Multiplicative derivations and d-filters of commutative residuated lattices

Abstract

In this paper, we consider some properties of multiplicative derivations and d-filters of commutative residuated lattices and show that, for an ideal derivation d of a residuated lattice $L=(L,\wedge , \vee , \odot , \rightarrow , 0,1)$, (1) the set $\hbox {Fix}_d(L)= (\hbox {Fix}_d(L), \wedge , \vee , \odot , \mapsto , 0, d1)$ of all fixed points of d forms a residuated lattice and d is a homomorphism from L to $\hbox {Fix}_d(L)$, (2) for a d-filter F, a map $d/F: L/F \rightarrow L/F$ defined by $(d/F)(x/F) = dx/F$ is also an ideal derivation of L/F and (3) two quotient residuated lattices $\hbox {Fix}_{d/F}(L/F)$ and $\hbox {Fix}_d (L)/d(F)$ are isomorphic as residuated lattices, that is, $\hbox {Fix}_{d/F}(L/F) \cong \hbox {Fix}_d (L)/d(F)$.

1 Introduction

A notion of derivation whose origin is in analysis has been applied to theory of algebras with two operations $+$ and $\cdot$, especially to the theory of rings (Posner 1957). For an algebra $A=(A,+,\cdot )$, a map $f:A\rightarrow A$ is called a derivation in Posner (1957) if it satisfies the conditions: For all $x,y\in A$,

$$\begin{aligned}&f(x+y) = f(x) + f(y) \\&f(x\cdot y) = f(x) \cdot y + x\cdot f(y). \end{aligned}$$

It was applied to the theory of lattices (Ferrari 2001; Szász 1975; Xin et al. 2008), where operations $+$ and $\cdot$ were interpreted as lattice operations $\vee$ and $\wedge$, respectively. In particular, it has been proven in Szász (1975) that if f is a derivation of a bounded lattice L, that is, f satisfies two conditions:

$$\begin{aligned} f(x\vee y)&= f(x) \vee f(y) \\ f(x\wedge y)&= (f(x) \wedge y) \vee (x\wedge f(y)), \end{aligned}$$

then f just has a form $f(x) = x \wedge f(1)$. Since then, every type of derivations is defined as a map $f: X \rightarrow X$ satisfying only a condition:

$$\begin{aligned} f(x\cdot y) = f(x) \cdot y + x\cdot f(y), \end{aligned}$$

where $(X,+, \cdot )$ is an algebra. Thus, in the case of lattices, a derivation f of a lattice L is defined a map $f:L \rightarrow L$ satisfying the condition:

$$\begin{aligned} f(x\wedge y) = (f(x) \wedge y) \vee (x\wedge f(y)). \end{aligned}$$

By the use of derivations of lattices, characterization theorems of distributive lattices and of modular lattices were proven in [6]:

Let L be a lattice and f a derivation.

• (1) The condition that f is monotone $\Leftrightarrow f(f(x)\vee y) = f(x) \vee f(y)\ (\forall x,y\in L)$ holds if and only if L is a modular lattice.

• (2) The condition that f is monotone $\Leftrightarrow f(x\vee y) = f(x) \vee f(y) \ (\forall x,y\in L)$ holds if and only if L is a distributive lattice.

Further, it is also applied to other algebras, such as MV algebras (Alshehri 2010; Ghorbani et al. 2013; Yazarli 2013), where operations $+$ and $\cdot$ were interpreted as $\oplus$ and $\odot$, respectively. In He et al. (2016), another type of derivations, multiplicative derivations, are defined on residuated lattices, where operations $+$ and $\cdot$ are interpreted as $\vee$ and $\odot$, respectively. By this definition of derivations, we have several interesting properties about residuated lattices. For example, in He et al. (2016), it is shown that if d is an ideal derivation of a residuated lattice L, then the set $\hbox {Fix}_d(L)$ of all fixed elements formed a residuated lattice. Thus, it is expectable for residuated lattices with multiplicative derivations to have other deeper properties. Indeed, we show some interesting properties here, for example, for every residuated lattice L with an ideal multiplicative derivation d, the set $\hbox {Fix}_d(L)$ is isomorphic to $L/\ker (d)$ and hence $\hbox {Fix}_d(L)$ is a residuated lattice.

