Derivational Modal Logics with the Difference Modality

Abstract

In this chapter we study modal logics of topological spaces in the combined language with the derivational modality and the difference modality. We give axiomatizations and prove completeness for the following classes: all spaces, $T_1$-spaces, dense-in-themselves spaces, a zero-dimensional dense-in-itself separable metric space, $\mathbf R^n~ (n\ge 2).$ We also discuss the correlation between languages with different combinations of the topological, derivational, universal and difference modalities in terms of definability.

In memory of Leo Esakia

Appendix

Finally let us give technical details of the proofs of Propositions 11, 19.

Proposition 11 Let $\mathfrak X$ be a dense-in-itself separable metric space, $B\subset X$ a closed nowhere dense set. Then there exists a d-morphism $g: \mathfrak X\twoheadrightarrow ^d \Phi _{ml}$ with the following properties:

• (1) $B \subseteq g^{-1}(b_1)$;

• (2) every $g^{-1}(a_i)$ $(for i\le l)$ is a union of a set $\alpha _i$ of disjoint open balls, which is dense at any point of $g^{-1}(\{ b_1,\ldots ,b_m\})$.

The frame $\Phi _{ml}$ is shown in Fig. 11.1.

Proof

Let $X_1,\dots , X_n,\dots$ be a countable base of $\mathfrak X$ consisting of open balls. We construct sets $A_{ik}, ~B_{jk}$ for $1\le i\le l, ~1\le j\le m, ~k\in \omega$, with the following properties:

• (1) $A_{ik}$ is the union of a finite set $\alpha _{ik}$ of nonempty open balls whose closures are disjoint;

• (2) $\mathbf{C}A_{ik} \cap \mathbf{C}A_{i'k} = \varnothing$ for $i\not = i'$;

• (3) $\alpha _{ik} \subseteq \alpha _{i,k+1};~A_{ik} \subseteq A_{i,k+1}$;

• (4) $B_{jk}$ is finite;

• (5) $B_{jk} \subseteq B_{j,k+1};$

• (6) $A_{ik} \cap B_{jk} = \varnothing ;$

• (7) $X_{k+1} \subseteq \bigcup \limits _{i=1}^l A_{ik}\Rightarrow \alpha _{i,k+1} = \alpha _{ik}, ~B_{j,k+1} = B_{jk};$

• (8) if $X_{k+1}\not \subseteq \bigcup \limits _{i=1}^l A_{ik}$, there are closed nontrivial balls $P_1,\dots ,P_l$ such that for any $i$, $j$

  $$P_i \subseteq X_{k+1}-A_{ik},~ \alpha _{i,k+1} = \alpha _{ik} \cup \{ \mathbf{I}P_i\},~ (B_{j,k+1} - B_{jk}) \cap X_{k+1}\not = \varnothing ; $$

• (9) $A_{ik} \subseteq X-B$;

• (10) $B_{jk} \subseteq X-B$;

• (11) $j\not = j'\Rightarrow B_{j'k} \cap B_{jk} = \varnothing$ .

We carry out both the construction and the proof by induction on $k$.

Let $k=0$; $(X-B)$ is infinite since it is nonempty and open in a dense-in-itself $\mathfrak X$. Take distinct points $v_1,\dots ,v_l \not \in B$ and disjoint closed nontrivial balls $Z_1,\dots ,Z_l \subset X-B$ with centers at $v_1,\dots ,v_l$, respectively (see Fig.11.6).

Fig. 11.6

Case $k=0$

Put

$$\alpha _{i0}: =\{ \mathbf{I}Z_i\}; ~A_{i0}: = \mathbf{I}Z_i;$$

then $Z_i = \mathbf{C}A_{i0}$. As above, since $(X-B) -\bigcup \limits ^l_{i=1}Z_i$ is nonempty and open, it is infinite. Pick distinct $w_1,\dots ,w_m\in X-B$ and put $B_{j0}: = \{ w_j\}$. Then the required properties hold for $k=0$.

At the induction step we construct $A_{i,k+1}, B_{j,k+1}$. Put $Y_k :=\bigcup \limits ^l_{i=1}A_{ik}$ and consider two cases.

