Proof. Note that C is also M-separated by inclusion. Let $C^\times = \{c\in C : F\cap Mc \neq \varnothing \}$ . Observe that $C^{\times }$ is finite, as $C^\times \subset M^{-1}F$ . By the fact that C is M-separated, we have

$$ \begin{align*} |F\cap MC| \leq \sum_{c\in C^\times}|M| = |M||C^\times|. \end{align*} $$

Now let $C^\circ = \{c\in C : Lc\subset F\}$ . Observe that $C^\circ \subset C^\times $ . We emphasize that $C^\times $ is given in terms of M, while $C^\circ $ is given in terms of L. By definition, we have $LC^\circ \subset F$ , in which case

$$ \begin{align*} |F| \geq |LC^\circ| = |L||C^\circ|, \end{align*} $$

where the equality is a consequence of the fact that C is L-separated. Hence ${|C^\circ | \leq |F|/|L|}$ .

Let $c \in C^\times \setminus C^\circ $ be fixed, in which case $F\cap Mc \neq \varnothing $ and $Lc \not \subset F$ . Therefore, if ${c\in F}$ , then $c \in \partial _L F$ , while if $c \notin F$ , then $c \in M^{-1}F\setminus F$ . This demonstrates that

$$ \begin{align*} C^\times \setminus C^\circ \subset (\partial_L F) \cup (M^{-1}F\setminus F). \end{align*} $$

From this, we see that

$$ \begin{align*} |F\cap MC| &\leq |M||C^\times|\\ &= |M||C^\circ| + |M||C^\times \setminus C^\circ|\\ &\leq |M|\frac{|F|}{|L|} + |M||\partial_L F| + |M| |M^{-1}F\setminus F|. \end{align*} $$

After dividing by $|F|$ , we obtain the conclusion.

Proof. In [Reference Downarowicz and Huczek6], $n_1$ is chosen as $n_0 + 1$ and $n_i$ is inductively chosen so that $F_{n_i}$ is $(F_{n_j}, \delta _j)$ -invariant for every $j < i$ , where $\delta _j> 0$ is specified in the construction. From this, we infer that $n_i - n_{i-1}$ may be arbitrarily large for each $i = 1, \ldots , r$ . We therefore additionally assume that for each $i = 1, \ldots , r$ , we have the following properties.

1. (P1) $|F_{n_i} \cap F_{n_i}\ell | \geq (1-\varepsilon )|F_{n_i}|$ for every $\ell \in L^{-1}L$ .

2. (P2) $F_{n_{i-1}} L^{-1} L \subset F_{n_i}$ .

This is possible because the sequence $(F_n)_n$ was chosen to be both left and right Følner, to be ascending in n, and to satisfy $\bigcup _n F_n = G$ (Definition 2.1).

Let $S_i = F_{n_i}$ for each $i = 1, \ldots , r$ and let ${\mathcal S} = \{S_1, \ldots , S_r\}$ . Our modifications to the choice of ${\mathcal S}$ preserve the invariance conditions assumed by [Reference Downarowicz and Huczek6]. We now define and construct everything else as in [Reference Downarowicz and Huczek6], summarized below.

Let $x \in X$ be fixed. In [Reference Downarowicz and Huczek6], the quasi-tiling $t = {\mathcal T}(x) \in \Lambda ({\mathcal S})^G$ is constructed inductively, with the tiles of shape $S_r$ chosen first, then $S_{r-1}$ , and so on, down to $S_1$ . Consequently, for each $c\in C(t)$ , there is a well-defined subset $C(t)_{<c} \subset C(t)$ that denotes the centers of all the tiles laid before the tile at c in the inductive process. Some tiles of the same shape are laid simultaneously, so the implied ordering given by $c_1 < c_2$ if $c_1 \in C(t)_{<c_2}$ is not total. However, tiles laid simultaneously in the construction of [Reference Downarowicz and Huczek6] are necessarily disjoint.

This in hand, the retraction map $t\mapsto \operatorname {\mathrm {ret}}(t)$ is given by

$$ \begin{align*} \operatorname{\mathrm{ret}}(t)(c) = t(c) \setminus \bigg( \bigcup_{c_0 \in C(t)_{<c}} t(c_0)c_0\bigg)c^{-1} \end{align*} $$

for each $c \in C(t)$ , and $\operatorname {\mathrm {ret}}(t)(g) = \varnothing $ otherwise. In [Reference Downarowicz and Huczek6], it is shown by induction that this is a disjoint retraction satisfying $|t(c)\setminus \operatorname {\mathrm {ret}}(t)(c)| < \varepsilon |t(c)|$ for each $c\in C(t)$ . Moreover, it is quick to check that the map $t\mapsto \operatorname {\mathrm {ret}}(t)$ is continuous and shift-commuting, as the elements $c_0 \in C(t)_{<c}$ with $t(c)c \cap t(c_0)c_0 \neq \varnothing $ are determined by $\sigma ^c(t)(F)$ for some (possibly very large) finite subset $F \subset G$ . The construction in [Reference Downarowicz and Huczek6] also gives that t is $(1-\varepsilon )$ -covering.

Let $T = {\mathcal T}(X) \subset \Lambda ({\mathcal S})^G$ . From the observations of the previous paragraph, we see that properties (1) through (4) hold. For the theorem, it remains to check property (5): that $C(t)$ is L-separated for each $t \in T$ . Let $t \in T$ be fixed and suppose in contrast that ${Lc_1 \cap Lc_2 \neq \varnothing} $ for some distinct $c_1 \neq c_2 \in C(t)$ , in which case $c_1c_2^{-1} \in L^{-1}L$ . Let $t(c_1) = S_i$ and $t(c_2) = S_j$ and suppose without loss of generality that $i \leq j$ .

If $i = j$ , then let $S = S_i = S_j$ , in which case $|Sc_1 \cap Sc_2| = |S\cap Sc_1c_2^{-1}| \geq (1-\varepsilon )|S|$ by the assumed property (P1). Note that $R_1 = \operatorname {\mathrm {ret}}(t)(c_1) \subset S$ and $R_2 = \operatorname {\mathrm {ret}}(t)(c_2) \subset S$ are subsets such that $|S \setminus R_1| < \varepsilon |S|$ and $|S \setminus R_2| < \varepsilon |S|$ by construction. Moreover, by construction, $R_1c_1 \cap R_2c_2 = \varnothing $ , in which case $Sc_1 \cap Sc_2 \subset (Sc_1 \setminus R_1 c_1) \cup (Sc_2 \setminus R_2 c_2)$ . This implies that

$$ \begin{align*} |Sc_1 \cap Sc_2| \leq |Sc_1 \setminus R_1c_1| + |Sc_2 \setminus R_2 c_2| < 2\varepsilon|S|. \end{align*} $$

We thus obtain $(1-\varepsilon )|S| \leq |Sc_1\cap Sc_2| < 2\varepsilon |S|$ , which contradicts the assumption that $\varepsilon < 1/3$ .

In the second case, suppose $i < j$ . Then, $c_1 c_2^{-1} \in L^{-1} L$ implies that ${S_ic_1c_2^{-1} \subset S_iL^{-1}L \subset S_j}$ by the assumed property (P2). Hence, $t(c_1)c_1 \subset t(c_2)c_2$ . This does not immediately contradict the $\varepsilon $ -disjointness of t, as it may be the case that $t(c_1)c_1 \subset t(c_2)c_2 \setminus \operatorname {\mathrm {ret}}(t)(c_2)c_2$ . However, here we appeal to a property of the retraction map that is easily checked by induction. For each $c \in C(t)$ , we have that

(R1) $$ \begin{align} \bigcup_{c_0 \in C(t)_{<c}} t(c_0)c_0 = \bigcup_{c_0 \in C(t)_{<c}} \operatorname{\mathrm{ret}}(t)(c_0)c_0. \end{align} $$

Moreover, the assumption that $i < j$ implies that $c_2 \in C(t)_{<c_1}$ , as the tiles are placed in order of largest to smallest. In that case, we have that

$$ \begin{align*} \operatorname{\mathrm{ret}}(t)(c_1)c_1 \subset t(c_1)c_1 \subset t(c_2)c_2 \subset \bigcup_{c_0 \in C(t)_{<c_1}} t(c_0)c_0. \end{align*} $$

This, together with property (R1), implies that $\operatorname {\mathrm {ret}}(t)(c_1)c_1$ intersects $\operatorname {\mathrm {ret}}(t)(c_0)c_0$ for some $c_0 \in C(t)_{<c_1}$ , which contradicts the disjointness of $\operatorname {\mathrm {ret}}(t)$ .

