Proof. Let

$$ \begin{align*} C := A \cup \bigg( B \setminus \bigcup_{\gamma \in P} \sigma_{\gamma}(A)\bigg). \end{align*} $$

Clearly, $C \subseteq (A \cup B)$ . The set C is clopen because it is contained in the Boolean algebra spanned by $\{A,B\}\cup \{\sigma _\gamma (A) : \gamma \in P\}$ . It is clear that $A \subseteq C$ .

Next, we check that $B \subseteq C \cup \bigcup _{\gamma \in P}\sigma _{\gamma }C$ : Otherwise, suppose that $x \in B \setminus (C \cup \bigcup _{\gamma \in P} \sigma _\gamma (C))$ , then, in particular, $x \not \in C$ so $x \not \in A$ because $A \subseteq C$ . Also, because $x \not \in B \setminus C$ , we have that $x \in \sigma _\gamma (A)$ for some $\gamma \in P$ . However, since $A \subseteq C$ , it follows that $\sigma _\gamma (A) \subseteq \sigma _\gamma (C)$ , so $x \in \sigma _\gamma C$ for this $\gamma \in P$ , which contradicts the assumption that $x \in B \setminus (C \cup \bigcup _{\gamma \in P} \sigma _\gamma (C))$ . This shows that $B \subseteq C \cup \bigcup _{\gamma \in P} \sigma _\gamma (C)$ .

It remains to check that C is a P-marker. For any $\gamma \in P$ , we have

$$ \begin{align*}C \subseteq A \cup ( B \cap \sigma_\gamma (A^c) ).\end{align*} $$

Because P is symmetric, also $-\gamma \in P$ , so

$$ \begin{align*}C \subseteq A \cup ( B \cap \sigma_\gamma^{-1} A^c ).\end{align*} $$

Applying $\sigma _\gamma $ to both sides, we get

$$ \begin{align*}\sigma_\gamma (C) \subseteq \sigma_\gamma(A) \cup ( \sigma_\gamma(B) \cap A^c ).\end{align*} $$

It follows that $C \cap \sigma _\gamma C$ is contained in the union of the following four sets:

1. (1) $E_1 := A \cap \sigma _\gamma A$ ;

2. (2) $E_2:= A \cap ( \sigma _\gamma B \cap A^c )$ ;

3. (3) $E_3 := ( B \cap \sigma _\gamma A^c ) \cap \sigma _\gamma A$ ;

4. (4) $E_4:= ( B \cap \sigma _\gamma A^c ) \cap ( \sigma _\gamma B \cap A^c )$ .

By assumption, A is a P-marker, so $E_1 \ = \emptyset $ . Clearly, $E_2 \subseteq A \cap (A^c) = \emptyset $ and ${E_3 \subseteq \sigma _\gamma (A^c) \cap \sigma _{\gamma }(A) = \emptyset} $ . Also, $E_4 \subseteq B \cap \sigma _\gamma (B) = \emptyset $ by assumption that B is a P-marker. This shows that $C\cap \sigma _\gamma C =\emptyset $ for all $\gamma \in P$ .

Proof. For every $x \in V$ and $\gamma \in P$ , we have that $x \ne \sigma _{\gamma }(x)$ . We can thus find for every $x \in V$ , a clopen set $E_x \subseteq V$ such that $x \in E_x$ and $E_x \cap \sigma _{\gamma }(E_x) = \emptyset $ for every $\gamma \in P$ (that is, $E_x$ is a P-marker).

Since V is compact, we can find a finite subcover $\mathcal {U} = \{D_1,\ldots ,D_n\}$ of $\{E_x : x \in V\}$ . Thus, each $D_j \subseteq V$ is a clopen P-marker and $V \subseteq \bigcup _{j=1}^n D_j$ . We sequentially construct clopen P-markers $C_1,\ldots ,C_n \subseteq V$ such that $\bigcup _{k=1}^j D_k \subseteq C_j \cup \bigcup _{\gamma \in P} \sigma _{\gamma }(C_j)$ . We start by setting $C_1 = D_1$ . After $C_1,\ldots ,C_{j-1}$ have been constructed, apply Lemma 3.2 with $A := C_{j-1}$ and $B:=D_j$ to obtain a clopen P-marker $C_j$ such that $A \subseteq C_j$ and ${B \subseteq C_j \bigcup _{\gamma \in P}\sigma _\gamma (C_j)}$ . Then, $C= C_n$ is a clopen set that satisfies the conclusion of the lemma.

Proof. Suppose that Y has the map extension property. Consider the subshift $X = \{0,1\}^\Gamma \times Y \times Y$ . Clearly, Y is a factor of X and so, in particular, $X \rightsquigarrow Y$ .

Consider the subshifts

$$ \begin{align*}\tilde X_1 = (\{\overline{0},\overline{1}\} \times Y \times Y) \subseteq X,\end{align*} $$

$$ \begin{align*} \tilde X_2 = \{ (z,y^{(1)},y^{(2)}) \in \{0,1\}^\Gamma \times Y \times Y: y^{(1)} = y^{(2)} \}, \end{align*} $$

and

$$ \begin{align*} \tilde X = \tilde X_1 \cup \tilde X_2, \end{align*} $$

and the map $\tilde \pi : \tilde X \to Y$ given by

$$ \begin{align*} \tilde \pi(z,y,y) = y \quad\text{for every } z\in \{0,1\}^\Gamma \text{ and every } y \in Y, \end{align*} $$

$$ \begin{align*} \tilde \pi(\overline{0},y^{(0)},y^{(1)}) = y^{(0)} \quad\text{and}\quad\tilde \pi(\overline{1},y^{(0)},y^{(1)}) = y^{(1)} \end{align*} $$

for every $y^{(0)},y^{(1)} \in Y$ . By the map extension property, $\tilde \pi $ extends to a map $\psi :\{0,1\}^\Gamma \times Y \times Y \to Y$ . It is easily verified that the map $\psi $ is a strong contraction homotopy. Hence, Y is indeed strongly contractible.

