq I'$ or $I'\subseteq I$. If without loss of generality we have $I' \subsetneq I$ we select $\lambda \in I \setminus I'$. Let $\underline{\iota}: I \rightarrow I'$ be an order isomorphism witnessing $W\upharpoonright I \equiv W \upharpoonright I'$. Then we have $\lambda \lt \underline{\iota}(\lambda) \lt \underline{\iota}(\underline{\iota}(\lambda)) \lt \cdots$ and $W(\lambda) = W(\underline{\iota}(\lambda)) = W(\underline{\iota}(\underline{\iota}(\lambda))) = \cdots$. Either $W(\lambda) = 1$, which is impossible since W is reduced, or $W(\lambda) \neq 1$, say $W(\lambda) = g \in G_p \setminus \{1\}$, which is impossible since W is a word and cannot attain infinitely many outputs in G_p. Thus we know $I = I'$, from which we have $U\upharpoonright \alpha(I, \iota) \equiv U\upharpoonright \alpha(I', \iota)$ and $[[U\upharpoonright \alpha(I, \iota)]] = [[U\upharpoonright \alpha(I', \iota)]]$.

Next, we know that no nonempty terminal subword of U is an element of

\begin{equation*}\operatorname{Fine}(\{U_x\}_{x \in X} \cup \{U_n'\}_{n \in \omega} \cup \mathcal{F}),\end{equation*}

and therefore if interval $I \subseteq \overline{U}$ is such that $U \upharpoonright I \in \operatorname{Fine}(\{U_x\}_{x \in X} \cup \{U_n'\}_{n \in \omega} \cup \mathcal{F})$ there exists some $N \in \omega$ such that $I \subseteq \overline{\prod_{n = 0}^N U_n'k_n^{r_n}}$. Thus, if $y \in X \cup \omega$, I and I′ are intervals in $\overline{U}$ and $\overline{U_y}$ respectively, and $\epsilon \in \{-1, 1\}$ with $U \upharpoonright I \equiv (U_y \upharpoonright I')^{\epsilon}$, we have $U \upharpoonright I \in \operatorname{Fine}(\{U_x\}_{x \in X})$, so $I \subseteq \overline{\prod_{n = 0}^N U_n'k_n^{r_n}}$ for some N. Then $[[W \upharpoonright \alpha(I, \iota^{-1})]] = [[(W_y\upharpoonright \alpha(I', \iota_y^{-1}))^{\epsilon}]]$ follows as in the proof of Lemma 4.1 as the collection

\begin{equation*} \{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x \in X} \cup \{\operatorname{coi}(W_n, \iota_n, U_n')\}_{n \in \omega}\\ \cup \{\operatorname{coi}(\prod_{n = 0}^NW_n, \bigcup_{n = 0}^N \iota_n , U_0'k_0^{r_0}\cdots U_N'k_N^{r_N})\} \end{equation*}

is coherent.

Finally, we know that all nonempty terminal subwords of U are not elements of $\operatorname{Fine}(\{U_x\}_{x \in X} \cup \{U_n'\}_{n \in \omega} \cup \mathcal{F})$, but it is also clear that any proper initial subword of U is an element of $\operatorname{Fine}(\{U_x\}_{x \in X} \cup \{U_n'\}_{n \in \omega} \cup \mathcal{F})$, so for any nonempty interval $I \subseteq \overline{U}$ we have $U \upharpoonright I \in \operatorname{Fine}(\{U_x\}_{x \in X} \cup \{U_n'\}_{n \in \omega} \cup \mathcal{F})$ if and only if I is not terminal.

Suppose that $I, I'$ are intervals in $\overline{U}$ and $\epsilon \in \{-1, 1\}$ is such that $U \upharpoonright I \equiv (U \upharpoonright I')^{\epsilon}$. If $U \upharpoonright I \in \operatorname{Fine}(\{U_x\}_{x \in X} \cup \{U_n'\}_{n \in \omega} \cup \mathcal{F})$ then both I and I are empty or nonterminal, and we argue as before that $[[W \upharpoonright \alpha(I, \iota^{-1})]] = [[(W \upharpoonright \alpha(I', \iota^{-1}))^{\epsilon}]]$ using the fact that

\begin{equation*} \{\operatorname{coi}(W_x, \iota_x, U_x)\}_{x \in X} \cup \{\operatorname{coi}(W_n, \iota_n, U_n')\}_{n \in \omega}\\ \cup \{\operatorname{coi}(\prod_{n = 0}^NW_n, \bigcup_{n = 0}^N \iota_n , U_0'k_0^{r_0}\cdots U_N'k_N^{r_N})\} \end{equation*}

is coherent for an appropriate value of N. If $U \upharpoonright I \notin \operatorname{Fine}(\{U_x\}_{x \in X} \cup \{U_n'\}_{n \in \omega} \cup \mathcal{F})$ then both I and I′ are nonempty terminal and as before we argue that ϵ = 1 and in fact $I = I'$, so that $[[W \upharpoonright \alpha(I, \iota^{-1})]] = [[(W \upharpoonright \alpha(I', \iota^{-1}))^{\epsilon}]]$ is immediate. The lemma is proved.