In this paper, we show the following results. Let L be a commutative residuated lattice and d be an ideal derivation of it. Then, we have

• (1) The set $\hbox {Fix}_d(L)$ of all fixed points of d forms a residuated lattice and $L/{\ker (d)} \cong \hbox {Fix}_d(L)$;

• (2) A map $d/F: L/F \rightarrow L/F$ defined by $(d/F)(x/F) = dx/F$ is also an ideal derivation of L/F;

• (3) The quotient residuated lattices $\hbox {Fix}_{d/F}(L/F)$ and $\hbox {Fix}_d(L)/d(F)$ are isomorphic, namely $\hbox {Fix}_{d/F}(L/F) \cong \hbox {Fix}_d(L)/d(F)$.

We also show the characterization theorem of d-filters, which says that for an ideal derivation d and a non-empty subset S, the smallest d-filter containing S is identical with the filter containing $S\cup d(S)$, that is, $[S)_d = [S\cup d(S))$.

2 Derivations of residuated lattices

We recall a definition of bounded integral commutative residuated lattices (Galatos et al. 2007; Ward and Dilworth 1939). An algebraic structure $(L, \wedge , \vee , \odot , \rightarrow , 0, 1)$ is called a bounded integral commutative residuated lattice (simply called a residuated lattice) if

• (1) $(L, \wedge , \vee , 0, 1)$ is a bounded lattice;

• (2) $(L, \odot , 1)$ is a commutative monoid with unit element 1;

• (3) For all $x, y, z \in L$, $x \odot y \le z$ if and only if $x \le y \rightarrow z$.

For all $x \in L$, by $x'$, we mean $x' = x \rightarrow 0$, which is a negation in a sense.

In what follows, let $L=(L,\wedge , \vee , \odot , \rightarrow ,0,1)$ be a residuated lattice. An element $x\in L$ is called complemented if there exists an element $y\in L$ such that $x \wedge y=0$ and $x\vee y=1$. By B(L), we mean the set of all complemented elements of L, i.e.,

$$\begin{aligned} B(L) = \{ x\in L \,|\, \exists y\in L \ \mathrm{s.t.} \ x\wedge y=0, x\vee y=1 \}. \end{aligned}$$

It is easy to show the following results, so we omit their proofs.

Proposition 1

(Galatos et al. 2007) For a residuated lattice L, we have

• (1) $x\in B(L)$ if and only if $x\vee x'=1$;

• (2) If $x\in B(L)$, then $x\wedge y= x\odot y$ for all $y\in L$;

• (3) If $x\oplus y= x\vee y$ for all $y\in L$, then $x\vee x'=1$, where $x\oplus y = (x'\odot y')'$;

• (4) B(L) is a Boolean subalgebra of L.

We note that $a\odot (a\rightarrow x)=a\odot x$ for all $a\in B(L)$ and $x\in L$, because, since $a\in B(L)$ and $a\odot (a\rightarrow x)\le x$, we have $a\odot (a\rightarrow x)=a\odot a\odot (a\rightarrow x) \le a\odot x$. Conversely, it follows from $x\le a\rightarrow x$ that $a\odot x\le a\odot (a\rightarrow x)$. Therefore, we get $a\odot (a\rightarrow x)=a\odot x$ for all $a\in B(L)$ and $x\in L$.

We have the following basic properties of residuated lattices (Galatos et al. 2007).

Proposition 2

For all $x, y, z \in L$, we have

• (1) $0'= 1, 1'= 0$;

• (2) $x \odot x'= 0$;

• (3) $x \le y \ \Longleftrightarrow \ x \rightarrow y = 1$;

• (4) $x \odot (x \rightarrow y) \le y$;

• (5) $x \le y \ \Longrightarrow \ x \odot z \le y \odot z, z \rightarrow x \le z \rightarrow y, y \rightarrow z \le x \rightarrow z$;

• (6) $1 \rightarrow x = x$;

• (7) $(x \vee y) \odot z = (x \odot z) \vee (y \odot z)$;

• (8) $(x \vee y)' = x' \wedge y'$;

• (9) $(x''\odot y'')^{''} =(x\odot y'')^{''}= (x\odot y)^{''}$.