Proof. Let $\phi :\Gamma \to \mathbb {Z}$ be a surjective homomorphism. Denote $N_0= \max |\phi (W)|$ . Choose $v_0 \in \Gamma $ such that $\phi (v_0)=N_0$ . Let $v \in \Gamma $ satisfy $\phi (v)> 4 N_0$ .

Denote

$$ \begin{align*}N= \phi(v),\quad N_1 = \lfloor N/2 \rfloor\end{align*} $$

and choose $v_1 \in \Gamma $ such that $\phi (v_1)= N_1$ .

Let $z \in \{0,1\}^\Gamma $ be given by

$$ \begin{align*} z_w = \begin{cases} 0, & \phi(w) \le 0,\\ 1, & \phi(w)> 0. \end{cases} \end{align*} $$

We record a few simple observations that follow from the fact that $\psi $ is a strong contraction homotopy with coding window W.

1. (1) For any $\tilde y \in Y$ , the point

   (4) $$ \begin{align} \hat y:= \psi(z,\tilde y,\sigma_{-v}(\tilde y)) \end{align} $$
   satisfies $\hat y_w = \hat y_{w+v}$ for every $w \in \Gamma $ such that $ -N+N_0 < \phi (w) < -N_0$ and $\hat y_w =\tilde y_w$ for every $w \in \Gamma $ such that $\phi (w) < -N_0$ .

2. (2) If $\tilde y \in Y$ satisfies $\tilde y_w = \tilde y_{w+v}$ for every $w \in \Gamma $ such that $|\phi (w)| < N_0$ , then the point $\hat y \in Y$ given by equation (4) satisfies $\hat y_w = \tilde y_{w+v}$ whenever $-\phi (v) < \phi (w) <0$ and also $\hat y_w= \tilde y_w$ for all $w \in \Gamma $ such that $\phi (w) < 0$ .

3. (3) Suppose $\tilde y \in Y$ satisfies $\tilde y_w =\tilde y_{w+v}$ for every $w \in \Gamma $ such that $|\phi (w)| < kN_1+N_0$ for some $k \in \mathbb {N}$ . Then,

   $$ \begin{align*} \hat y = \sigma_{-kv_1}\phi(z,\sigma_{kv_1}(\tilde y),\sigma_{kv_1-v}(\tilde y)) \end{align*} $$
   satisfies $\hat y_w =\tilde y_{w+v}$ for every $w \in \Gamma $ such that $|\phi (w)| < (k+1)N_1+N_0$ and also $\hat y_w= \tilde y_w$ for all $w \in \Gamma $ such that $\phi (w) < kv_1$ .

Using the above observations, starting with any $y^{(0)} \in Y$ , we can construct by induction a sequence of points $y^{(1)},\ldots , y^{(n)},\ldots \in Y$ such that $y^{(n)}_w = y^{(n)}_{w+v}$ for any $w \in \Gamma $ satisfying $|\phi (w)| < nN_1 +N_0$ , and also $y^{(n+1)}_w = y^{(n)}_{w}$ for any $w \in \Gamma $ satisfying $|\phi (w)| < n N_1$ . Thus, the limit $y = \lim _{n \to \infty } y^{(n)}$ exists and satisfies $\sigma _v(y)=y$ . It also satisfies $\mathrm {stab}(y^{(0)}) \cap \mathrm {Ker}(\phi ) \subseteq \mathrm {stab}(y)$ .

Proof. Let $\psi : \{0,1\}^\Gamma \times Y \times Y \to Y$ be a contraction homotopy with coding window $W \Subset \Gamma $ and let $\Delta $ be the subgroup of $\Gamma $ generated by W. In this case, we claim that $Y = Y^{[\Delta ]}$ . Indeed, for any $y^{(1)},y^{(2)} \in Y$ and any $A \subseteq \Gamma /\Delta $ , we can find a point $y \in Y$ that agrees with $y^{(1)}$ on the cosets of $\Delta $ that belong to A and agrees with $y^{(2)}$ on the cosets of $\Delta $ that do not belong to A. Let $z \in \{0,1\}^\Gamma $ be the point satisfying that $z_w=1$ if the $\Delta $ -coset of w is in A and $0$ otherwise. Then, the point $y= \psi (z,y^{(1)},y^{(2)})$ agrees with $y^{(1)}$ on the $\Delta $ -cosets that belong to A and agrees with $y^{(2)}$ on the cosets of $\Delta $ that do not belong to A. By repeated applications, for any tuple of distinct $\Delta $ -cosets $(v_1+\Delta ,\ldots ,v_k +\Delta )$ with $v_1,\ldots , v_k \in \Gamma $ and any points $y^{(1)},\ldots , y^{(k)}$ , we can find $y \in Y$ that agrees with $y^{(i)}$ on $v_i + \Delta $ . This proves that $Y = Y^{[\Delta ]}$ .

Proof. By Lemma 4.7, we can assume without loss of generality that $\Gamma $ is a finitely generated abelian group. In this case, by the structure theorem for finitely generated abelian groups, we can assume that $\Gamma = \mathbb {Z}^d \times G$ , where G is a finite abelian group. Let Y be a contractible $\Gamma $ -subshift and let $\psi :\{0,1\}^\Gamma \times Y \times Y \to Y$ be a strong contraction homotopy with coding window $W \Subset \Gamma $ .

We first prove the following auxiliary statement. There exists a finite set $F \Subset \Gamma $ (that depends on the set W) such that for any subgroup $\Gamma _0 \le \Gamma $ such that $F \cap \Gamma _0 = \emptyset $ , there exists an integer $0 \le n \le d$ , surjective homomorphisms $\phi _1,\ldots ,\phi _n : \Gamma \to \mathbb {Z}$ and $v_1,\ldots ,v_n \in \Gamma _0$ such that $v_1,\ldots ,v_n$ is a generating set for $\Gamma _0$ and so that:

• • $\phi _i(v_j)=0$ for all $0 \le i < j \le n$ ;

• • $\phi _i(v_i)> 2(\max \phi _i(W))$ for every $1\le i \le n$ .