Proof. Suppose for contradiction that U is not reduced. We point out that there cannot be adjacent elements $i_0, i_1 \in \overline{U}$ such that $U(i_0)$ and $U(i_1)$ are in the same group K_N, since each of the words U_s is nonempty reduced and $\mathbb{Q}$ is order dense. Then as U is not reduced, there exists a nonempty interval $I \subseteq \overline{U}$ such that $U \upharpoonright I \sim E$. Then by Proposition 2.6 (1) we have a reduction scheme $\mathcal{S}$ on $U \upharpoonright I$ such that $\bigcup_{C \in \mathcal{S}} \operatorname{set}(C) = I$ and $\pi(U \upharpoonright I, C) \equiv E$ for all $C \in \mathcal{S}$.

Let $N_0' = \min\{P(s):(\exists s \in \mathbb{Q}) \overline{U_s} \cap I \neq \emptyset\}$. We claim that there exists a component $C = (i_0; \ldots; i_k) \in \mathcal{S}$ such that $d(C) = N_0'$ and for each $0 \leq j \leq k$ we have $U(i_j) \in \{h_{N_0'}^{R(N_0')}, h_{N_0'}^{-R(N_0')}\}$. To see this, take $s' \in \mathbb{Q}$ such that $\overline{U_{s'}} \cap I \neq \emptyset$ and $P(s') = N_0'$. Take $C' = (i_0'; \ldots; i_m') \in \mathcal{S}$ such that $\operatorname{set}(C') \cap \overline{U_{s'}} \neq \emptyset$, say $0 \leq j' \leq m$ has $i_{j'} \in \operatorname{set}(C') \cap \overline{U_{s'}}$. If $d(C') = N_0'$ we will take $C = C'$, so suppose that this is not the case. As $\pi(U \upharpoonright I, C) = E$ and the word U is simple, we know m > 0. Let without loss of generality $j' + 1 \leq m$ (otherwise $0 \leq j' - 1$ and the proof will be similar). As $U_{s'}$ is reduced, we know $i_{j' + 1} \notin \overline{U_{s'}}$, say $i_{j' + 1} \in \overline{U_{s^{\prime\prime}}}$ and $s' \lt s^{\prime\prime}$ in $\mathbb{Q}$. As $\mathcal{S}$ is a reduction scheme and $i_{j'} \lt \max(\overline{U_{s'}}) \lt i_{j' + 1}$ there exists $C = (i_0; \ldots; i_k) \in \mathcal{S}$ such that $\max(\overline{U_{s'}}) = i_j \in \operatorname{set}(C)$. As $P(s') = N_0'$ we know $U(i_j) \in \{h_{N_0'}^{R(N_0')}, h_{N_0'}^{-R(N_0')}\}$ so in particular $d(C) = N_0'$. Thus in either case we have found a component $C \in \mathcal{S}$ with $d(C) = N_0'$ and $\operatorname{set}(C) \cap \overline{U_{s'}} \neq \emptyset$. Letting $J = \{s \in \mathbb{Q} \mid \operatorname{set}(C) \cap \overline{U_s} \neq \emptyset\}$, by minimality of $N_0'$ we know that $P(s) = N_0'$ for each $s \in J$. Then as $d(C) = N_0'$, each element of $\operatorname{set}(C)$ is either a $\max(\overline{U_s})$ or a $\min(\overline{U_s})$ for some $s \in J$. Therefore $U(i_j) \in \{h_{N_0'}^{R(N_0')}, h_{N_0'}^{-R(N_0')}\}$ for each $i_j \in \operatorname{set}(C)$.