(a) $X_{k+1}\subseteq Y_k$. Then put:

$$\alpha _{i,k+1}: = \alpha _{ik};~ A_{i,k+1}: = A_{ik}; ~B_{j,k+1}: = B_{jk}.$$

(b) $X_{k+1}\not \subseteq Y_k$. Then $X_{k+1} \not \subseteq \mathbf{C}Y_k$. Indeed, $X_{k+1} \subseteq \mathbf{C}Y_k$ implies $X_{k+1} \subseteq \mathbf{I}\mathbf{C}Y_k=Y_k$ since $X_{k+1}$ is open and by (1) and (2). So we put

$$W_0 := X_{k+1} - \mathbf{C}Y_k -\bigcup ^m_{j=1}B_{jk},~ W := W_0 - B.$$

Since $(X_{k+1} - \mathbf{C}Y_k )$ is nonempty and open and every $B_{jk}$ is finite by (4), $W_0$ is also open and nonempty (by the density of $\mathfrak X$). By the assumption of Proposition 11, $B$ is closed, and thus $W$ is open.

$W$ is also nonempty. Otherwise $W_0 \subseteq B$, and then $W_0 \subseteq \mathbf{I}B=\varnothing$ (since $B$ is nowhere dense by the assumption of Proposition 11).

Now we argue similarly to the case $k=0$. Take disjoint closed nontrivial balls $P_1,\dots ,P_l\subset W$. Then $W - \bigcup \limits ^l_{i=1}P_i$ is infinite, so we pick distinct $b_{1,k+1},\dots , b_{m,k+1}$ in this set and put

$$B_{j,k+1}: = B_{jk} \cup \{ b_{j,k+1}\},~ \alpha _{i,k+1}: = \alpha _{ik} \cup \{ \mathbf{I}P_i\},~ A_{i,k+1}: = A_{ik} \cup \mathbf{I}P_i.$$

In case (a) all the required properties hold for $(k+1)$ by the construction.

In case (b) we have to check only (1), (2), (6), (8)–(11).

(8) holds since by construction we have

$$\begin{aligned} P_i\subset W\subset&X_{k+1}-\mathbf{C}Y_k\subset X_{k+1}-A_{ik};\\ b_{j,k+1}\in W \subseteq&X_{k+1},~ b_{j,k+1}\in (B_{j,k+1}-B_{jk}). \end{aligned}$$

(1): From IH it is clear that $\alpha _{i,k+1}$ is a finite set of open balls and their closures are disjoint; note that $P_i \cap \mathbf{C}A_{ik} = \varnothing$ since $P_i \subseteq W \subseteq -\mathbf{C}A_{ik}.$

(2): We have

$$\begin{aligned} \mathbf{C}A_{i,k+1} \cap \mathbf{C}A_{i^\prime ,k+1}&= (\mathbf{C}A_{ik} \cup P_i) \cap (\mathbf{C}A_{i^\prime k}\cup P_{i^\prime }) \\&=(\mathbf{C}A_{ik} \cap \mathbf{C}A_{i^\prime k}) \cup (\mathbf{C}A_{ik} \cap P_{i'}) \cup (\mathbf{C}A_{i^\prime k} \cap P_i) \cup (P_i \cap P_{i'})\\ {}&=\mathbf{C}A_{ik} \cap \mathbf{C}A _{i^\prime k} = \varnothing \end{aligned}$$

by IH and by the construction; note that $P_i, P_i' \subseteq W \subseteq -\mathbf{C}Y_k$.

(6): We have

$$ A_{i,k+1} \cap B_{j,k+1} = (A_{ik} \cap B_{jk} ) \cup (\mathbf{I}P_i \cap \left\{ b_{j,k+1}\right\} ) \cup ( A_{ik} \cap \left\{ b_{j,k+1}\right\} ) \cup (\mathbf{I}P_i \cap B_{jk}) = \varnothing $$

by IH and since $b_{j,k+1}\not \in P_i,~ b_{j,k+1}\in W \subseteq X-Y_k$, $P_i \subset W \subseteq X-B_{jk}.$

(9): We have $A_{i,k+1} = A_{ik} \cup \mathbf{I}P_i \subseteq -B$ since $A_{ik} \subseteq -B$ by IH, and $P_i \subset W \subseteq -B$ by the construction.

Likewise, (10) follows from $B_{jk} \subseteq -B$ and $b_{j,k+1}\in W \subseteq -B$.