Indeed, by taking F to be a set that contains the non-identity elements of the torsion subgroup of $\Gamma $ , we get that any subgroup $\Gamma _0 < \Gamma $ that does not intersect F is torsion free, and hence the projection of $\Gamma _0$ onto $\mathbb {Z}^d$ is injective. So, without loss of generality, we can prove the statement under the assumption that $\Gamma = \mathbb {Z}^d$ . We choose $R>0$ large enough, in a manner that depends on the set W and the integer d (the precise conditions of R will be clarified soon). Let F denote the intersection of the Euclidean ball of radius R approximately $0$ with $\mathbb {Z}^d \setminus \{0\}$ and let $\Gamma _0 < \Gamma = \mathbb {Z}^d$ be a subgroup that does not intersect F. We can assume without loss of generality that $\Gamma _0$ is a finite index subgroup of $\mathbb {Z}^d$ . In other words, $\Gamma _0$ is a lattice whose shortest non-zero element has length at least R. Let $v_1,\ldots ,v_d \in \Gamma _0$ be a Korkine–Zolotarev reduced basis of $\Gamma _0$ as in [Reference Korkine17]. Then, there exists a constant $c_1>0$ depending only on d such that for any $1\le i \le d$ , the orthogonal projection of $v_i$ onto the orthogonal complement of the span of $v_1,\ldots ,v_{i-1}$ has length at least $c_1 R$ . Thus, there exists another constant $c_2>0$ that depends only on d, and primitive vectors $w_1,\ldots ,w_d \in \mathbb {Z}^d$ such that $w_i$ belongs to the orthogonal complement of the span of $v_1,\ldots ,v_{i-1}$ and $\langle v_i,w_i \rangle> c_2 R \|w_i \|$ . Define $\phi _i:\mathbb {Z}^d \to \mathbb {Z}$ by $\phi _i(v)=\langle v,w_i \rangle $ . Then, for every $i=1,\ldots ,d$ , we have that $\phi _i$ is a surjective homomorphism that satisfies $\phi _i(v_i) \ge c_2 R \|w_i\|$ . Since $\max \phi _i(W) \le \| w_i \| \max _{w \in W}\| w\|$ , if $R> 2 c_2^{-1} \max _{w \in W}\| w\|$ , we have that $\phi _i(v_i)> 2(\max \phi _i(W)$ . This proves the auxiliary statement.

Now, repeatedly apply Lemma 4.6 to produce a sequence of points $y^{(1)},\ldots ,y^{(d)} \in~Y$ such that $v_1,\ldots ,v_i \in \mathrm {stab}(y^{(i)})$ . In particular, $\Gamma _0$ is contained in $\mathrm {stab}(y^{(d)})$ . This completes the proof.

Proof. The classes of sofic shifts, topological mixing subshifts, and strongly irreducible subshifts are each closed under taking (subshift) factors. In particular, they are closed under taking retracts.

Let us show that a retract of a subshift with the map extension property also has the map extension property. Suppose that $\hat Y$ is a subshift with the map extension property, that $Y \subseteq \hat Y$ , and that $r:\hat Y \to Y$ is a retraction map. Let $\tilde X \subseteq X$ be subshifts so that $X \rightsquigarrow Y$ and $\tilde \pi :\tilde X \to Y$ be a map. Since $Y \subseteq \hat Y$ , it follows that $X \rightsquigarrow \hat Y$ . Since $\hat Y$ has the map extension property, $\tilde \pi $ extends to a map $\hat \pi :X \to \hat Y$ . Then, $\pi = r \circ \hat \pi : X \to Y$ is a map that extends $\tilde \pi $ .

It remains to show that a retract of any retract of a shift of finite type is also a shift of finite type. Let $X \subseteq A^\Gamma $ be a shift of finite type with a defining set of forbidden patterns $\mathcal {F}_1 \Subset A^{W_1}$ . Suppose that $Y \subseteq X$ is a retract, with $r:X \to Y$ a retraction. Let $W_0 \Subset \Gamma $ be a coding window for r such that $0 \in W_0$ and let $\Phi :A^{W_0} \to A$ be a sliding block code for r. Let

$$ \begin{align*} \mathcal{F}_0 = \{ w \in A^{W_0} :\Phi(w)_0 \ne w_0 \}. \end{align*} $$

Now suppose that $x \in X$ satisfies that $\sigma _v(x)_{W_0} \not \in \mathcal {F}_0$ for all $v \in \Gamma $ . It follows that $r(x)=x$ , so $x \in Y$ . This means that

$$ \begin{align*} Y = \{ x \in X: \sigma_v(x)_{W_0} \not \in \mathcal{F}_0 \text{ for all } v \in \Gamma \}. \end{align*} $$

Denote $W = W_0 \cup W_1$ and let

$$ \begin{align*}\mathcal{F} = \{ w \in A^W : w_{W_0} \in \mathcal{F}_0 \text{ or } w_{W_1} \in \mathcal{F}_1 \}.\end{align*} $$

Then, $\mathcal {F}$ is a defining set of forbidden patterns for Y. This proves that Y is a shift of finite type.

Proof. If $K \Subset \Gamma $ witnesses that Y is $\mathcal {G}$ -free, then each $\Gamma _0 \in \mathcal {G}$ admits a finite generating set $F_{\Gamma _0} \Subset \Gamma $ so that for every $y \in Y$ and every $\Gamma _0 \in \mathcal {G}$ , there exists $v \in K$ and $\gamma \in F_{\Gamma _0}$ such that $y_v \ne y_{v + \gamma }$ , and so $\Gamma _0 \not \subseteq \mathrm {stab}(y)$ . This proves that Y is $\mathcal {G}$ -free.