We define derivations of residuated lattices according to He et al. (2016). A map $d: L \rightarrow L$ is called a multiplicative derivation (or simply derivation here) of L if it satisfies the condition:

$$\begin{aligned} d(x\wedge y) = (dx \odot y) \vee (x \odot dy) \ \ \ (\forall x,y\in L). \end{aligned}$$

We simply denote dx instead of d(x). A derivation d is called monotone; if $x\le y$, then $dx\le dy$, that is, d is order-preserving. A derivation d is good when $d1\in B(L)$. If a derivation d is monotone and good, then it is said to be ideal. We note that any multiplicative derivation d is contractive, $dx\le x$ for all $x\in L$, because we have $dx = d(x\wedge x) = (dx\odot x) \vee (x\odot dx) = dx\odot x \le x$. This result was not referred in He et al. (2016).

Example

Let $X=\{0,a,1\}$ with $0<a<1$ be a residuated lattice if we define $x\wedge y=x\odot y=\min \{x,y\}, x\vee y=\max \{x,y\}$ and

$$\begin{aligned} x\rightarrow y=\left\{ \begin{array}{ll} 1 &{}\quad (\mathrm{if} \ x\le y) \\ y &{}\quad (\mathrm{otherwise}) \\ \end{array}. \right. \end{aligned}$$

A map $d_a$ defined by $d_a(x)=x\wedge a$ for all $x\in X$ is a monotone derivation, but not good. As another example of derivation, we have $f:X\rightarrow X$ defined by $f1=0=f0, fa=a$. It is easy to show that f is a good derivation, but it is not monotone.

We have fundamental results about derivations of residuated lattices.

Proposition 3

(He et al. 2016) Let d be a derivation of L. For all $x,y\in L$,

• (1) $d0=0$;

• (2) $dx \ge x\odot d1$;

• (3) $dx^n = x^{n-1}\odot dx$ for all $n \ge 1$;

• (4) If $x\odot y=0$ then $dx \odot y = x\odot dy = dx \odot dy =0$;

• (5) $d(x') \le (dx)'$.

In He et al. (2016), the following characterization theorem about an ideal derivation was proven.

Theorem 1

(He et al. 2016) Let d be a derivation of L and $d1\in B(L)$. Then, the following are equivalent: for all $x,y\in L$,

• (1) d is an ideal derivation;

• (2) $dx\le d1$;

• (3) $dx= x\odot d1= x\wedge d1$;

• (4) $d(x\wedge y) = dx \wedge dy$;

• (5) $d(x\vee y) = dx\vee dy$;

• (6) $d(x\odot y) = dx \odot dy$.

From the above, we see that $dx=d(dx)$ for all $x\in L$ for an ideal derivation d, that is, $d=d^2$, because, since d is the ideal derivation, we have $dx= x\wedge d1$ and thus $d^2x = d(dx) =d(x\wedge d1) = (x\wedge d1)\wedge d1 = x\wedge d1 = dx$ for all $x\in L$. This means that $d^2 = d$.

For a derivation d of L, we denote by $\hbox {Fix}_d(L)$ the set of all fixed elements of L for d, that is, $\hbox {Fix}_d(L) =\{x\in L\,|\, dx=x \}$.

It is easy to show the next result; hence, we omit its proof.

Proposition 4

For an ideal derivation d of a residuated lattice L, we have $\hbox {Fix}_d(L) = d(L)$.

Lemma 1

Let d be an ideal derivation. Then, we have $d(dx\rightarrow dy) = d(x\rightarrow y)$ for all $x,y\in L$.

Proof

Suppose that d is an ideal derivation. Since $dz = z\odot d1=z\wedge d1$ for all $z\in L$, we have

$$\begin{aligned} d(dx\rightarrow dy)&= (dx\rightarrow dy) \odot d1 \\&= (x\wedge d1 \rightarrow y\wedge d1)\odot d1 \\&= \{(x\wedge d1\rightarrow y) \wedge (x\wedge d1 \rightarrow d1)\} \odot d1 \\&= (x\wedge d1 \rightarrow y)\odot d1 \\&= d1\odot (d1\odot x \rightarrow y) \\&= d1 \odot (d1 \rightarrow (x\rightarrow y)) \\&= d1 \odot (x\rightarrow y) \ \ \ (d1 \in B(L))\\&= d(x\rightarrow y). \end{aligned}$$