To check (11), assume $j \not = j'$. We have $B_{j',k+1} \cap B_{j,k+1} = B_{j'k} \cap B_{jk}$ since $b_{j^\prime ,k+1} \not = b_{j,k+1} , ~ b_{j,k+1}\in W \subseteq -B_{j^\prime k}$ and $b_{j^\prime ,k+1}\in W\subseteq -B_{jk}$. Then apply IH.

Therefore, the required sets $A_{ik}, B_{jk}$ are constructed. Now put

$$\alpha _i: =\bigcup _k \alpha _{ik},~ A_i:= \bigcup \alpha _i=\bigcup _k A_{ik}, ~ B_j: =\bigcup _k B_{jk},$$

$$ B^\prime _1: = X - (\bigcup _i A_i \cup \bigcup _j B_j),$$

and define a map $g: X\longrightarrow \Phi _{ml}$ as follows:

$$ g(x):=\left\{ \begin{array}{ll} a_i &{} \text{ if } x\in A_i, \\ b_j &{} \text{ if } x\in B_j,~j\ne 1,\\ b_1 &{} \text{ otherwise } \text{(i.e., } \text{ for } x\in B^\prime _1). \\ \end{array} \right. $$

By (2), (3), (5), (6), (11), $g$ is well defined; by (9), (10), $B \subseteq g^{-1}(b_1)$.

To prove that $g$ is a d-morphism, we check some other properties.

$$(12)\quad X - \bigcup \limits ^l_{i=1}A_i \subseteq \mathbf{d}B_j.$$

Indeed, take an arbitrary $x\not \in \bigcup \limits ^l_{i=1}A_i$ and show that $x\in \mathbf{d}B_j$, i.e.,

$$(13)\quad (U-\{ x\}) \cap B_j \not = \varnothing $$

for any neighbourhood $U$ of $x$. First assume that $x\not \in B_j$. Take a basic open $X_{k+1}$ such that $x\in X_{k+1} \subseteq U$. Then $X_{k+1}\not \subseteq \bigcup \limits ^l_{i=1}A_i$, and (8) implies $B_{j,k+1} \cap X_{k+1} \not = \varnothing .$ Thus $B_j \cap U \not = \varnothing$. So we obtain (13).

Suppose $x\in B_j$; then $x\in B_{jk}$ for some $k$. Since $\mathfrak X$ is dense-in-itself and $\left\{ X_1,\,X_2,\dots \right\}$ is its open base, $\left\{ X_{s+1}\,\left| \,s\ge k \right. \right\}$ is also an open base (note that every ball in $\mathfrak X$ contains a smaller ball). So $x\in X_{s+1}\subseteq U$ for some $s\ge k$. Since $x\not \in \bigcup \limits ^l_{i=1}A_i$, we have $X_{s+1}\not \subseteq \bigcup \limits ^l_{i=1}A_i$, and so $(B_{j,s+1}-B_{js}) \cap X_{s+1} \not = \varnothing$ by (8); thus $(B_j-B_{js}) \cap U \not = \varnothing$. Now $x\in B_{jk} \subseteq B_{js}$ implies (13).

$$(14)\, \mathbf{d}B_j \subseteq X -\bigcup \limits ^l_{i=1}A_i.$$

Indeed, $B_j\subseteq -A_i$ by (3), (5), (6). So $\mathbf{d}B_j \subseteq \mathbf{d}(-A_i) \subseteq -A_i$ since $A_i$ is open.

Similarly we obtain

$$(15) \, \mathbf{d}B^\prime _1\subseteq X -\bigcup \limits ^l_{i=1}A_i, \qquad \mathbf{d}A_i \subseteq X- \bigcup \limits _{r\not =i}A_r.$$

Also note that

$$(16)\, A_i \subseteq \mathbf{d}A_i$$

since $A_i$ is open, $\mathfrak X$ is dense-in-itself. As in (12) we have

$$(17)\, \alpha _i \text{ is } \text{ dense } \text{ at } \text{ every } \text{ point } \text{ of } B_j, B'_1\ \text{(and } \text{ thus } B_j,\ B'_1 \subseteq \mathbf{d}A_i).$$