For the converse direction, for each $\Gamma _0 \in \mathcal {G}$ , fix a finite generating set $F_{\Gamma _0} \Subset \Gamma $ . Given $K \Subset \Gamma $ , say that $w \in \mathcal {L}_K(Y)$ is $F_{\Gamma _0}$ -periodic if $w_v = w_{v + \gamma }$ whenever $\gamma \in F_{\Gamma _0}$ , and $v,v + \gamma \in K$ . Suppose that Y is $\mathcal {G}$ -free. Then, for any $y \in Y$ and every $\Gamma _0 \in \mathcal {G}$ , there exists a finite set $K_{y,\Gamma _0} \Subset \Gamma $ such that the restriction of y to $K_{y,\Gamma _0}$ is not $F_{\Gamma _0}$ periodic. Take $K_y = \bigcup _{\Gamma _0 \in \mathcal {G}}K_{y,\Gamma _0}$ , then the restriction of y to $K_{y}$ is not $F_{\Gamma _0}$ periodic for all $\Gamma _0 \in \mathcal {G}$ . For each $y \in Y$ , let

$$ \begin{align*} U_y = \{ \tilde y \in Y: \tilde y_{K_y} = y_{K_y} \}.\end{align*} $$

Then, $U_y \subseteq Y$ is an open neighborhood of y. By compactness of Y, the open cover $\{U_y\}_{y \in Y}$ admits a finite subcover $U_{y_1},\ldots ,U_{y_N}$ . Let $K =\bigcup _{j=1}^N K_{y_j}$ , then for every $y \in Y$ , we have that $y_K$ is not $F_{\Gamma _0}$ periodic for all $\Gamma _0 \in \mathcal {G}$ . This proves that K witnesses that Y is $\mathcal {G}$ -free.

Proof. It is clear that $Y \subseteq Y^{(K,D)}$ , because for any $y \in Y$ , any $v \in \Gamma $ , and $w \in D$ , we have that $\sigma _{v+w}(y)_K \in \mathcal {L}_K(Y)$ . Because K witness that Y is $\mathcal {G}$ -free, it follows that $D+K$ witnesses that $Y^{(K,D)}$ is $\mathcal {G}$ -free.

Proof. Let $\mathcal {G} \Subset \mathrm {Sub}(\Gamma )$ be a finite set and let Y be a $\mathcal {G}$ -free $\Gamma $ -subshift.

Choose a finite symmetric generating set $F_{\Gamma _0} \Subset \Gamma $ for each $\Gamma _0 \in \mathcal {G}$ . By Lemma 4.15, there exists a finite set $K_0 \Subset \Gamma $ with $0 \in K_0$ that witnesses that Y is $\mathcal {G}$ -free. Let

$$ \begin{align*}P= \bigcup_{\Gamma_0 \in \mathcal{G}}F_{\Gamma_0}\end{align*} $$

and let $\mathcal {Q}$ denote the collection of symmetric subsets $Q \subseteq P$ that satisfy $Q \cap (\Gamma _0 \setminus \{0\}) \ne~\emptyset $ for all $\Gamma _0 \in \mathcal {G}$ .

We first prove part (i). We are about to define a $\Gamma $ -subshift $\hat Y \subseteq \hat A^{\Gamma }$ and show that $\hat Y$ is $\mathcal {G}$ -free, contains Y, and has the map extension property. Let us first define the ‘alphabet’ $\hat A$ :

$$ \begin{align*} \hat A = A \uplus \mathcal{Q} \uplus \{\hat a\}, \end{align*} $$

where $Y \subseteq A^\Gamma $ .

We regard the finite set $\mathcal {Q}$ as another alphabet disjoint for A, and $\hat a$ as a new ‘symbol’ not contained in A or in $\mathcal {Q}$ .

Let $\hat Y \subseteq \hat A^{\Gamma }$ be the $\Gamma $ -subshift of finite type defined by the following constraints. For every $y \in \tilde Y$ and any $v \in \Gamma $ :

1. (1) there exists $u \in K_0$ such that either $\sigma _{v-u}(y)_{K_0} \in \mathcal {L}_{K_0}(Y)$ or there exists $Q \in \mathcal {Q}$ and $\gamma \in Q \cup \{0\}$ such that $\sigma _{v-u+ \gamma }(y)=Q$ ;

2. (2) if there exists $Q \in \mathcal {Q}$ such that $y_v = Q$ , then for every $\gamma \in Q$ , we have that $y_{v+\gamma } \ne~Q$ .

It is clear that $Y \subseteq \hat Y$ . If $y \in Y$ , then for every $v \in \Gamma $ and every $u \in K_0$ , we have that $\sigma _{v-u}(y)_{K_0} \in \mathcal {L}_{K_0}(Y)$ and $y_v \not \in Q$ , so $y \in \hat Y$ . This shows that $Y \subseteq \hat Y$ . Let us prove that $\hat Y$ is $\mathcal {G}$ -free. Given $y \in \hat Y$ and $\Gamma _0 \in \mathcal {G}$ , we need to show that $\Gamma _0 \not \le \mathrm {stab}(y)$ . Consider an arbitrary point $y \in \hat Y$ . If there exists $v \in \Gamma $ such that $\sigma _v(y)_{K_0} \in \mathcal {L}_{K_0}(Y)$ , then $\Gamma _0 \not \le \mathrm {stab}(y)$ for any $\Gamma _0 \in \mathcal {G}$ , because $K_0$ witnesses that Y is $\mathcal {G}$ -free. Otherwise, there exists $v \in \Gamma $ and $Q \in \mathcal {Q}$ such that $\sigma _v(y)=Q$ . In this case, for every $\gamma \in Q$ , we have $y_{v+\gamma } \ne y_v$ , so $\mathrm {stab}(y) \cap Q =~\emptyset $ . Since $(\Gamma _0 \setminus \{0\}) \cap Q \ne \emptyset $ this shows that $\Gamma _0 \not \le \mathrm {stab}(y)$ .