$\square$

It follows from the characterization theorem about ideal derivations and above that we define some operations in $\hbox {Fix}_d(L)=d(L)$ by

$$\begin{aligned} dx \sqcap dy&= d(x\wedge y); \\ dx\sqcup dy&= d(x\vee y); \\ dx\boxdot dy&= d(x\odot y); \\ dx \mapsto dy&= d(dx\rightarrow dy). \end{aligned}$$

Then, we have

Theorem 2

Let L be a residuated lattice and d be an ideal derivation of L. Then, $\hbox {Fix}_d(L) = (\hbox {Fix}_d(L), \sqcap , \sqcup , \boxdot , \mapsto , 0, d1)$ is a residuated lattice. However, it is not a subalgebra of L in general.

Proof

We only show that $dx \odot dy \le dz$ if and only if $dx \le dy \mapsto dz$ for all $dx, dy, dz\in d(L)=\hbox {Fix}_d(L)$. If $dx\odot dy \le dz$, then we have $dx\le dy \rightarrow dz$ and thus $dx = d(dx)\le d(dy\rightarrow dz) = dy \mapsto dz$. Conversely, suppose that $dx\le dy\mapsto dz=d(dy\rightarrow dz)$. Since d is contractive, we have $d(dy\rightarrow dz)\le dy\rightarrow dz$ and hence $dx\le dy\rightarrow dz$. This yields $dx\odot dy\le dz$. $\square$

Remark 1

We note that the theorem above was already proven as Theorem 3.15 in He et al. (2016), where the meet operation $\sqcap$ is defined by $d(dx\wedge dy)$.

We have proved above that $\hbox {Fix}_d(L)$ is a residuated lattice for an ideal derivation d. Moreover, we see that any ideal derivation d is a homomorphism from L to $\hbox {Fix}_d(L)$, since $d(x\rightarrow y) = d(dx\rightarrow dy) = dx\mapsto dy$. The other cases such as $d(x\wedge y)$ also can be proved easily. Therefore, d is the homomorphism from L to $\hbox {Fix}_d(L)$. It follows from the homomorphism theorem of residuated lattices that $L/\ker (d) \cong \hbox {Fix}_d(L)$, where $\ker (d) = \{(x,y) \,|\, dx=dy \}$.

Theorem 3

For every ideal derivation d, it is a homomorphism from L to $\hbox {Fix}_d(L)$ and hence

$$\begin{aligned} L/\ker (d) \cong \hbox {Fix}_d(L). \end{aligned}$$

3 Galois connection of derivations

In this section, we consider Galois connections of derivations. Let P, Q be partially ordered sets and f, g be maps, $f:P \rightarrow Q$, $g: Q \rightarrow P$. A pair (f, g) of maps is called a Galois connection if

$$\begin{aligned} f(x)\le _{Q} y \ \Leftrightarrow \ x\le _{P} g(y) \ \ (\forall x\in P, \forall y\in Q). \end{aligned}$$

A Galois connection (f, g) is simply denoted by $f \dashv g$.

Let L be a residuated lattice and d be an ideal derivation of L. By the characterization theorem of ideal derivations, d has the form $dx=x\odot d1 = x\wedge d1$. In this case, we may ask

$$\begin{aligned} \text {``Is there a map}\ g\ \hbox {such that}\ d \dashv g?\text {''} \end{aligned}$$

We have an affirmative solution as follows.

Theorem 4

Let d be an ideal derivation. There exists an ideal derivation $g: L\rightarrow L$ such that $d \dashv g$. Moreover, g is idempotent.

Proof

We define $gx = d1 \rightarrow x$ for all $x\in L$. It is obvious that $dx=x\odot d1 \le y$ if and only if $x\le d1 \rightarrow y= gy$, that is, $dx\le y \ \Leftrightarrow x\le gy$ for all $x,y\in L$. Moreover, since $g(gx)=d1 \rightarrow (d1 \rightarrow x) = d1 \odot d1 \rightarrow x = d1 \rightarrow x = gx$, g is idempotent. $\square$

From the general theory of idempotent Galois connection, that is, $f\dashv g$, $f^2 =f$ and $g^2=g$ that two subsets $F_f= \{x\in L \,|\, fx=x\}$ and $F_g= \{x\in L \,|\, gx=x\}$ are isomorphic as partially ordered sets. Hence, we have the following result.