To conclude that $g$ is a d-morphism, note that

$$g^{-1}(a_i) = A_i, ~g^{-1}(b_j) = B_j~ (\text{ for } j\not = 1),~ g^{-1}(b_1) = B^\prime _1,$$

and so by (15), (16), (17)

$$\begin{aligned} \mathbf{d}g^{-1}(a_i)&= \mathbf{d}A_i = X - \bigcup \limits _{r\not =i}A_r = g^{-1}(R^{- 1}(a_i)), \end{aligned}$$

and by (12), (14), (15)

$$\begin{aligned} \mathbf{d}g^{-1}(b_j)&= \mathbf{d}B_j = X - \bigcup \limits ^l_{i=1}A_i = g^{-1}(R^{- 1}(b_j))\ \ \text{(for } j\ne 1\text{) },\\ \qquad \qquad \qquad \mathbf{d}g^{-1}(b_1)&= \mathbf{d}B^\prime _1= X - \bigcup \limits ^l_{i=1}A_i = g^{-1}(R^{- 1}(b_1)). \qquad \qquad \qquad \square \end{aligned}$$

Proposition 19 For a finite rooted $\mathbf {DT_1CK}$ -frame $F = (W, R, R_D)$ and $R$ -reflexive points $w', w'' \in W$, the following holds.

• (a) If $X=\left\{ x\in \mathbf{R}^n\mid ||x|| \le r \right\}$, $n \ge 2$, then there exists $f: X \twoheadrightarrow ^{dd}F$ such that $f(\partial X) = \left\{ w'\right\} ;$

• (b) If $0\le r_1<r_2$ and

  $$\begin{aligned} X&= \left\{ x\in \mathbf{R}^n\,\left| \,r_1\le ||x||\le r_2 \right. \right\} , \\ Y'&=\left\{ x\in \mathbf{R}^n\,\left| \,||x||=r_1 \right. \right\} ,\ Y'' =\left\{ x\in \mathbf{R}^n\,\left| \,||x||=r_2 \right. \right\} , \end{aligned}$$

  then there exists $f: X \twoheadrightarrow ^{dd}F$ such that $f(Y') = \left\{ w'\right\}$, $f(Y'') = \left\{ w''\right\} .$

Proof By induction on $\left| W\right|$. Let us prove (a) first. There are five cases:

(a1) $W = R(b)$ (and hence $b R b$) and $b = w'$. Then there exists $f:X\twoheadrightarrow ^d (W,R)$. Indeed, let $C$ be the cluster of $b$ (as a subframe of $(W,R)$). Then $(W,R)=C$ or $(W,R)=C\cup F_1\cup \ldots \cup F_l$, where the $F_i$ are generated by the successors of $C$. If $(W,R)=C$, we apply Proposition 11; otherwise we apply Lemma 15 and IH.

By Proposition 8 it follows that $R_D$ is universal. And so by Proposition 12(3) $f$ is a dd-morphism.

(a2) $W = R(b)$ and not $w'Rb$. We may assume that $r=3$. Put

$$ X_1 := \left\{ x \mid ||x|| \le 1\right\} ,~ Y := \left\{ x \mid 1\le ||x|| \le 2 \right\} ,~ X_2 := \left\{ x \mid 2 \le ||x|| \le 3 \right\} . $$

By the case (a1), there is $f_1: X_1 \twoheadrightarrow ^{dd}F$ with $f_1(\partial X_1) = \left\{ b\right\}$. Let $C$ be a maximal cluster in $R(w')$. By Proposition 11 there is $g: \mathbf{I}Y \twoheadrightarrow ^d C$. Since $R(w')\ne W$, we can apply IH to the frame $F':=F^{w'}_{\forall }$ and construct a dd-morphism $f_2: X_2 \twoheadrightarrow ^{dd}F'$ with $f_2(\partial X_2) = \left\{ w'\right\}$. Now since $f_i(\partial X_i)\subseteq R^{-1}(C)$, the Glueing lemma 16 is applicable. Thus $f:X\twoheadrightarrow ^d F$ for $f:=f_1\cup f_2\cup g$ [See Fig. 11.7, Case (a2)]. Note that $\partial X\subset \partial X_2$, so $f(\partial X)=f_2(\partial X)=\{w'\}$.

Fig. 11.7

dd-morphism f

As in case (a1), $f$ is a dd-morphism by Proposition 12.