This covers all cases, so we conclude that $C(t)$ must be L-separated for every $t \in T$ . With this, we have verified all properties for the theorem not already proved in [Reference Downarowicz and Huczek6].

Proof. Let $K\subset G$ be a finite subset containing e that witnesses the strong irreducibility of Y and write $K = \{e,k_1,k_2,\ldots ,k_n\}$ . Because Y has no global period, for each $i \leq n$ , there exists a point $y_i \in Y$ and an element $g_i \in G$ such that $\sigma ^{k_i}(y_i)(g_i) \neq y_i(g_i)$ .

Let $F = \{e, g_1, g_2, \ldots , g_n\}$ and let $g\in G\setminus \{e\}$ be chosen arbitrarily. Because Y is non-trivial and strongly irreducible, if $g \in G \setminus K$ , then we may construct a point $y \in Y$ such that $y(e) \neq y(g)$ , in which case $y(F) \neq \sigma ^g(y)(F)$ . If instead $g = k_i$ for some $i\leq n$ , then the fact that $y_i(g_i) \neq \sigma ^{k_i}(y_i)(g_i)$ implies that $y_i(F) \neq \sigma ^g(y_i)(F)$ .

We have shown that for every $g\neq e$ , there is a point $y\in Y$ for which $y(F) \neq \sigma ^g(y)(F)$ , in which case Y is conjugate to a subshift that separates elements of G by the observation in the paragraph following Definition 3.1, and the lemma is completed.

Proof. Let ${\mathcal A}$ be a finite alphabet such that $Y \subset {\mathcal A}^G$ . Let $K \subset G$ be a finite subset with ${e \in K}$ chosen to witness Y as a strongly irreducible SFT. We shall abbreviate $\operatorname {\mathrm {int}}^n F = \operatorname {\mathrm {int}}_{K^n} F$ and $\partial ^n F = \partial _{K^n} F$ for each natural $n \in \mathbb {N}$ and finite subset $F \subset G$ for the remainder of this proof. We shall also abbreviate $\partial ^n p = p(\partial ^n F)$ for each pattern p of shape F.

Choose $\varepsilon> 0$ such that

$$ \begin{align*} \varepsilon < \min\bigg(\frac{b-a}{3+\log 2 + 2\log\!|{\mathcal A}|}, \frac{b-a}{5+4\log\!|{\mathcal A}|}\bigg). \end{align*} $$

It is a theorem of Frisch and Tamuz [Reference Frisch and Tamuz10, Theorem 2.1] that for any $(K,\delta )$ , there exists a strongly irreducible system of disjoint, $(1-\delta )$ -covering quasi-tilings of G whose every shape is $(K,\delta )$ -invariant and whose entropy is less than $\delta $ . This, in combination with Theorem 2.11 and the fact that strong irreducibility is preserved under factor maps, implies the following. There exists a finite collection of shapes ${\mathcal S}$ and a strongly irreducible system $T \subset \Lambda ({\mathcal S})^G$ of exact tilings of G such that $h(T) < \varepsilon $ and every shape $S\in {\mathcal S}$ satisfies the following:

1. (S1) $K \subset \operatorname {\mathrm {int}}^2 S$ ;

2. (S2) $|S|> \varepsilon ^{-1}$ and $2|S| < e^{\varepsilon |S|}$ ;

3. (S3) $|\partial ^2 S| < \varepsilon |S|$ ; and

4. (S4) $h(S,\tilde {Y}) < h(\tilde {Y}) + \varepsilon $ .

For the majority of this proof, we operate primarily in the product system $Z_0 = Y \times T$ . For each finite subset $F \subset G$ and pattern $p \in {\mathcal P}(F,Z_0)$ , we shall write $p = (p^Y,p^T)$ , where $p^Y \in {\mathcal P}(F,Y)$ and $p^T \in {\mathcal P}(F,T)$ . If $F = S$ for some $S \in {\mathcal S}$ , then we shall describe p as a ‘block’.

A block $b \in {\mathcal P}(S,Z_0)$ is called aligned if $b^T(e) = S$ (note $e\in K$ and $K\subset S$ by property (S1)). Note that if b is aligned, then $b^T(s) = \varnothing $ for each $s \in S\setminus \{e\}$ , by the fact that every tiling $t\in T$ is disjoint. For a given subshift $Z \subset Z_0$ , we denote the subcollection of all aligned blocks of shape S allowed in Z by

$$ \begin{align*} {\mathcal P}^a(S,Z) \subset {\mathcal P}(S,Z) \subset ({\mathcal A} \times \Lambda({\mathcal S}))^S, \end{align*} $$

where the superscript a identifies the subcollection.

Let $\pi : Z_0 \to Y$ be the projection map defined by $\pi (y,t) = y$ for each $(y,t) \in Z_0$ , which is a homomorphism. For each subshift $Z \subset Z_0$ and each fixed $z = (y,t) \in Z$ , we have $y = \pi (z) \in \pi (Z)$ and $t \in T$ , thus $Z \subset \pi (Z) \times T$ . Consequently, for each subshift $Z \subset Z_0$ , it holds that

$$ \begin{align*} h(Z) \leq h(\pi(Z) \times T) = h(\pi(Z)) + h(T) < h(\pi(Z)) + \varepsilon, \end{align*} $$

where above we have used Lemma 2.7 and the fact that $h(T) < \varepsilon $ .

Let $S \in {\mathcal S}$ be fixed. Here we choose and fix a collection of aligned blocks $\mathcal {W}(S) \subset {\mathcal P}^a(S,Z_0)$ . We shall refer to these as ‘witness patterns’, because they will later allow us to demonstrate (‘witness’) for each $g \in G$ a point $z \in Z_0$ such that $z^Y(e) \neq z^Y(g)$ . These witness patterns shall be of three types.

For the first type, let $s\in S\setminus \{e\}$ be arbitrary. Because Y separates elements of G, there is a pattern $b^Y \in {\mathcal P}(S,Y)$ such that $b^Y(e) \neq b^Y(s)$ . If $b^T \in {\mathcal P}(S,T)$ is given by $b^T(e) = S$ and $b^T(s) = \varnothing $ for each $s \in S\setminus \{e\}$ , then $b = (b^Y, b^T)$ is an allowed block in $Z_0$ that satisfies $b^Y(e) \neq b^Y(s)$ . For each $s \in S\setminus \{e\}$ , pick and save one such block b to the collection $\mathcal {W}(S)$ .

For the second type, let $0 \in {\mathcal A}$ be a distinguished symbol of the alphabet. For every $y\in Y$ , we can find a block $b \in {\mathcal P}(S,Z_0)$ such that $b(e) = (0,S)$ and $b^Y(\partial ^2 S) = y(\partial ^2 S)$ . This is because Y is strongly irreducible of parameter K and property (S1). For each pattern $y(\partial ^2 S) \in {\mathcal P}(\partial ^2 S, Y)$ , pick and save one such block b to the collection $\mathcal {W}(S)$ .