As $\pi(U, C) \equiv E$ we have that there exist some $0 \leq j \leq k$ for which $U(i_j) = h_{N_0'}^{R(N_0')}$ and there also exists some $0 \leq j \leq k$ for which $U(i_j) = h_{N_0'}^{-R(N_0')}$. Then there exists some $0 \leq j' \leq k$ for which $U(i_{j'}) = (U(i_{j'+1}))^{-1}$. Then the reduction scheme $\{C' \in \mathcal{S}: \operatorname{set}(C') \cap (i_{j'}, i_{j'+1}) \neq \emptyset\}$ witnesses that $U \upharpoonright (i_{j'}, i_{j'+1}) \sim E$. Letting $i_{j'} \in \overline{U_{s_0'}}$ and $i_{j'+1} \in \overline{U_{s_1'}}$ we have $P(s_0') = P(s_1') = N_0'$ and $i_{j'} \in \{\min\overline{U_{s_0'}}, \max\overline{U_{s_0'}}\}$ and similarly $i_{j'+1} \in \{\min\overline{U_{s_1'}}, \max\overline{U_{s_1'}}\}$. If $i_{j'} = \min\overline{U_{s_0'}}$ then we claim that $i_{j'+1} = \max\overline{U_{s_1'}}$, for otherwise we have $(i_{j'}, i_{j'+1}) \cap \{i \in \overline{U}: d(U(i)) = N_0'\}$ is odd and so the word $p_{N_0'}(U \upharpoonright (i_{j'}, i_{j'+1}))$ does not represent the trivial element in $G_{N_0'}$, since $h_{N_0'}$ is of infinite order, a contradiction. By the same token, if $i_{j'} = \max\overline{U_{s_0'}}$ then $i_{j'+1} = \min\overline{U_{s_1'}}$. In either case, we see that $U \upharpoonright (\max\overline{U_{s_0'}}, \min\overline{U_{s_1'}}) \sim E$, for even if $i_{j'} = \min\overline{U_{s_0'}}$ and $i_{j'+1} = \max\overline{U_{s_1'}}$ we have that

\begin{equation*} E \sim (U_{s_0'}\upharpoonright (\overline{U_{s_0'}} \setminus \{\min\overline{U_{s_0'}}\}))^{-1} E (U_{s_0'}\upharpoonright (\overline{U_{s_0'}} \setminus \{\min\overline{U_{s_0'}}\})) \end{equation*}

\begin{equation*} \equiv (U_{s_0'}\upharpoonright \overline{U_{s_0'}} \setminus \{\min\overline{U_{s_0'}}\})^{-1} E (U_{s_1'}\upharpoonright( \overline{U_{s_1'}} \setminus \{\max\overline{U_{s_1'}}\}))^{-1} \end{equation*}

\begin{equation*} \sim (U_{s_0'}\upharpoonright \overline{U_{s_0'}} \setminus \{\min\overline{U_{s_0'}}\})^{-1} (U\upharpoonright (i_{j'}, i_{j'+1})) (U_{s_1'}\upharpoonright( \overline{U_{s_1'}} \setminus \{\max\overline{U_{s_1'}}\}))^{-1} \end{equation*}

\begin{equation*} \sim U\upharpoonright (\max\overline{U_{s_0'}}, \min\overline{U_{s_1'}}). \end{equation*}

In particular we may replace the interval I with the nonempty interval

\begin{equation*}(\max\overline{U_{s_0'}}, \min\overline{U_{s_1'}})\end{equation*}

and thus get that I is an open interval such that $I = \bigcup_{I \cap \overline{U_s} \neq \emptyset} \overline{U_s}$ and also replace the old reduction scheme $\mathcal{S}$ with $\{C' \in \mathcal{S}: \operatorname{set}(C) \cap (\max\overline{U_{s_0'}}, \min\overline{U_{s_1'}}) \neq \emptyset\}$. Henceforth in the proof we will therefore assume that $I = \bigcup_{I \cap \overline{U_s} \neq \emptyset} \overline{U_s}$ and that $\mathcal{S}$ is a reduction scheme on $U\upharpoonright I$ such that $\pi(U \upharpoonright I, C) \equiv E$ for all $C \in \mathcal{S}$ and $\bigcup_{C\in \mathcal{S}}\operatorname{set}(C) = I$. Let $\mathcal{I} \subseteq \mathbb{Q}$ be the open interval $\{s \in \mathbb{Q}: I \cap \overline{U_s} \neq \emptyset\}$. We let $Q: I \rightarrow \mathcal{I} \subseteq \mathbb{Q}$ be the surjective function defined by $Q(i)= s$, where $i \in \overline{U_s}$.