Let us show that $\hat Y$ has the map extension property. Suppose that X is a $\Gamma $ -subshift such that $X \rightsquigarrow \hat Y$ . In particular, $X \subseteq B^\Gamma $ is a $\mathcal {G}$ -free subshift and suppose that $\tilde \pi : \tilde X \to \hat Y$ is a map from a $\Gamma $ -subshift $\tilde X \subseteq X$ . Let $W \Subset \Gamma $ be a coding window for $\tilde \pi $ and let $K = K_0 + W$ . We will now describe a map $\pi :X \to \hat Y$ that extends $\tilde \pi $ . As before, since X is compact and $\mathcal {G}$ -free, there exists a finite set $K_2 \Subset \Gamma $ that witnesses that X is $\mathcal {G}$ -free. For every $Q \in \mathcal {Q}$ , let $L_Q \subseteq B^{K_2}$ be the set of patterns $w \in \mathcal {L}_{K_2}(X)$ such that:

• • for every $\gamma \in Q$ , there exists $v \in K_2$ such that $v+\gamma \in K_2$ and $w_v \ne v_{v+\gamma }$ ;

• • for every $\gamma \in P\setminus Q$ and every $v \in K_2$ such that $v+\gamma \in K_2$ , we have $w_v = v_{v+\gamma }$ .

The sets $\{L_Q\}_{Q \in \mathcal {Q}}$ are mutually disjoint and $\mathcal {L}_{K_2}(X)) \subseteq \biguplus _{Q \in \mathcal {Q}}L_Q$ . For every $Q \in \mathcal {Q}$ , let $V_Q =\{ x \in X: x_{K_2} \in L_Q\}$ . Then, $V_Q$ is a clopen set in X. Furthermore, for every $x \in V_Q$ and every $\gamma \in Q$ , we have $\sigma _\gamma (x) \ne x$ . By Lemma 3.3, there exists a clopen set $C_Q \subseteq V_Q$ such that

$$ \begin{align*} C_Q \cap \sigma_\gamma(C_Q) = \emptyset \quad\text{for every } \gamma \in Q \text{ and } V_Q \subseteq C_Q \cup \bigcup_{\gamma \in Q}\sigma_\gamma(C_Q). \end{align*} $$

Define a map $\pi :X\to \hat Y$ as follows:

$$ \begin{align*} \pi(x)_v = \begin{cases} \tilde \pi(\sigma_v(x))_0 & \text{if } \text{ there exists } u \in K_0 \text{ such that } \sigma_{v-u}(x)_{K_0+W} \in \mathcal{L}_{W}(\tilde X),\\ Q & \text{if } \sigma_v(x)\in\!C_Q \text{ for some } Q \in\! \mathcal{Q} \text{ and we are not in the previous case},\\ \hat{a} & \text{otherwise.} \end{cases} \end{align*} $$

Notice that if there exists $u \in K_0$ such that $\sigma _{v-u}(x)_{K_0+W} \in \mathcal {L}_{K_0+W}(\tilde X)$ , then it follows that $\sigma _v(x)_{W} \in \mathcal {L}_W(\tilde X)$ and so $\tilde \pi (\sigma _v(x))_0$ is well defined, because W is a coding window for $\tilde \pi $ . Moreover, if $\sigma _v(x)_K \in \mathcal {L}_{K_0+W}(\tilde X)$ , then there exists $\tilde x \in \tilde X$ such that $\pi (x)_{K_0}= \tilde \pi (\tilde x)_{K_0}$ and, in particular, $\pi (x)_0 = \tilde \pi (x)_0$ . Since $X \subseteq \bigcup _{Q \in \mathcal {Q}} V_Q$ and $V_Q \subseteq C_Q \cup \bigcup _{\gamma \in Q}\sigma _\gamma (C_Q)$ for every $Q \in \mathcal {Q}$ , it follows that for every $x \in X$ and $v \in \Gamma $ , either there exists $u \in K_0$ such that $\sigma _{v-u}(x)_{K} \in \mathcal {L}_{K_0+W}(X)$ , in which case $\sigma _{v-u}(\pi (x)) \in \mathcal {L}_{K_0}(Y)$ , or there exists $Q \in \mathcal {Q}$ and $\gamma \in Q \cup \{0\}$ such that $\sigma _{v+\gamma }(\hat \pi (y))=Q$ . We have thus shown that $\pi (x) \in \hat Y$ for every $x \in X$ , completing the proof of part (i).

We now prove part (ii). Let Y be an absolute retract for the class of $\mathcal {G}$ -free subshifts and let $\hat Y$ be the subshift defined in the first part of the proof. Since $Y \subseteq \hat Y$ and $\hat Y$ is $\mathcal {G}$ -free, there exists a retraction map $\hat r:\hat Y \to Y$ . Let $W \Subset \Gamma $ be a coding window for $\hat r$ . Let X be a $\mathcal {G}$ -free subshift that contains Y. Again, choose a finite set $K_2 \Subset \Gamma $ that witnesses that X is $\mathcal {G}$ -free. For every $Q \in \mathcal {Q}$ , let $L_Q$ , $V_Q \subset X$ , and $C_Q$ be as in the previous part of the proof. Define a map $\tilde r:X\to \hat Y$ as follows:

$$ \begin{align*}\tilde r(x)_v = \begin{cases} \sigma_v(x)_0 & \text{if } \text{ there exists } u \in K_0 \text{ such that } \sigma_{v-u}(x)_{K_0} \in \mathcal{L}_{K_0}(Y),\\ Q & \text{if } \sigma_v(x) \in C_Q \text{ for some } Q \in \mathcal{Q} \text{ and we are not in the previous case},\\ \hat{a} & \text{otherwise.} \end{cases} \end{align*} $$

As in the previous part, it follows that indeed $\hat r(x) \in \hat Y$ for every $x \in X$ , so $\tilde r:X\to \hat Y$ is well defined. Clearly, $\tilde r(x)_0 = x_0$ whenever $x_{K_0} \in \mathcal {L}_{K_0}(Y)$ . Let $K = K_0 +W$ . Let $r = \hat r \circ \tilde r$ . Then, $r: X \to Y$ is a retraction map, and $r(y)_0= y_0$ whenever $x \in X$ and $x_{K} \in \mathcal {L}_K(Y)$ . This completes the proof of part (ii).