For the third type, let $1 \in {\mathcal A}$ be a distinguished symbol different from $0$ . For each $s \in S$ , we can find a block $b \in {\mathcal P}(S,Y)$ such that $b^T(e) = S$ and $b^Y(s) = 1$ . For each $s \in S$ , pick and save one such block b to the collection $\mathcal {W}(S)$ . This completes the description of the collection $\mathcal {W}(S)$ . Note that $|\mathcal {W}(S)| \leq 2|S| + {\mathcal A}^{|\partial ^2 S|} < e^{\varepsilon |S|} \cdot |{\mathcal A}|^{\varepsilon |S|}$ by properties (S2) and (S3).

We now construct a descending chain of subshifts $(Z_n)_n$ of $Z_0$ and claim that the subshift $Y_0$ desired for the theorem shall be given by $Y_0 = \pi (Z_n)$ for some $n \leq N$ . Suppose for induction that $Z_n \subset Z_0$ has been constructed for $n \geq 0$ . If there exists a shape $S_n \in {\mathcal S}$ and an aligned block $\beta _n \in {\mathcal P}^a(S_n, Z_n)$ such that:

1. (B1) $\beta _n^Y$ does not appear in $\tilde {Y}$ ( $\beta _n^Y \notin {\mathcal P}(S_n, \tilde {Y})$ );

2. (B2) $\beta _n$ is not one of the reserved witness patterns ( $\beta _n \notin \mathcal W(S_n)$ ); and

3. (B3) there exists an aligned block $b \in {\mathcal P}^a(S_n, Z_n)$ with $b \neq \beta _n$ and $\partial ^2 b = \partial ^2 \beta _n$ ,

then let $Z_{n+1} = \langle Z_n \mid \beta _n\rangle $ . If no such block exists for any shape, then the descending chain is finite in length and $Z_n$ is the terminal subshift.

In fact, the chain must be finite in length. This is because $Z_{n+1} \subset Z_n$ implies ${\mathcal P}(S,Z_{n+1}) \subset {\mathcal P}(S,Z_n)$ for each $S \in {\mathcal S}$ , and ${\mathcal P}(S_n,Z_{n+1}) \sqcup \{\beta _n\} \subset {\mathcal P}(S_n, Z_n)$ for each $n \geq 0$ . Hence, we have that

$$ \begin{align*} \sum_{S\in{\mathcal S}} |{\mathcal P}(S,Z_n)| \end{align*} $$

is a non-negative integer sequence that strictly decreases with n, and therefore must terminate. Let $N \geq 0$ be the index of the terminal subshift, and note by construction that for each $S \in {\mathcal S}$ and each aligned block $b \in {\mathcal P}^a(S, Z_N)$ , either b is uniquely determined by $\partial ^2 b$ , or $b \in \mathcal {W}(S)$ , or $b^Y \in {\mathcal P}(S,\tilde {Y})$ .

We note the following intermediate lemma that shall be referenced multiple times in the remainder of this proof.

Now continuing the proof of Theorem 3.2, we claim that for each n, the subshift $\pi (Z_n)$ is strongly irreducible, separates elements of G, and satisfies $\tilde {Y} \subset \pi (Z_n) \subset Y$ . Moreover, we claim that $h(\pi (Z_n)) - h(\pi (Z_{n+1})) < b-a$ for each $n < N$ , and $h(\pi (Z_N)) - h(\tilde {Y}) < b-a$ .

For $\tilde {Y} \subset \pi (Z_n) \subset Y$ , let $\tilde {y} \in \tilde {Y}$ and $t \in T$ be arbitrary, in which case $(\tilde {y},t) \in Z_0$ . Note that for each $n < N$ , the block $\beta _n$ cannot appear in $(\tilde {y},t)$ , else that would imply that $\beta _n^Y$ appears in $\tilde {y}$ , contradicting property (B1). Thus, $(\tilde {y},t) \in Z_n$ for each ${n \leq N}$ . This demonstrates that $\tilde {Y} \times T \subset Z_n$ , and in particular $\tilde {Y} \subset \pi (Z_n) \subset Y$ , for each $n \leq N$ .

For the strong irreducibility, note that $Z_0 = Y \times T$ is strongly irreducible because both Y and T are strongly irreducible. For induction, suppose $Z_n$ is strongly irreducible of parameter $K_n \subset G$ (with $e \in K_n$ ) for a fixed $n < N$ . Let $U = \bigcup _{S\in {\mathcal S}} S$ . We claim that $Z_{n+1}$ is strongly irreducible of parameter $UU^{-1} K_n UU^{-1}$ .

Indeed, let $(y_1,t_1)$ , $(y_2,t_2) \in Z_{n+1}$ be arbitrary points, and let $F_1$ , $F_2 \subset G$ be arbitrary finite subsets with $K_nUU^{-1} F_1 \cap UU^{-1} F_2 = \varnothing $ . As $Z_{n+1} \subset Z_n$ and $Z_n$ is strongly irreducible of parameter $K_n$ , there is a point $(y,t) \in Z_n$ with $(y,t)(UU^{-1}F_1) = (y_1,t_1)(UU^{-1}F_1)$ and $(y,t)(UU^{-1}F_2) = (y_2,t_2)(UU^{-1}F_2)$ .

Now consider the forbidden block $\beta _n$ of shape $S_n \in {\mathcal S}$ . Suppose $\sigma ^g(y,t)(S_n) = \beta _n$ for some $g \in G$ with $S_ng \cap F_1 \neq \varnothing $ . It follows that $g \in S_n^{-1} F_1$ , and hence $S_ng \subset S_nS_n^{-1} \subset UU^{-1}F_1$ , and hence $\sigma ^g(y_1,t_1)(S_n) = \sigma ^g(y,t)(S_n) = \beta _n$ , contradicting the fact that $(y_1,t_1) \in Z_{n+1}$ and $\beta _n$ is forbidden in $Z_{n+1}$ . From this observation (and by an identical argument for $F_2$ ), we see that if $\beta _n$ appears anywhere in $(y_0,t)$ , then it does not appear on any tile of t that intersects either $F_1$ or $F_2$ .

Let $(y^{*},t) \in Z_{n+1}$ be the point delivered by Lemma 3.3 as applied to $(y,t)$ . From the previous paragraph, we have $(y^{*},t)(F_i) = (y,t)(F_i) = (y_i,t_i)(F_i)$ for each $i = 1,2$ . We conclude that $Z_{n+1}$ is strongly irreducible of parameter $UU^{-1}K_n UU^{-1}$ , therefore completing the induction. As strong irreducibility is preserved under taking factors, it follows that $\pi (Z_n)$ is strongly irreducible for each $n \leq N$ .

To see that each $\pi (Z_n)$ separates elements of G, let $n \leq N$ and $g \neq e$ be fixed. We proceed by cases on g.

If $g \in S$ for some $S \in {\mathcal S}$ , then there is a witness block $b \in \mathcal {W}(S)$ such that $b^Y(e) \neq b^Y(g)$ . Choose and fix $(y_0, t) \in Z_0$ such that $(y_0,t)(S) = b$ and apply Lemma 3.3 at most N times to produce the point $(y_n,t) \in Z_n$ . Note $(y_n,t)(S) = (y_0,t)(S) = b$ , as $b \neq \beta _n$ for every $n < N$ by property (B2). Consequently, $y_n \in \pi (Z_n)$ satisfies $y_n(e) = b^Y(e) \neq b^Y(g) = y_n(g)$ .