(a3) $(W,R)$ is not rooted. By Lemma 33 there is a global path $\alpha$ in $F$ with a single occurrence of every $R_D$-irreflexive point. We may assume that $\alpha = b_0 c_0 b_1 c_1\ldots c_{m-1} b_m$, $b_m = w'$ and for any $i<m$, $c_i \in C_i \subseteq R(b_i) \cap R(b_{i+1})$, where $C_i$ is an $R$-maximal cluster. Such a path is called reduced. For $0 \le j \le m$ we put $F_j := F|\overline{R}(b_j)$.

Since $(W,R)$ is not rooted, each $F_j$ is of smaller size than $F$, so we can apply the induction hypothesis to $F_j$. We may assume that

$$X = \left\{ x \mid ||x||\le 2m+1\right\} , \ Y=\left\{ x \mid ||x||=2m+1\right\} .$$

Then put

$$ X_i := \left\{ x \mid ||x|| \le i+1 \right\} \text{ for } 0 \le i \le 2m, $$

$$ Y_i := \partial X_i,~ {\Delta }_i := \mathbf{C}(X_{i} -X_{i-1}) \text{ for } 0 \le i \le 2m. $$

By IH and Proposition 11 there exist

$$\begin{aligned} f_0:&X_0 \twoheadrightarrow ^{dd} F_0 \hbox { such that } f_0(Y_0) = \{c_0\},\\ f_{2j}:&{\Delta }_{2j} \twoheadrightarrow ^{dd}F_{j} \text{ such } \text{ that } f_{2j}(Y_{2j}) = \{c_j\}, \ f_{2j}(Y_{2j-1}) = \{c_{j-1}\}\ \text{ for } 1 \le j \le m,\\ f_{2j-1}:&\mathbf{I}{\Delta }_{2j+1} \twoheadrightarrow ^d{}C_j \text{ for } 0 \le j \le m-1. \end{aligned}$$

One can check that $f:X\twoheadrightarrow ^{dd} F$ for $f:=\bigcup \limits _{j=0}^{2m}f_j$ (Fig. 11.7).

(a4) $W = \overline{R}(b)$, $\lnot b R_D b$ (and so $\lnot b R b$). We may assume that

$$X = \left\{ x \mid ||x||\le 2\right\} , \ Y=\left\{ x \mid ||x||=2\right\} .$$

Then similar to case (a3) put

$$X_0 := X, ~Y_0 := Y, ~ X_i := \left\{ x \mid ||x|| \le \frac{1}{i} \right\} ,~ Y_i := \partial X_i,~ {\Delta }_i := \mathbf{C}(X_{i} -X_{i+1}), ~ (i>0). $$

Consider the frame $F' {:}= F|W'$, where $W' = W - \left\{ b\right\}$. Note that $w'\in W'$ since $w'Rw'$ by the assumption of Proposition 19. By Lemma 30 $F'$ is connected, and thus $F'\vDash \mathbf {DT_1CK}$. By Lemma 33 there is a reduced global path $\alpha = a_1 \ldots a_m$ in $F'$ such that $a_1 = w'$. Let

$$ \upgamma = a_1 a_2 \ldots a_{m-1} a_m a_{m-1} \ldots a_2 a_1 a_2 \ldots $$

be an infinite path shuttling back and forth through $\alpha$. Rename the points in $\upgamma$:

(11.6)

Again as in case (a3) we put $F_j {:}= F | \overline{R} (b_j)$, and assume that $c_j \in C_j$ and $C_j$ is an $R$-maximal cluster. By IH there exist

$$\begin{aligned} f_0:&{\Delta }_0 \twoheadrightarrow ^{dd} F_0 \hbox { such that } f_0(Y_0) = \{b_0\} = \{w'\}, \ f_1(Y_1) =\{ c_0\},\\ f_{2j}:&{\Delta }_{2j} \twoheadrightarrow ^{dd}F_{j} \hbox { such that } f_{2j}(Y_{2j}) = \{c_{j-1}\}, \ f_{2j}(Y_{2j+1}) = \{c_{j}\}\hbox { for }j>0, \end{aligned}$$

and by Proposition 11 there exist $f_{2j+1}: \mathbf{I } {\Delta }_{2j+1} \twoheadrightarrow ^d{}C_j$. Put

$$ f(x) := \left\{ \begin{array}{ll} b &{}\hbox { if } x = \mathbf{0},\\ f_{2j}(x)&{}\hbox { if } x \in \Delta _{2j},\\ f_{2j+1}(x)&{}\hbox { if } x \in \mathbf{I}\Delta _{2j+1}, \end{array} \right. $$

One can check that $f$ is a d-morphism (Fig. 11.8).