Proof. Suppose that Y has the map extension property. By Proposition 4.8, there exists a finite set $F \Subset \Gamma \setminus \{0\}$ such that for every $\Gamma _0 \le \Gamma $ with $\Gamma _0 \cap F = \emptyset $ , there exists $y \in Y$ such that $\Gamma _0 \le \mathrm {stab}(y)$ . Let $\mathcal {G} \Subset \mathrm {Sub}(\Gamma )$ denote the collection of subgroups $\Gamma _0 \le \Gamma $ that are generated by a subset of F and so that $\Gamma _0 \not \le \mathrm {stab}(y)$ for every $y \in Y$ . It follows that Y is $\mathcal {G}$ -free and that $X \rightsquigarrow Y$ for any $\mathcal {G}$ -free $\Gamma $ -subshift X. Let X be a $\mathcal {G}$ -free subshift that contains Y. Since Y has the map extension property and $Y \subseteq X$ , it follows that then the identity map on Y extends to a retraction from X to Y. This shows that Y is an absolute retract for the class of $\mathcal {G}$ -free subshifts.

For the converse direction, suppose that there exists a finite set $\mathcal {G} \Subset \mathrm {Sub}(\Gamma )$ such that Y is an absolute retract for the class of $\mathcal {G}$ -free subshifts. Let X be a $\Gamma $ -subshift such that $X \rightsquigarrow Y$ and let $\pi :\tilde X \to Y$ be a map from a closed, $\Gamma $ -invariant subset $\tilde X \subseteq X$ . Our goal is to show that $\tilde \pi $ extends to a map $\pi :X \to Y$ . By Lemma 4.18, since X is $\mathcal {G}$ -free, there exists a $\mathcal {G}$ -free subshift $\hat Y$ that contains Y and has the map extension property. Since $Y \subseteq \hat Y$ , it follows that $\tilde \pi $ extends to a map $\hat \pi :X \to \hat Y$ . Since Y is an absolute retract for the class of $\mathcal {G}$ -free subshifts, there exists a retraction map $r:\hat Y \to Y$ . Setting $\pi = r \circ \hat \pi $ , we see that $\pi :X \to Y$ extends $\tilde \pi $ .

Proof. Suppose that $Y \subseteq A^{\Gamma }$ is a $\Gamma $ -subshift with the map extension property. The inequality

$$ \begin{align*}\limsup_{\Gamma_0 \to \{0\}}h(\overline{Y}_{[\Gamma_0]}) \le h(Y)\end{align*} $$

holds by Proposition 2.23 for any subshift, regardless of the map extension property. So, to conclude the first part of the statement, we only need to prove the inequality

$$ \begin{align*}\liminf_{\Gamma_0 \to \{0\}}h(\overline{Y}_{[\Gamma_0]}) \ge h(Y).\end{align*} $$

By Lemma 4.7, it suffices to prove this under the additional assumption that $\Gamma $ is a finitely generated abelian group. Furthermore, using Proposition 2.23, it suffices to prove that for any $\epsilon>0$ , there exists a finite set $K \Subset \Gamma $ such that for any K-separated subgroup $\Gamma _0 < \Gamma $ of finite index, we have $h(\overline {Y}_{[\Gamma _0]}) \ge h(Y) - \epsilon $ . By Proposition 4.19, there exists a finite set $\mathcal {G} \Subset \mathrm {Sub}(\Gamma )$ such that Y is an absolute retract for the class of $\mathcal {G}$ -free subshifts. In particular, Y is $\mathcal {G}$ -free and so by Lemma 4.15, there exists a finite set $K_0 \Subset \Gamma $ that witnesses that Y is $\mathcal {G}$ -free. By part (ii) of Lemma 4.18, there exists another finite set $K_1 \Subset \Gamma $ such that for any $\mathcal {G}$ -free subshift $\hat Y$ that contains Y, there exists a retraction map $r:\hat Y \to Y$ so that $r(y)_0=r_0$ for any $y \in \hat Y$ with $y_{K_1} \in \mathcal {L}_{K_1}(Y)$ .

For a finitely generated abelian group $\Gamma $ , it is not difficult to show that for any $\delta>0$ and any finite set $F \subset \Gamma $ , there exists a finite set $K \Subset \Gamma $ so that any K-separated subgroup finite-index subgroup $\Gamma _0$ admits a fundamental domain D which is $(F,\delta )$ -invariant. This can be deduced from Lemma 6.14 in §6, by taking D to be the disjointified Voronoi cell of $0$ with respect to $\Gamma _0$ (see Definition 6.7). By Lemma 2.22, for any $\epsilon>0$ , by choosing F and $\delta $ appropriately as functions of $K_1$ and $\delta $ , for any $(F,\delta )$ -invariant, we have

$$ \begin{align*} \frac{1}{|D|} \log | \mathcal{L}_{D \setminus \partial_{K_1} D}(Y) |> h(Y) -\epsilon. \end{align*} $$

So we can complete the proof by showing that for any finite index subgroup $\Gamma _0 < \Gamma $ that admits a fundamental domain $D \Subset \Gamma $ with $K_0 \subseteq D$ and $D \setminus \partial _{K_1} D \ne \emptyset $ , we have

(5) $$ \begin{align} h(\overline{Y}_{[\Gamma_0]}) \ge \frac{1}{|D|} \log | \mathcal{L}_{D \setminus \partial_{K_1} D}(Y) |. \end{align} $$

We assume that $Y \subseteq A^\Gamma $ . Given a finite index subgroup $\Gamma _0 < \Gamma $ that admits a fundamental domain $D \Subset \Gamma $ such that $K_0 \subseteq D$ and $D \setminus \partial _{K_1} D \ne \emptyset $ , consider the subshift

$$ \begin{align*} \hat Y = \{ y \in A^\Gamma: \text{ there exists } v \in \Gamma \text{ such that for every } w \in \Gamma_0, ~\sigma_{v+w}(y)_{D} \in \mathcal{L}_D(Y) \}. \end{align*} $$