If $g \notin S$ for any $S \in {\mathcal S}$ , then pick $S_0 \in {\mathcal S}$ arbitrarily and fix $t \in T$ with $t(e) = S_0$ . Because t is an exact tiling, there is a unique $c \neq e$ such that $g\in t(c)c$ . Let $S_1 = t(c)$ and write $g = s_1c$ for some $s_1 \in S_1 \in {\mathcal S}$ . Recall $0$ , $1 \in {\mathcal A}$ are two distinguished symbols determined in the construction of $\mathcal {W}(S)$ . By construction, there is a witness block ${b_1 \in \mathcal {W}(S_1)}$ with $b_1^Y(s_1) = 1$ . Pick any $y \in Y$ with $\sigma ^c(y)(S_1) = b_1$ . By construction, there is a witness block $b_0 \in \mathcal {W}(S_0)$ with $b_0^Y(e) = 0$ and $\partial ^2 b_0^Y = y(\partial ^2 S_0)$ . Because Y is an SFT of parameter K and by property (S1) and Lemma 2.5, there is a point $y^{*} \in Y$ given by $y^{*}(S_0) = b_0^Y$ and $y^{*}(g) = y(g)$ otherwise. In particular, $\sigma ^c(y^{*})(S_1) = b_1^Y$ . Then, apply Lemma 3.3 at most N times to the initial point $(y^{*},t) \in Z_0$ , thus yielding $(y_n, t) \in Z_n$ . Observe that $(y_n,t)(S_0) = (y^{*},t)(S_0) = b_0$ and $\sigma ^c(y_n,t)(S_1) = \sigma ^c(y^{*},t)(S_1) = b_1$ , because $b_0 \neq \beta _n \neq b_1$ for every $n < N$ by property (B2). Consequently, $y_n \in \pi (Z_n)$ satisfies $y_n(e) = 0 \neq 1 = y_n(g)$ . This finishes all cases and demonstrates that $\pi (Z_n)$ separates elements of G for each $n \leq N$ .

The proofs that $h(\pi (Z_n)) - h(\pi (Z_{n+1})) < b-a$ for every $n < N$ and $h(\pi (Z_N)) - h(\tilde {Y}) < b-a$ are nearly identical to arguments appearing in [Reference Bland, McGoff and Pavlov2, Theorem 4.1]. We proceed quickly through the argument here. Choose a finite subset $F\subset G$ such that $|h(Z_n) - h(F,Z_n)| < \varepsilon $ for every $n\leq N$ , $|h(F,T) - h(T)| < \varepsilon $ (this implies in particular that $h(F,T) < 2\varepsilon $ by choice of T), and $|F \setminus U^{-1}F| < \varepsilon |F|$ , where $U = \bigcup _{S\in {\mathcal S}} S$ . For each $n \leq N$ , let

$$ \begin{align*} P(n) = \sum_{t(F)} \prod_c |{\mathcal P}^a(\operatorname{\mathrm{int}} t(c), Z_n)|, \end{align*} $$

where the sum is taken over all patterns $t(F) \in {\mathcal P}(F,T)$ and the product is taken over all $c\in C(t) \cap U^{-1}F \cap F$ . For each $n \leq N$ , it holds that

$$ \begin{align*} P(n) \leq |{\mathcal P}(F,Z_n)| \leq |{\mathcal A}|^{2\varepsilon|F|}\cdot P(n). \end{align*} $$

The first inequality follows from the strong irreducibility of Y in combination with Lemma 3.3. The latter inequality follows from projecting a pattern $p\in {\mathcal P}(F,Z_n)$ to the interiors of the tiles described by $p^T = t(F) \in {\mathcal P}(F,T)$ that are contained in F. This projection determines p up to the portion of F not covered by those tile interiors, a subset of F of size at most $2\varepsilon |F|$ by the combination of the fact that the tiling is exact, the assumed invariance condition on F, and property (S3).

Moreover, for each $n < N$ , it holds that $P(n) \leq 2^{\varepsilon |F|}\cdot P(n+1)$ . This follows from the fact that $|{\mathcal P}^a(\operatorname {\mathrm {int}} S,Z_n)| - |{\mathcal P}^a(\operatorname {\mathrm {int}} S, Z_{n+1})| \leq 1$ (at most one aligned block is removed as one passes from $Z_n$ to $Z_{n+1}$ ), in combination with the assumed invariance condition on F.

The above, in combination with the assumed entropy estimating properties of F, implies that

$$ \begin{align*} h(Z_n) < h(Z_{n+1}) + 2\varepsilon + \varepsilon\log 2 + 2\varepsilon\log\!|{\mathcal A}|. \end{align*} $$

This, together with the fact that $h(Z_n) < h(\pi (Z_n)) +\varepsilon $ for each $n \leq N$ and the choice of $\varepsilon $ , finally gives that $h(\pi (Z_n)) - h(\pi (Z_{n+1}) < b-a$ holds for each $n < N$ .

Next, consider the terminal subshift $Z_N$ . Recall that for every shape $S\in {\mathcal S}$ , every aligned block $b\in {\mathcal P}^a(S,Z_N)$ either belongs to $\mathcal {W}(S)$ , is uniquely determined by $\partial ^2 b$ , or satisfies $b^Y \in {\mathcal P}(S,\tilde {Y})$ by construction. Recall also that $|\mathcal {W}(S)| \leq e^{\varepsilon |S|} \cdot |{\mathcal A}|^{\varepsilon |S|}$ . This, together with property (S4), implies that for every shape $S\in {\mathcal S}$ , it holds that

$$ \begin{align*} |{\mathcal P}^a(\operatorname{\mathrm{int}} S, Z_N)| \leq e^{h(\tilde{Y})|S|}\cdot e^{2\varepsilon|S|} \cdot |{\mathcal A}|^{2\varepsilon|S|}. \end{align*} $$

This, together with the facts that $h(F,T) < 2\varepsilon $ and $|{\mathcal P}(F,Z_N)| \leq |{\mathcal A}|^{2\varepsilon |F|}\cdot P(N)$ argued earlier, gives us that

$$ \begin{align*} |{\mathcal P}(F,Z_N)| \leq e^{h(\tilde{Y})|F|}\cdot e^{4\varepsilon|F|} \cdot |{\mathcal A}|^{4\varepsilon|F|}, \end{align*} $$

in which case it follows that $h(Z_N) < h(\tilde {Y}) + 5\varepsilon + 4\varepsilon \log \!|{\mathcal A}|$ . This and our choice of $\varepsilon $ finally give that $h(\pi (Z_N)) - h(\tilde {Y}) < b-a$ .

Recall that $h(\tilde {Y}) < a < b <\! h(Y)$ , and recall also that $\pi (Z_0)\!=\!Y$ and thus ${h(\pi (Z_0))\!>\!b}$ . We have demonstrated that $h(\pi (Z_n)) - h(\pi (Z_{n+1})) < b-a$ for every $n < N$ and $h(\pi (Z_N)) - h(\tilde {Y}) < b-a$ . There must therefore exist at least one $n\leq N$ for which $\pi (Z_n)$ satisfies $a < h(\pi (Z_n)) < b$ , thus completing the proof.

Proof. Because $h(Y_1) < h(Y)$ , there exists a finite subset $F \subset G$ with $e \in F$ such that $|{\mathcal P}(F,Y) \setminus {\mathcal P}(F,Y_1)|> r$ . Choose and fix r distinct patterns $a_1, \ldots , a_r \in {\mathcal P}(F,Y) \setminus {\mathcal P}(F,Y_1)$ .

Choose a finite subset $K \subset G$ with $e \in K$ to witness the strong irreducibility of Y and $Y_0$ . Let $\{e, g_1, \ldots , g_N\}$ be an enumeration of $F^{-1}KF$ . Let $M_0 = KF$ and suppose for induction that $M_{n-1} \subset G$ has been constructed for a fixed $n \in [1,N]$ . Because G is infinite, by the pigeonhole principle, there exists an element $h_n \in G$ such that $Kh_n \cup Kh_ng_n$ is disjoint from $M_{n-1}$ . Let $M_n = M_{n-1} \sqcup (Kh_n \cup Kh_ng_n)$ . We claim that $M = M_N$ is the subset desired for the theorem. Note that

$$ \begin{align*} M = KF \sqcup \bigg(\bigsqcup_{n=1}^{N} Kh_n \cup Kh_ng_n\bigg) \end{align*} $$

by induction.