Fig. 11.8

dd-morphism f

(a5) $W = \overline{R}(b)$, $\lnot b R b$ and $b R_D b$. Then $R_D$ is universal, $w'\ne b$. Put

$$ X' := \left\{ x \mid ||x|| < 1\right\} , X_4 := \left\{ x \mid 1 \le ||x|| \le 2\right\} , $$

and let $X_1,X_2$ be two disjoint closed balls in $X'$, $X_3 := X'- X_1 - X_2$.

Let $C$ be a maximal cluster in $R(w')$, $F' := F|R(w')$. Then there exist:

$$\begin{aligned} f_i:&X_i \twoheadrightarrow ^d (W,R) \text{ for } i=1,2 \text{ such } \text{ that } \ f_i(\partial X_i) = \left\{ w'\right\} , \hbox { by case (a4),}\\ f_3:&X_3 \twoheadrightarrow ^d C, \hbox { by Proposition 11,}\\ f_4:&X_4 \twoheadrightarrow ^{dd}F' \text{ such } \text{ that } \ f_4(\partial X_4) = \left\{ w'\right\} , \hbox { by the induction hypothesis.} \end{aligned}$$

Put $f:=f_1\cup f_2\cup f_3\cup f_4$ (Fig. 11.8). Then $f(\partial X) = \left\{ w'\right\}$.

By Lemma 16 (b), $f_1\cup f_2:X_1\cup X_2\twoheadrightarrow ^d F$, and hence $f:X\twoheadrightarrow ^d F$ by Lemma 16 (a). $f$ is manifold at $b$, thus it is a dd-morphism by Lemma 18.

Now we prove (b). There are three cases.

(b1) $w' = w'' = b$ and $W = R(b)$. The argument is the same as in case (a1), using Proposition 11, Lemma 15, the induction hypothesis, and Proposition 12.

(b2) $w' = w'' = b$, but $W \ne R(b)$. Consider a maximal cluster $C \subseteq R (b)$. Since all spherical shells for different $r_1$ and $r_2$ are homeomorphic, we assume that $r_1 = 1$, $r_2 = 4$. Consider the sets

$$\begin{aligned} X_1 := \left\{ x \mid 1 \le ||x|| \le 2\right\} ,\ \ X' := \left\{ x \mid 2 < ||x|| < 3\right\} ,\ \ X_3 := \left\{ x \mid 3 \le ||x|| \le 4\right\} , \end{aligned}$$

and let $X_0\subset X'$ be a closed ball, $X_2 := X' - X_0$. Let $F' :=F| R (b)$. There exist

$$\begin{aligned} f_1:&X_1 \twoheadrightarrow ^{dd}F' \text{ such } \text{ that } f_1(\partial X_1) = \left\{ b\right\} \!, \hbox { by case (b1)},\\ f_2:&X_2 \twoheadrightarrow ^d C, \hbox { by Proposition 11},\\ f_3:&X_3 \twoheadrightarrow ^{dd}F' \text{ such } \text{ that } \ f_3(\partial X_3) = \left\{ b\right\} \!, \hbox { by case (b1)},\\ f_0:&X_0 \twoheadrightarrow ^{dd}F \text{ such } \text{ that } \ f_4(\partial X_0) = \left\{ b\right\} \!, \hbox { by statement (a) for }F. \end{aligned}$$

One can check that $f:X\twoheadrightarrow ^{dd}F$ for $f:=f_0\cup f_1\cup f_2\cup f_3$.