Let $\{N_0, N_1, \ldots\} = \{P(s):(\exists s \in \mathbb{Q}) \overline{U_s} \cap I \neq \emptyset\}$, with $N_k \lt N_{k+1}$, so in particular $N_0 = \min\{P(s):(\exists s \in \mathbb{Q}) \overline{U_s} \cap I \neq \emptyset\}$. Suppose first that $C = (i_0; \ldots; i_k) \in \mathcal{S}$ and $s_0, s_1 \in \mathbb{Q}$ are such that $\overline{U_{s_0}} \cap \operatorname{set}(C) \neq \emptyset$ and $\overline{U_{s_1}} \cap \operatorname{set}(C) \neq \emptyset$ and $P(s_0) = N_0$. We will show that $P(s_1) = N_0$. If this is not the case, there exist $i_l, i_{l+1} \in \operatorname{set}(C)$ with $i_l \in \overline{U_{s_0'}}$ and $i_{l+1} \in \overline{U_{s_1'}}$ such that either $P(s_0') = N_0$ and $P(s_1') \gt N_0$ or $P(s_0') \gt N_0$ and $P(s_1') = N_0$. If without loss of generality $P(s_0') = N_0$ and $P(s_1') \gt N_0$, we know by Definition 2.5 condition (2) and Proposition 2.6 part (1) that $U \upharpoonright (i_l, i_{l+1}) \sim E$; however, $(i_l, i_{l+1}) \cap \{i \in \overline{U}: d(U(i)) = N_0\}$ is odd and so the word $p_{N_0}(U \upharpoonright (i_l, i_{l+1}))$ does not represent the trivial element in $G_{N_0}$, since $h_{N_0}$ is of infinite order, a contradiction. Now, suppose that it is the case that whenever $k \leq K$ and $C \in \mathcal{S}$ and $s_0, s_1 \in \mathbb{Q}$ are such that $\overline{U_{s_0}} \cap \operatorname{set}(C) \neq \emptyset$ and $\overline{U_{s_1}} \cap\operatorname{set}(C) \neq \emptyset$ and $P(s_0) = N_k$, then $P(s_1) = N_k$. Let $C = (i_0; \ldots; i_p) \in \mathcal{S}$ and $s_0, s_1 \in \mathbb{Q}$ be such that $\overline{U_{s_0}} \cap \operatorname{set}(C) \neq \emptyset$ and $\overline{U_{s_1}} \cap\operatorname{set}(C) \neq \emptyset$ and $P(s_0) = N_{K+1}$. We will show $P(s_1) = N_{K+1}$. If this is not the case then there exist $i_l, i_{l+1} \in \operatorname{set}(C)$ with $i_l \in \overline{U_{s_0'}}$ and $i_{l+1} \in \overline{U_{s_1'}}$ such that either $P(s_0') = N_{K+1}$ and $P(s_1') \gt N_{K+1}$ or $P(s_0') \gt N_{K+1}$ and $P(s_1') = N_{K+1}$. Without loss of generality $P(s_0') = N_{K+1}$ and $P(s_1') \gt N_{K+1}$. Letting Y be the finite set $\{s \in \mathbb{Q}: P(s) \leq K \wedge \overline{U_s} \subseteq (i_l, i_{l+1})\}$, we have $U \upharpoonright ((i_0, i_1) \setminus \bigcup_{s \in Y} \overline{U_s}) \sim E$ as witnessed by the reduction scheme $\mathcal{S}' = \{C' \in \mathcal{S}: \operatorname{set}(C') \cap (i_l, i_{l+1}) \neq \emptyset \wedge \operatorname{set}(C) \cap \bigcup_{s \in Y}\overline{U_s} = \emptyset\}$ (here we are using the fact that our induction hypothesis implies that $\bigcup_{s \in Y}\overline{U_s} = \bigcup_{C' \in \mathcal{S}, \operatorname{set}(C') \cap \bigcup_{s \in Y}\overline{U_s} \neq \emptyset} \operatorname{set}(C')$). However the set $M = (i_0, i_1) \cap \{i \in \overline{U}: d((U(i))) = N_{K+1}\} \setminus \bigcup_{s \in Y}\overline{U_s}$ is of odd cardinality, and we have $U(i) \in \{h_{N_{K+1}}^{R(N_{K+1})}, h_{N_{K+1}}^{-R(N_{K+1})}\}$ for each $i \in M$, and since $h_{N_{K+1}}$ is of infinite order we get, in particular, that $p_{N_{K+1}}(U \upharpoonright (i_0, i_1) \setminus \bigcup_{s \in Y} \overline{U_s}) = \sum_{i \in M} U(i)$ is not trivial, contradiction. What we have just shown is that for each $C \in \mathcal{S}$, the function $P\circ Q \upharpoonright \operatorname{set}(C)$ is constant. We also know that for each $C \in \mathcal{S}$, the function $Q \upharpoonright \operatorname{set}(C)$ is injective, since each U_s is reduced.