Let $y_0 \in Y_0$ be arbitrary. Let $y_0^{(0)} = y_0$ and suppose for induction that $y_0^{(n-1)} \in Y_0$ has been constructed for a fixed $n \in [1,N]$ . Because $Y_0$ separates elements of G, there exists a point $y^{*} \in Y_0$ with $y^{*}(h_n) \neq y^{*}(h_ng_n)$ . Because $Y_0$ is strongly irreducible of parameter K, we may find a point $y_0^{(n)} \in Y_0$ such that $y_0^{(n)}(\{h_n,h_ng_n\}) = y^{*}(\{h_n,h_ng_n\})$ and $y_0^{(n)}(g) = y_0^{(n-1)}(g)$ for every $g \notin Kh_n \cup Kh_ng_n$ .

By induction, the point $y_0^{(N)} \in Y_0$ satisfies $y_0^{(N)}(h_n) \neq y_0^{(N)}(h_ng_n)$ for each ${n = 1, \ldots , N}$ and $y_0^{(N)}(g) = y_0(g)$ for every $g \notin \bigcup _{n=1}^N Kh_n \cup Kh_ng_n$ .

For each $i = 1, \ldots , r$ , because Y is strongly irreducible of parameter K, we may find a point $y_i \in Y$ such that $y_i(F) = a_i$ and $y_i(g) = y^{(N)}(g)$ for each $g \notin KF$ . Hence, by construction, $y_i(h_n) \neq y_i(h_ng_n)$ for each $n = 1, \ldots , N$ , and $y_i(g) = y_0(g)$ for each ${g \notin M}$ . Finally, let $m_i = y_i(M)$ . We claim that these are the patterns and points desired for the theorem. See Figure 1 for an illustration of the construction.

Figure 1

An illustration of the construction of the marker pattern $m_i$ . The pattern $a_i$ is selected from Y to be forbidden in $Y_1$ . Thus, $y_i$ has no element of $G \setminus F^{-1}KF$ as a period. Then, small pairs of differing symbols are inductively mixed in (via $Y_0$ ) to prevent $y_i$ from having any period $g \in F^{-1}KF \setminus \{e\}$ . The shaded exterior is the base point $y_0 \in Y_0$ .

For property (1), let $i \leq r$ be fixed. Observe that there is no $g \in F^{-1}KF\setminus \{e\}$ such that $\sigma ^g(y_i)(M) = y_i(M)$ , as otherwise would contradict the fact that every $g \in F^{-1}KF$ has a corresponding $h \in M \cap Mg^{-1}$ such that $y_i(h) \neq y_i(hg)$ . Moreover, if $g \notin F^{-1}KF$ , then $Fg$ is disjoint from $KF$ , in which case $\sigma ^g(y_i)(F) = \sigma ^g(y_0^{(N)})(F)$ . We must therefore have $\sigma ^g(y_i)(F) \neq a_i$ , as otherwise would contradict the fact that $a_i$ is forbidden in $Y_1$ and hence also forbidden in $Y_0$ . Consequently, $\sigma ^g(y_i)(M) = y_i(M)$ only when $g = e$ .

For property (2), let $i \neq j$ be fixed. Observe that $a_j \neq a_i$ implies that $y_j(M) \neq y_i(M)$ . Moreover, $\sigma ^g(y_j)(M) \neq y_i(M)$ for any $g \neq e$ by an identical argument as in the previous paragraph. This is because $y_j(g) = y_0^{(N)}(g) = y_i(g)$ for every $g \notin KF$ , and both $a_i$ and $a_j$ are forbidden in $Y_1$ , and hence also forbidden in $Y_0$ .

Property (3) is true by construction.

Proof. Let ${\mathcal A}_X$ and ${\mathcal A}_Y$ be finite alphabets such that $X\subset {\mathcal A}_X^G$ and $Y\subset {\mathcal A}_Y^G$ . Suppose ${\phi : X \to Y}$ is a homomorphism, not necessarily injective, in which case $\tilde {Y} = \phi (X) \subset Y$ is a factor of X. Note that $h(\tilde {Y}) \leq h(X) < h(Y)$ . Without loss of generality (by Lemma 3.1), we may assume that Y separates elements of G (Definition 3.1). In that case, by Theorem 3.2, there exists a strongly irreducible subshift $Y_0$ that separates elements of G and satisfies $\tilde {Y} \subset Y_0 \subset Y$ and $h(X) < h(Y_0) < h(Y)$ . By [Reference Bland, McGoff and Pavlov2, Theorem 4.2], there exists an SFT $Y_1$ such that $Y_0 \subset Y_1 \subset Y$ and $h(Y_1) < h(Y)$ . It is significant for our proof that these inequalities are strict.

Choose a finite subset $K \subset G$ with $e\in K$ such that $K^{-1} = K$ , K witnesses Y and $Y_1$ as SFTs, and K witnesses Y and $Y_0$ as strongly irreducible. As in the proof of Theorem 3.2, we shall abbreviate $\operatorname {\mathrm {int}}^n F = \operatorname {\mathrm {int}}_{K^n} F$ and $\partial ^n F = \partial _{K^n} F = F \setminus \operatorname {\mathrm {int}}^n F$ for each natural $n \in \mathbb {N}$ and finite subset $F \subset G$ for the remainder of this proof. We shall also abbreviate $\partial ^n p = p(\partial ^n F)$ for each pattern p of shape F.

Choose $\varepsilon> 0$ such that $\varepsilon < 1/3$ and

$$ \begin{align*} \varepsilon < \frac{h(Y_0) - h(X)}{1 + 5\log\!|{\mathcal A}_X| + (5+4|K|^6)\log\!|{\mathcal A}_Y|}. \end{align*} $$

Choose $r = 1 + \lceil (2/\varepsilon )\log (1/\varepsilon )\rceil $ , in which case, $(1-\varepsilon /2)^r < \varepsilon $ . Let $M \subset G$ be the subset delivered by Theorem 3.4 for $Y_0$ , $Y_1$ , Y, and r as chosen here. By passing to a superset of M if necessary, we may assume that $K \subset M$ and that $M^{-1} = M$ . Choose a finite subset $L \subset G$ such that $M^6 \subset L$ and $|M^6|/|L| < \varepsilon $ .

Let $T_0 \subset \Lambda ({\mathcal S}_0)^G$ be the quasi-tiling system delivered by Theorem 2.9 for X, $\varepsilon $ , r, and L as chosen here, with $n_0$ chosen so that every $S_0 \in {\mathcal S}_0$ satisfies the following hypotheses.

1. (H1) $1/|S_0| < \varepsilon (1-\varepsilon )$ .

2. (H2) $(|K^3||M^6||L|)|LS_0 \setminus S_0| < \varepsilon |S_0|$ .

3. (H3) $h(S_0, X) < h(X) + \varepsilon $ .

Note that Theorem 2.9 gives that $|{\mathcal S}_0| \leq r$ , so choose and fix an enumeration $(S_0^{(i)})_{i}$ of ${\mathcal S}_0$ where $i = 1, \ldots , r$ . Theorem 2.9 also gives that every quasi-tiling $t_0 \in T_0$ is $\varepsilon $ -disjoint, as witnessed by a continuous and shift-commuting retraction map $t_0 \mapsto \operatorname {\mathrm {ret}}(t_0)$ . Let ${\mathcal S}_1$ be the set of all shapes realized as tiles of $\operatorname {\mathrm {ret}}(t_0)$ for each $t_0 \in T_0$ . Each shape $S_0 \in {\mathcal S}_0$ gives rise to a subcollection of shapes $S_1 \in {\mathcal S}_1$ , all of which satisfy $S_1 \subset S_0$ and $|S_0 \setminus S_1| < \varepsilon |S_0|$ . Let $T_1 = \operatorname {\mathrm {ret}}(T_0) \subset \Lambda ({\mathcal S}_1)^G$ be the system of all quasi-tilings obtained by taking retractions of the quasi-tilings in $T_0$ , in which case the map $\operatorname {\mathrm {ret}}$ is a factor map from $T_0$ to $T_1$ . Note that every $t_1 \in T_1$ is disjoint by Theorem 2.9. Moreover, every $t_1 \in T_1$ is $(1-\varepsilon )$ -covering, since the retraction map $\operatorname {\mathrm {ret}}$ has the property that $\bigcup _g t_0(g)g = \bigcup _g \operatorname {\mathrm {ret}}(t_0)(g)g$ for each $t_0 \in T_0$ , and every $t_0 \in T_0$ is $(1-\varepsilon )$ -covering by Theorem 2.9.