Because $K_0 \subseteq D$ and $K_0$ witnesses that Y is $\mathcal {G}$ -free, it follows that $\hat Y$ is also $\mathcal {G}$ -free. Clearly, $Y \subseteq \hat Y$ . Thus, there exists a retraction map $r:\hat Y \to Y$ so that $r(y)_0=r_0$ for any $y \in \hat Y$ with $y_{K_1} \in \mathcal {L}_{K_1}(Y)$ . For any $p \in \mathcal {L}_D(Y)$ , we can define a point $y \in \hat Y_{[\Gamma _0]}$ such that $\sigma _w(y)_D = p$ for every $w \in \Gamma _0$ . It follows that $r(y)_{D \setminus \partial _{K_1} D} = y_{D \setminus \partial _{K_1} D}$ . Since $r(y) \in Y_{[\Gamma _0]}$ , it follows that

$$ \begin{align*} |\mathcal{L}_{D}(Y_{[\Gamma_0]})| \ge |\mathcal{L}_{D \setminus \partial_{K_1} D}(Y)|. \end{align*} $$

Since $h(Y_{[\Gamma _0]}) = {1}/{|D|} \log |\mathcal {L}_{D}(Y_{[\Gamma _0]})|$ , we have proven equation (5).

The proof that

$$ \begin{align*} \lim_{\Gamma_0 \to \{0\}}h(\overline{Y}_{\Gamma_0}) = h(Y)\end{align*} $$

under the assumption that $\mathrm {Ker}(Y) = \{0\}$ is similar, so we omit the details.

Proof. Let $Y \subseteq A^\Gamma $ be a $\mathcal {G}$ -free subshift. By Lemma 4.15, there exists a finite set $W \Subset \Gamma $ that witnesses that Y is $\mathcal {G}$ -free. Replacing W by $W \cup (-W) \cup \{0\}$ , we can assume that $0 \in W$ and that W is symmetric. Let $\hat A$ be a finite set with $|\hat A| \ge |W|$ such that $A \subseteq \hat A$ and let $\mathcal {F} \subset \hat A^W$ denote the set of patterns that are not $\mathcal {G}$ -free in the sense that there exists $\Gamma _0 \in~\mathcal {G}$ such that for every $v \in \Gamma _0$ , if $u \in W$ and $u+v \in W$ , then $w_u=w_{u+v}$ . Let $\hat Y \subseteq \hat A^\Gamma $ be the subshift of finite type defined by the set of forbidden words $\mathcal {F}$ . Then, $\hat Y$ is $\mathcal {G}$ -free, as witnessed by the set W. Clearly, $Y \subseteq \hat Y$ . We claim that $\hat Y$ is strongly irreducible. To see this, it suffices to show that for any $F \subseteq \Gamma \setminus \{0\}$ , any pattern $w \in \hat A^F$ which is $\mathcal {F}$ -free (in the sense that sub-patterns from $\mathcal {F}$ do not occur in it) can be extended to a $\mathcal {F}$ -free pattern $\tilde w \in \hat A^{F \cup \{0\}}$ .

Let $F \subseteq \Gamma \setminus \{0\}$ and we have an $\mathcal {F}$ -free pattern $w \in A^F$ . Since $|W \setminus \{0\}| < |\hat A|$ , there exists $\hat a \in \hat A$ such that $w_{v} \ne \hat a$ for any $v \in (W \cap F) \setminus \{0\}$ . Let $\tilde w \in \hat A^{F \cup \{0\}}$ be defined by

$$ \begin{align*} \tilde w_v = \begin{cases} \hat a & \text{if } v= 0,\\ w_v & \text{otherwise.} \end{cases} \end{align*} $$

Then, $\tilde W$ is also $\mathcal {F}$ -free $\tilde w_F =w$ .

Proof. Suppose that Y has the map extension property. Then, by Proposition 4.19, there exists a finite set $\mathcal {G} \Subset \mathrm {Sub}(\Gamma )$ such that Y is an absolute retract for the class of $\mathcal {G}$ -free subshifts.

Then by Lemma 4.23, Y embeds in a $\mathcal {G}$ -free strongly irreducible SFT $\hat Y$ . By Proposition 4.10, it follows that Y is also a strongly irreducible subshift of finite type.

Proof. Let $\Gamma $ be a countable abelian group. Suppose that Y is a $\Gamma $ -subshift with the map extension property and $\Gamma _0 < \Gamma $ . We want to prove that $\overline {Y}_{[\Gamma _0]}$ also has the map extension property. Suppose that $\tilde X \subseteq X$ are $\Gamma /\Gamma _0$ subshifts with $X \rightsquigarrow \overline {Y}_{[\Gamma _0]}$ and that $\tilde \pi :\tilde X \to \overline {Y}_{[\Gamma _0]}$ is a map of $\Gamma /\Gamma _0$ -subshifts. For any set finite set A, there is a natural continuous injective map $\psi _{\Gamma _0}: A^{\Gamma /\Gamma _0} \to A^{\Gamma _0}$ given by $\psi _{\Gamma _0}(x)_v = x_{v +\Gamma _0}$ for any $x \in A^{\Gamma /\Gamma _0}$ , $v \in \Gamma $ , and $x \in A^{\Gamma /\Gamma _0}$ . Then, $\psi _{\Gamma _0}^{-1}(\tilde X) \subseteq \psi _{\Gamma _0}^{-1}(X)$ are $\Gamma $ -subshifts with $\Gamma _0 < \ker (\psi _{\Gamma _0}^{-1}(X))$ and $\psi _{\Gamma _0}^{-1}(X) \rightsquigarrow Y$ because $X \rightsquigarrow \overline {Y}_{[\Gamma _0]}$ . The map $\tilde \pi $ naturally lifts to a map $\overline {\pi }:\psi _{\Gamma _0}(\tilde X) \to Y$ . By the map extension property, the map $\overline {\pi }$ extends to a map $\hat \pi :\psi _{\Gamma _0}(X) \to Y$ . Since $\Gamma _0 < \ker (\psi _{\Gamma _0}^{-1}(X))$ , it follows that the image of $\psi _{\Gamma _0}(X))$ under $\hat \pi $ is contained in $Y_{[\Gamma _0]}$ . Hence, $\hat \pi $ induces a well-defined map $\pi :X \to \overline {Y}_{[\Gamma _0]}$ that extends $\tilde \pi $ .