(b3) $w' \ne w''$ and for some $b\in W$, $W = R(b)$, so $F$ has an $R$-reflexive root. Let

$$ F_1 := F|R(w'), ~ F_2 := F|R(w''), $$

and let $C_i$ be an $R$-maximal cluster in $F_i$ for $i\in \left\{ 1,2\right\} .$

We assume that $r_1 = 1$, $r_2 = 6$ and consider the sets

$$\begin{aligned} X_i&:= \left\{ x \mid i \le ||x|| \le i+1\right\} , \ i \in \left\{ 1, \ldots , 5.\right\} \end{aligned}$$

By case (b1) and Proposition 11 we have

$$\begin{aligned} f_1:&~ X_1 \twoheadrightarrow ^{dd}F_1 \text{ such } \text{ that } f_1(\partial X_1) = \left\{ w'\right\} \!,&f_2:&~ \mathbf{I}X_2 \twoheadrightarrow ^d C_1,\\ f_3:&~ X_3 \twoheadrightarrow ^{dd}F \text{ such } \text{ that } \ f_3(\partial X_3) = \left\{ b\right\} \!,&f_4:&~ \mathbf{I}X_4 \twoheadrightarrow ^d C_2, \\ f_5:&~ X_5 \twoheadrightarrow ^{dd}F_2 \text{ such } \text{ that } f_1(\partial X_5) = \left\{ w''\right\} \!. \end{aligned}$$

One can check that $f:X\twoheadrightarrow ^{dd}F$ for $f:= \bigcup \limits _{i=1}^5 f_i$ [Fig. 11.9, Case (b3)].

Fig. 11.9

dd-morphism f

(b4) $w' \ne w''$ and $W \ne R(b)$ for any $b\in W$. By Lemma 32 there is a reduced path $\alpha = b_0 c_0 b_1 \ldots c_{m-1} b_m$ from $b_0 = w'$ to $b_m = w''$ that does not contain $R_D$-irreflexive points, $c_i \in C_i$, where $C_i$ is an $R$-maximal cluster. We may also assume that

(11.7)

Indeed, if the frame $(W,R)$ is not rooted, then (11.7) obviously holds. If $(W,R)$ is rooted, then its root $r$ is irreflexive and by Lemma 30, $R(r)$ is connected, so there exists a path $\alpha$ in $R(r)$ satisfying (11.7). Put

$$ F_0 := F,~ F_j := F|R(b_j), 1 \le j \le m. $$

Assuming that $r_1 = 1$, $r_2 = 2m+1$ we define

$$\begin{aligned} X_i&:= \left\{ x\,\left| \,||x|| \le i +1 \right. \right\} ,~ Y_i := \partial X_i~ (\text{ for } 0 \le i \le 2m+1),\\ {\Delta }_i&:= \mathbf{C}(X_{i+1} -X_{i}) \ (\text{ for } 0\le i\le 2m). \end{aligned}$$

By cases (b2), (b1), Proposition 11, and the induction hypothesis there exist

$$\begin{aligned}&f_0: {\Delta }_0 \twoheadrightarrow ^{dd}F=F_0 \hbox { such that } f_0(Y_0) = f_0(Y_1) = \left\{ w'\right\} \!; \\&f_{2j}: {\Delta }_{2j} \twoheadrightarrow ^{dd}F_{j} \hbox { such that } f_{2j}(Y_{2j+1}) = \left\{ c_{j}\right\} , \ f_{2j}(Y_{2j}) = \{c_{j-1}\}\ (1 \le j \le m);\\&f_{2j-1}: \mathbf{I}{\Delta }_{2j-1} \twoheadrightarrow ^d C_{j-1} \ (1 \le j \le m),\\&f_{2m}: {\Delta }_{2m} \twoheadrightarrow ^{dd}F_m \hbox { such that } f_{2m}(Y_{2m}) = \left\{ c_{m}\right\} , \ f_{2m}(Y_{2m+1}) = \left\{ w''\right\} \!. \end{aligned}$$

We claim that $f:X\twoheadrightarrow ^{dd}F$ for $f:= \bigcup \limits _{i=0}^{2m} f_i$ (Fig. 11.10). First, we prove by induction using Lemma 16 (see the previous cases) that $f$ is a d-morphism. Note that $f(Y')= f(Y_0) = \left\{ w'\right\}$ and $f(Y'')= f(Y_{2m+1}) = \left\{ w''\right\}$.

Fig. 11.10

dd-morphism f, case (b4)

Second, there are no $R_D$-irreflexive points in $\alpha$, so all preimages of $R_D$-irreflexive points are in $\Delta _0$; since $f_0$ is a dd-morphism, $f$ is 1-fold at any $R_D$-irreflexive point and manifold at all the others. Thus $f$ is a dd-morphism by Proposition 12. $\square$