We aim to go one step further and construct a factor $T_2$ of $T_1$ , which shall be a system of exact tilings. It is in this step that we appeal to the comparison property of G. To begin, note that for each pair of shapes $S_0 \in {\mathcal S}_0$ and $S_1 \in {\mathcal S}_1$ with $S_1 \subset S_0$ and $|S_0 \setminus S_1| < \varepsilon |S_0|$ , it holds that

$$ \begin{align*} |S_1| = |S_0| - |S_0 \setminus S_1|> (1-\varepsilon)|S_0| > 1/\varepsilon, \end{align*} $$

where above we invoke hypothesis (H1). This implies that there exists an integer in the interval $[2\varepsilon |S_1|, 3\varepsilon |S_1|)$ . Therefore, for each shape $S_1 \in {\mathcal S}_1$ , we may find and fix an arbitrary subset $B(S_1) \subset S_1$ such that $2\varepsilon |S_1| \leq |B(S_1)| < 3\varepsilon |S_1|$ .

To each $t_1 \in T_1$ , we assign two disjoint subsets $A_{t_1}$ , $B_{t_1}$ of G in the following way. Let $A_{t_1} = G \setminus \bigcup _g {t_1}(g)g$ and let $B_{t_1} = \bigcup _g B({t_1}(g))g$ . Observe that the assignments ${{t_1} \mapsto A_{t_1}}$ , $B_{t_1}$ are continuous and shift-commuting. Observe also that $\overline {D}(A_{t_1}) \leq \varepsilon $ because ${t_1}$ is $(1-\varepsilon )$ -covering, and that $\underline {D}(B_{t_1}) \geq 2\varepsilon (1-\varepsilon )$ by Lemma 2.8 (with $\rho _0 = 1-\varepsilon $ and $\rho _1 = 2\varepsilon $ ). Consequently,

$$ \begin{align*} \underline{D}(B_{t_1}) - \overline{D}(A_{t_1}) \geq 2\varepsilon(1-\varepsilon) - \varepsilon> 0, \end{align*} $$

where above we have used the fact that $\varepsilon \in (0,1/2)$ .

Therefore, by Theorem 2.10, there exists a family of injections $\phi _{t_1} : A_{t_1} \to B_{t_1}$ that is induced by a block code, in the sense that there is a finite subset $F \subset G$ and a function $\Phi : {\mathcal P}(F, T_1) \to F$ such that for every $t_1\in T_1$ and $g\in A_{t_1}$ , it holds that

$$ \begin{align*} \phi_{t_1}(g) = \Phi(\sigma^g({t_1})(F))g. \end{align*} $$

With this, we are ready to construct $T_2$ . For each $t_1 \in T_1$ , let $t_2 = \operatorname {\mathrm {ex}}(t_1)$ be the quasi-tiling obtained according to the rule

$$ \begin{align*} t_2(c) = t_1(c) \sqcup \phi_{t_1}^{-1}(B(t_1(c))c)c^{-1} \end{align*} $$

for every $c \in C(t_1)$ , and $t_2(g) = \varnothing $ otherwise. In words, we expand each tile $t_1(c)c$ by including all group elements that map into $B(t_1(c))c \subset B_{t_1}$ under the injective map $\phi _{t_1}$ . Consequently, $|t_2(c) \setminus t_1(c)| \leq |B(t_1(c))| < 3\varepsilon |t_1(c)|$ for every $c \in C(t_2) = C(t_1)$ .

Let ${\mathcal S}_2$ be the set of all shapes realized as tiles of $\operatorname {\mathrm {ex}}(t_1)$ for each $t_1 \in T_1$ . There are at most finitely many because $\operatorname {\mathrm {ex}}(t_1)(g) \subset F^{-1}t_1(g)$ for every $g\in G$ . Each shape $S_1 \in {\mathcal S}_1$ gives rise to a subcollection of shapes $S_2 \in {\mathcal S}_2$ , all of which satisfy $S_1 \subset S_2$ and $|S_2 \setminus S_1|< 3\varepsilon |S_1|$ .

Let $T_2 = \operatorname {\mathrm {ex}}(T_1) \subset \Lambda ({\mathcal S}_2)^G$ . We claim that the map $\operatorname {\mathrm {ex}}$ is a homomorphism and that every $t_2 \in T_2$ is an exact tiling of G. To see that the map $\operatorname {\mathrm {ex}}$ is continuous and shift-commuting, let $t_1 \in T_1$ be arbitrary and let $t_2 = \operatorname {\mathrm {ex}}(t_1)$ . We note that for each $g \in G$ and $c \in C(t_2) = C(t_1)$ , the definition given above implies that $g\in t_2(c)$ if and only if

$$ \begin{align*} g\in t_1(c)\quad \text{or}\quad \Phi(\sigma^{gc}(t_1)(F))g \in B(t_1(c)). \end{align*} $$

Moreover, for each $f \in F$ , we have $\sigma ^{gc}(t_1)(f) = t_1(fgc) = \sigma ^c(t_1)(fg)$ . Because $t_2(c) \subset F^{-1}t_1(c)$ , we see that $t_2(c)$ depends only on $\sigma ^c(t_1)(FF^{-1}U_1)$ , where $U_1 = \bigcup _{S_1\in {\mathcal S}_1} S_1$ . This demonstrates that the map $\operatorname {\mathrm {ex}}$ is induced by a block code, which is sufficient to demonstrate that $\operatorname {\mathrm {ex}}$ is continuous and shift-commuting.

For the claimed exactness, let $t_2 \in T_2$ and $g\in G$ be arbitrary. Choose $t_1 \in T_1$ such that $t_2 = \operatorname {\mathrm {ex}}(t_1)$ . If there exists a $c \in C(t_1)$ such that $g\in t_1(c)c$ , then the c is necessarily unique by the disjointness of $t_1$ and also it follows that $g \in t_2(c)c$ . Otherwise, $g\in A_{t_1}$ , in which case $\phi _{t_1}(g) \in B_{t_1} = \bigcup _c B(t_1(c))c$ . Therefore, there exists a $c \in C(t_1)$ such that $\phi _{t_1}(g)\in B(t_1(c))c \subset t_1(c)c$ . The c must again be unique by the disjointness of $t_1$ . Then $g\in \phi _{t_1}^{-1}(B(t_1(c)c)) \subset t_2(c)c$ . We see that for each $g\in G$ , there exists a unique $c \in C(t_2)$ such that $g\in t_2(c)c$ , and hence $t_2$ is an exact tiling of G.

There is one last quasi-tiling system we shall need. Let ${\mathcal S}_1^{*}$ denote the collection of all shapes obtained in the form $\operatorname {\mathrm {int}}^3(S_1 \setminus M^6C)$ for any $S_1 \in {\mathcal S}_1$ and any L-separated subset $C\subset G$ . Each shape $S_1 \in {\mathcal S}_1$ gives rise to a subcollection of shapes $S_1^{*} \in {\mathcal S}_1^{*}$ , all of which satisfy $S_1^{*} \subset S_1$ . For each $t_1 \in T_1$ , let $t_1^{*}$ be the retraction of $t_1$ such that, for each $c\in C(t_1)$ , it holds that

$$ \begin{align*} t_1^{*}(c) = \operatorname{\mathrm{int}}^3(t_1(c) \setminus M^6C(t_1)c^{-1}) \end{align*} $$

and $t_1^{*}(g) = \varnothing $ otherwise. Observe that each such quasi-tiling belongs to $\Lambda (S_1^{*})^G$ . Let $T_1^{*} = \{t_1^{*} \in \Lambda ({\mathcal S}_1^{*})^G : t_1 \in T_1\}$ . It is quick to see that the map $t_1 \mapsto t_1^{*}$ is a factor map from $T_1$ to $T_1^{*}$ , because the assignment $t_1 \mapsto C(t_1)$ is continuous and shift-commuting. Moreover, for any choice of $t_0 \in T_0$ with corresponding $t_1$ , $t_2$ , and $t_1^{*}$ , it holds that $C(t_0) = C(t_1) = C(t_2) = C(t_1^{*})$ , thus all of these subsets are L-separated.