Proof. From Proposition 4.24, it follows that any $\Gamma $ -subshift which has the map extension property is a strongly irreducible SFT. In particular, any $\mathbb {Z}$ -subshift which is a restricted absolute retract is a mixing shift of finite type. Now suppose $Y \subset A^{\mathbb {Z}}$ is a mixing SFT. Let $\mathcal {G} \subset \mathrm {Sub}(\mathbb {Z})$ denote the collection of subgroups $\Gamma _0 \le \mathbb {Z}$ such that for every $y \in Y$ , the subgroup $\Gamma _0$ is not contained in $\mathrm {stab}(x)$ . We claim that $\mathcal {G}$ is a finite set. Indeed, since Y is a mixing $\mathbb {Z}$ -SFT, there exists $n_0 \in \mathbb {N}$ such that for every $n \ge n_0$ , there exists $y \in Y$ with $\sigma ^n(y)=y$ . Thus, every subgroup $\Gamma _0 \in \mathcal {G}$ must be of the form $n \mathbb {Z}$ with $n < n_0$ . It follows from Boyle’s extension lemma (Lemma 4.26) that Y has the map extension property.

Proof. Let $X \subseteq A^\Gamma $ be a $\Gamma $ -SFT with a safe symbol $a \in A$ and suppose that $Y \subseteq \tilde A^\Gamma $ is another subshift such that $X \subseteq Y$ . Let $\mathcal {F} \subset A^W$ be a defining set of forbidden patterns for X. We can define a retraction $r:Y \to X$ as follows:

$$ \begin{align*}r(y)_v = \begin{cases} y_v & \text{if }(\sigma_{-v-w}(y))_W \not \in \mathcal{F} \text{ for all } w \in W,\\ a & \text{otherwise}. \end{cases}\\[-37pt]\end{align*} $$

Proof. Let $F \Subset \Gamma \setminus \{0\}$ be a finite symmetric set and $k>|F|$ . Let $\mathcal {G} \Subset \mathrm {Sub}(\Gamma )$ denote the set of cyclic subgroups generated by elements of F:

$$ \begin{align*} \mathcal{G} = \{ \langle v \rangle : v \in F \}. \end{align*} $$

We will prove that $X_{k,F}$ is an absolute retract for the class of $\mathcal {G}$ -free subshifts. Since $x_{0} \ne x_{v}$ for any $x \in X_{k,F}$ and $v \in F$ , it is clear that $v \not \in \mathrm {stab}(x)$ for any $x \in X_{k,F}$ , so $X_{k,F}$ is $\mathcal {G}$ -free. Let X be any $\mathcal {G}$ -free subshift such that $X_{k,F} \subseteq X$ . By Lemma 3.3, there exists a clopen set $C \subset X$ such that $C \cap \sigma _v(C) = \emptyset $ for every $v \in F$ and $X= C \bigcup _{v \in V} \sigma _v(C)$ . Define a map $\alpha :X \to \{0,1\}^\Gamma $ by declaring for $x \in X$ and $v \in \Gamma $ ,

$$ \begin{align*}\alpha(x)_v= \begin{cases} 1 & \text{if }\sigma_v(x) \in C,\\ 0 & \text{otherwise}. \end{cases} \end{align*} $$

For $x \in X$ , let

$$ \begin{align*}B(x) = \{ x_v: v \in F\}.\end{align*} $$

Let $F= \{ v_0,v_1,\ldots ,v_{|F|}\}$ be some enumeration of the elements of $F \cup \{0\}$ . For every $x \in X$ , let

$$ \begin{align*} t(x) = \min \{ 0\le j \le |F|: \sigma_v(x) \in C\}.\end{align*} $$

For every $0 \le j \le |F|$ , define a map $r^{(j)}:X \to X$ as follows by declaring for $x \in X$ and $v \in \Gamma $ :

$$ \begin{align*} r^{(j)}(x)_v = \begin{cases} x_v & \text{if } t(X) \ne j \text{ or } x_v \in \{1,\ldots,k\} \setminus B(\sigma_{v}(x)),\\ \min( \{1,\ldots,k\} \setminus B(\sigma_{v}(x)) ) & \text{otherwise. } \end{cases} \end{align*} $$

Note that for every $v \in F \cap \{0\}$ , we have that since $|B(\sigma _{v}(x)| \le |F| < k$ , it follows that $ \{1,\ldots ,k\} \setminus B(\sigma _{v}(x)) \ne \emptyset $ , so $r^{(j)}$ is well defined.

Define $\tilde r^{(0)} = r^{(0)}$ and $\tilde r^{(j)}= r^{(j)} \circ \tilde r^{(j-1)}$ for every $ 1\le j \le |F|$ . Let $r =\tilde r^{(|F|)}:X \to X$ . We claim that $r:X \to X_{k,F}$ is a retraction. By induction, it follows that if $x \in C \cup \bigcup _{\ell =1}^j \sigma _{v_\ell }^{-1}(C)$ , then $x_0 \in \{1\ldots ,k\} \setminus B(x)$ . In particular, since $X= \bigcup _{\ell =1}^{|F|} \sigma _{v_\ell }^{-1}(C)$ , it follows that $r(x)_0 \in \{1\ldots ,k\} \setminus B(x)$ for every $x \in X$ . So for every $x \in X$ , $r(x) \in \{1,\ldots ,l\}^\Gamma $ and $r(x)_v \ne r(x)_{v+W}$ for every $v \in \Gamma $ and $w \in F$ . This proves that $r(x) \in X_{k,F}$ for every $x \in X$ . It is easy to check that $r^{(j)}(x) = x$ for every $ 0 \le j \le |F|$ and $x \in X_{k,F}$ , so $r(x)=x$ for every $x \in X_{k,F}$ . This proves that $r:X \to X_{k,F}$ is indeed a retraction.