With our quasi-tiling systems constructed, we now aim to construct an injection that will carry patterns on tiles in X to patterns on tiles in $Y_0$ . On the side of X, we use patterns of those shapes belonging to ${\mathcal S}_2$ . On the side of $Y_0$ , we shall use patterns of those shapes belonging to ${\mathcal S}_1^{*}$ . We separate out the construction (and its rather involved estimations) into the following lemma.

Lemma 3.6. Let $(S_0, S_1, S_2, S_1^{*})$ be any tuple of shapes from ${\mathcal S}_0$ , ${\mathcal S}_1$ , ${\mathcal S}_2$ , and ${\mathcal S}_1^{*}$ , respectively, such that $S_1^{*} \subset S_1 \subset S_0 \cap S_2$ , $|S_0 \setminus S_1| < \varepsilon |S_0|$ , $|S_2 \setminus S_1| < 3\varepsilon |S_1|$ , and ${S_1^{*} = \operatorname {\mathrm {int}}^3(S_1 \setminus M^6C)}$ for some L-separated subset $C\subset G$ . Then

$$ \begin{align*} |{\mathcal P}(S_2, X)| < |{\mathcal P}(S_1^{*}, Y_0)|. \end{align*} $$

Proof. We begin on the side of X. By the hypotheses, a quick calculation shows that $|S_0 \triangle S_2| < 5\varepsilon |S_0|$ . It therefore holds that

(X1) $$ \begin{align} |{\mathcal P}(S_2, X)| \leq |{\mathcal P}(S_0, X)| \cdot |{\mathcal A}_X|^{|S_0 \triangle S_2|} < e^{(h(X)+\varepsilon)|S_0|} \cdot |{\mathcal A}_X|^{5\varepsilon|S_0|}, \end{align} $$

where above we invoke hypothesis (H3) in the latter inequality.

Now we shall estimate the number of patterns in $Y_0$ of shape $S_1^{*}$ . We begin by estimating the size of $S_1^{*}$ in terms of the size of $S_0$ . We proceed in stages, beginning with $S_1$ . We first note that

$$ \begin{align*} |S_1 \cap M^6C| & \leq |S_0 \cap M^6C|\\ &\leq \frac{|M^6|}{|L|}|S_0| + |M^6||\partial_L S_0| + |M^6||(M^6)^{-1}S_0 \setminus S_0|\\ &< \varepsilon|S_0| + \varepsilon|S_0| + \varepsilon|S_0|\\ & = 3\varepsilon|S_0|, \end{align*} $$

where above we have used the fact that $S_1 \subset S_0$ , Lemma 2.4, the choice of L, Lemma 2.3, and hypothesis (H2) in conjunction with the fact that $(M^6)^{-1} = M^6 \subset L$ . It follows that

$$ \begin{align*} |S_1 \setminus M^6C| = |S_1| - |S_1 \cap M^6C| \geq (1-\varepsilon)|S_0| - 3\varepsilon|S_0| = (1-4\varepsilon)|S_0|. \end{align*} $$

This, together with the fact that $S_1 \setminus M^6C \subset S_1 \subset S_0$ , gives us

$$ \begin{align*} |\partial^3(S_1 \setminus M^6C)| &\leq |K^3||K^3(S_1 \setminus M^6C) \setminus (S_1\setminus M^6C)|\\ &\leq |K^3||K^3S_0 \setminus S_0| + |K^3|^2|S_0 \setminus(S_1 \setminus M^6C)|\\ &< \varepsilon|S_0| + |K|^6 \cdot 4\varepsilon|S_0|\\ &= \varepsilon(1 + 4|K|^6)|S_0|, \end{align*} $$

where above we have used Lemmas 2.3, 2.1, hypothesis (H2) in conjunction with the fact that $K^3 \subset L$ , and the calculation of the previous display.

The combination of the previous two displays gives us

$$ \begin{align*} |\!\operatorname{\mathrm{int}}^3(S_1 \setminus M^6C)| &= |S_1 \setminus M^6C| - |\partial^3(S_1 \setminus M^6C)| \\ & \geq (1-4\varepsilon)|S_0| - \varepsilon(1+4|K|^6)|S_0| \\ &= |S_0| - \varepsilon(5+4|K|^6)|S_0|. \end{align*} $$

Recall that $S_1^{*} = \operatorname {\mathrm {int}}^3(S_1 \setminus M^6C)$ and note that $|{\mathcal P}(S_0, Y_0)| \leq |{\mathcal P}(S_1^{*}, Y_0)| \cdot |{\mathcal A}_Y|^{|S_0 \setminus S_1^{*}|}$ . This, in combination with the previous display, gives that

$$ \begin{align*} |{\mathcal P}(S_1^{*}, Y_0)| &\geq |{\mathcal P}(S_0, Y_0)| \cdot |{\mathcal A}_Y|^{-|S_0 \setminus S_1^{*}|}\\ &> e^{h(Y_0)|S_0|} \cdot |{\mathcal A}_Y|^{-\varepsilon(5+4|K|^6)|S_0|}, \end{align*} $$

where above we have also used the fact that $h(S_0,Y_0) \geq h(Y_0)$ (see Definition 2.11). Our choice of $\varepsilon $ gives us

$$ \begin{align*} h(X) + \varepsilon + 5\varepsilon\log\!|{\mathcal A}_X| < h(Y_0) - \varepsilon(5+4|K|^6)\log\!|{\mathcal A}_Y|, \end{align*} $$

in which case the previous calculation, in combination with (X1), finally gives that

$$ \begin{align*} |{\mathcal P}(S_2, X)| < |{\mathcal P}(S_1^{*}, Y_0)|.\\[-42pt]\end{align*} $$

The previous lemma implies that for each tuple $(S_0, S_1, S_2, S_1^{*})$ satisfying the hypotheses of the previous lemma, there exists an injective map

$$ \begin{align*} \Psi(S_2, S_1^{*}) : {\mathcal P}(S_2, X) \to {\mathcal P}(S_1^{*}, Y_0). \end{align*} $$

Let these injections be chosen and fixed. We now continue the proof of Theorem 3.5.

Here we choose our marker patterns. Pick a point $y^{(m)} \in Y_0$ arbitrarily to serve as the ‘substrate’ of the marker patterns. Let $y_1,\ldots , y_r \in Y$ and $m_i = y_i(M) \in {\mathcal P}(M,Y)\setminus {\mathcal P}(M,Y_1)$ be the points and patterns delivered by Theorem 3.4 for the choice of substrate  $y^{(m)}$ .

Now we appeal to the strong irreducibility of $Y_0$ to construct certain ‘mixing’ patterns. We do this twice: once for patterns of shape $M^6$ , and once for patterns of each shape $S_1^{*}\in {\mathcal S}_1^{*}$ .

Note that $K^2\cdot KM^3 \subset M^6$ because $K\subset M$ , thus $KM^3 \subset \operatorname {\mathrm {int}}^2(M^6)$ and thus $KM^3$ is disjoint from $\partial ^2M^6$ . The subshift $Y_0$ is strongly irreducible as witnessed by K; therefore, for any pair of patterns $u\in {\mathcal P}(M^3, Y_0)$ and $\partial ^2 v \in {\mathcal P}(\partial ^2M^6, Y_0)$ , there is a pattern $w\in {\mathcal P}(M^6,Y_0)$ such that $w(g) = u(g)$ for each $g\in M^3$ and $w(g) = \partial ^2v(g)$ for each $g\in \partial ^2M^6$ . Choose and fix one such pattern w for each choice of u and $\partial ^2v$ ; we shall denote it by $w = u\cup \partial ^2 v$ .

Let $S_1^{*} \in {\mathcal S}_1^{*}$ be arbitrary and suppose $S_1^{*} = \operatorname {\mathrm {int}}^3(S_1 \setminus M^6C)$ for some shape $S_1 \in {\mathcal S}_1$ and some L-separated subset $C \subset G$ . Observe that $K^2\cdot KS_1^{*} \subset S_1\setminus M^6C$ , and thus $KS_1^{*}$ is disjoint from $\partial ^2(S_1\setminus M^6C)$ . Therefore, for each pair of patterns $u\in {\mathcal P}(S_1^{*}, Y_0)$ and $\partial ^2 v\in {\mathcal P}(\partial ^2(S_1\setminus M^6C), Y_0)$ , there is a pattern $w \in {\mathcal P}(S_1\setminus M^6C, Y_0)$ such that ${w(g) = u(g)}$ for each $g \in S_1^{*}$ and $w(g) = \partial ^2v(g)$ for each $g\in \partial ^2(S_1\setminus M^6C)$ . Choose and fix one such pattern w for each choice of u and $\partial ^2v$ ; we shall again denote it by $w = u \cup \partial ^2 v$ .

We are now ready to construct the map $\psi : X \to Y$ desired for the theorem. To begin, let $x\in X$ be fixed, let $t_0 = {\mathcal T}(x)\in T_0$ , let $t_1 = \operatorname {\mathrm {ret}}(t_0) \in T_1$ , let $t_2 = \operatorname {\mathrm {ex}}(t_1) \in T_2$ , let $t_1^{*} \in T_1^{*}$ be delivered by $t_1$ , and let $y_0 = \phi (x)\in Y_0$ be the point delivered by the homomorphism $\phi : X \to Y$ assumed to exist at the beginning of the proof. Let $C = C(t_0) = C(t_1) = C(t_2) = C(t_1^{*}) \subset G$ .

We construct the point $y = \psi (x)\in Y$ in two stages, first by constructing a point $y_1 \in Y_1$ that locally looks like a point of $Y_0$ , and then modifying $y_1$ into a point of Y by placing down the marker patterns.

Note that for each $c \in C$ , the tuple of shapes $(t_0(c), t_1(c), t_2(c), t_1^{*}(c))$ satisfies the hypotheses of Lemma 3.6 by construction. Let $y_1$ be the point satisfying the following two conditions for every $c\in C$ when $(S_1, S_2, S_1^{*}) = (t_1(c), t_2(c), t_1^{*}(c))$ :

(C1) $$ \begin{align} \sigma^c(y_1)(M^6) = y^{(m)}(M^3) \cup \sigma^c(y_0)(\partial^2M^6); \end{align} $$

(C2) $$ \begin{align} \sigma^c(y_1)(S_1\setminus M^6Cc^{-1}) = \Psi(S_2, S_1^{*})(\sigma^c(x)(S_2)) \cup \sigma^c(y_0)(\partial^2(S_1\setminus M^6C)). \end{align} $$

Everywhere else, let $y_1(g) = y_0(g)$ . Here we use the block injection(s) $\Psi (S_2,S_1^{*})$ delivered by Lemma 3.6, as well as the mixing boundary patterns $w = u\cup \partial ^2 v$ constructed earlier. The construction of $y_1$ is illustrated in Figure 2.

Figure 2

An illustration of the construction of $y_1$ for a hypothetical tiling of $\mathbb {Z}^2$ using different sized circles. The largest solid circles indicate the tiles of $t_0$ (the overlap, indicated by dashed lines, is removed in $t_1$ ). The smaller solid circles indicate the translates of $M^6$ , wherein the marker patterns will later be placed. The stippled tile interiors are the patterns given by the block injections from Lemma 3.6. The darkened boundaries are delivered by the strong irreducibility of $Y_0$ , to mix each tile interior pattern with its respective boundary pattern from $y_0$ . The small hatched circles are each labeled with the pattern drawn from the marker substrate, $y^{(m)}(M^3)$ . The shaded exterior is the base point $y_0$ .

First we argue that $y_1$ is well defined. Let $g\in G$ be fixed; we split over three cases. If $g\in M^6C$ , then there is a unique $c \in C$ such that $g\in M^6c$ because C is L-separated and $M^6 \subset L$ . In this case, $y_1(g)$ is determined by the condition in equation (C1). If $g\in (\bigcup _c t_1(c)c)\setminus M^6C$ , then again there is a unique $c\in C$ such that $g\in t_1(c)c\setminus M^6C = (t_1(c)\setminus M^6Cc^{-1})c$ by the disjointness of $t_1$ . In this case, $y_1(g)$ is determined by the condition in equation (C2). Otherwise, $y_1(g) = y_0(g)$ is again uniquely determined.

Next we argue that $y_1$ belongs to $Y_1$ . Recall that $Y_1$ is an SFT as witnessed by K. Let $g\in G$ be arbitrary and consider the translate $Kg$ . If $Kg$ intersects $\operatorname {\mathrm {int}}^2(M^6)c$ for some $c\in C$ , then $Kg \subset M^6c$ by Lemma 2.2 and the fact that $K = K^{-1}$ . Moreover, the c must be unique. In this case, $\sigma ^g(y_1)(K)$ is given by the condition in equation (C1) and therefore $\sigma ^g(y_1)(K) \in {\mathcal P}(K,Y_0)$ . If $Kg$ intersects $\operatorname {\mathrm {int}}^2(t_1(c)c \setminus M^6C)$ for some $c\in C$ , then $Kg \subset t_1(c)c \setminus M^6C$ also by Lemma 2.2, in which case the c must again be unique. In this case, $\sigma ^g(y_1)(K)$ is given by the condition in equation (C2) and therefore $\sigma ^g(y_1)(K) \in {\mathcal P}(K,Y_0)$ again. If neither of these cases hold, then

$$ \begin{align*} Kg \subset (G \setminus M^6 C) \cup \partial^2(M^6)C \cup \bigg(G \setminus \bigcup_c (t_1(c)c \setminus M^6C)\bigg) \cup \bigcup_c \partial^2(t_1(c)c \setminus M^6C). \end{align*} $$

In this case, $\sigma ^g(y_1)(K) = \sigma ^g(y_0)(K)$ and therefore $\sigma ^g(y_1)(K) \in {\mathcal P}(K,Y_0)$ again. We see that every pattern of shape K appearing in $y_1$ is allowed in $Y_0$ . Since $Y_0 \subset Y_1$ and $Y_1$ is an SFT witnessed by K, we conclude that $y_1 \in Y_1$ .

Next we argue that the map $x\mapsto y_1$ is continuous and shift-commuting. For a fixed $g \in G$ , to determine the symbol $y_1(g)$ , one must know first the tiles from $t_0$ , $t_1$ , $t_2$ , and $t_1^{*}$ to which g belongs. This requires looking only at the symbols of the involved quasi-tilings within a finite neighborhood of g. Then, one either applies the condition in equation (C1) or the condition in equation (C2) or returns $y_0(g)$ . As every involved quasi-tiling and $y_0$ are derived from x in a continuous and shift-commuting manner, it is evident that $y_1(g)$ depends only on $\sigma ^g(x)(F_1)$ , where $F_1$ is some (possibly very large but) finite subset of G. We conclude that the map $x \mapsto y_1$ is continuous and shift-commuting (but possibly non-injective).