Theorem 5

For an ideal derivation d, $\hbox {Fix}_{d}(L) =F_{d} (L) \cong F_{g}(L) = \hbox {Fix}_{g} (L)$ as partially ordered sets.

4 d-filter and its characterization

We define a filter of a residuated lattice which plays an important role in this paper. Let F be a non-empty subset of L. We call F a filter of L if it satisfies the following conditions: For all $x,y\in L$,

• (F1) if $x,y\in F$, then $x\odot y\in F$;

• (F2) if $x\in F$ and $x\le y$, then $y\in F$.

Let d be an ideal derivation of L. A filter F of L is called an ideal derivation filter (simply d-filter here) if $x\in F$ implies $dx\in F$ for all $x\in L$. By ${\mathcal {F}}(L)$ (or ${{\mathcal {F}}}_d(L)$), we mean the set of all filters (or d-filters) of L. By [S) (or $[S)_d$), we mean the generated filter (or generated d-filter, respectively) by S. At first, we provide a characterization theorem about d-filters.

Theorem 6

Let d be an ideal derivation of L. For a non-empty subset S of L, we have $[S)_d = [S\cup d(S))$.

Proof

It is sufficient to show that $[S\cup d(S))$ is the least d-filter including S. It is obvious $S \subseteq [S\cup d(S))$. We show that $[S\cup d(S))$ is a d-filter, that is, if $x\in [S\cup d(S))$, then $dx\in [S\cup d(S))$. Suppose that $x\in [S\cup d(S))$. There exist $a_i, b_j\in S$ such that $a_1 \odot \cdots \odot a_m \odot db_1 \odot \cdots db_n \le x$. Since d is the ideal derivation, we have $da_1 \odot \cdots \odot da_m \odot db_1 \odot \cdots db_n =da_1 \odot \cdots \odot da_m \odot d(db_1) \odot \cdots d(db_n) =d(a_1 \odot \cdots \odot a_m \odot db_1 \odot \cdots db_n) \le dx$. It follows from $da_i, db_j \in d(S) \subseteq S \cup d(S)$ that $dx \in [S\cup d(S))$. Therefore, $[S\cup d(S))$ is the d-filter. Lastly, for any d-filter F including S, we have $d(S) \subseteq d(F) \subseteq F$ and hence $S\cup d(S) \subseteq F$. This implies $[S\cup d(S)) \subseteq F$. Therefore, $[S\cup d(S))$ is the least d-filter containing S and thus $[S)_d = [S\cup d(S))$. $\square$

Corollary 1

(Theorem 3.19 He et al. 2016) The class ${\mathcal {F}}_d(L)$ of all d-filters forms a complete Heyting subalgebra of the class ${\mathcal {F}}(L)$ of all filters of a residuated lattice L

Corollary 2

If F is a filter, then $[F)_d = [d(F))$.

Corollary 3

For any $F\in {\mathcal {F}}(L)$, F is a d-filter if and only if $F=[d(F))$.

Proof

It is easy to show that if F is a d-filter, then $F=[F)_d=[d(F))$. Conversely, suppose that $x\in F=[d(F))$. There exist $a_i \in F$ such that $da_1 \odot \cdots \odot da_n\le x$. If we take $a= a_1 \odot \cdots \odot a_n \in F$, then $da\in d(F)$. Moreover, we have $da = d(a_1 \odot \cdots \odot a_n) = da_1 \odot \cdots \odot da_n \le x$ and thus $da = d(da) \le dx$. This implies that $dx\in [d(F))=F$, namely F is the d-filter. $\square$

Proposition 5

Let d be an ideal derivation of L. Then, we have

• (1) $F\in {\mathcal {F}}_d(L) \Rightarrow d(F)\in {\mathcal {F}}(d(L))$;

• (2) $G\in {\mathcal {F}}(d(L)) \Rightarrow d^{-1}(G)\in {\mathcal {F}}_d(L)$, where $d^{-1}(G) = \{ x\in L\,|\, dx\in G\}$.

Proof

• (1) If $dx, dy\in d(F), (x,y\in F)$ then, since $x\odot y\in F$ and d is the ideal derivation, we have $dx\odot dy = d(x\odot y) \in d(F)$. Moreover, if $dx\in d(F)$ for $x\in F$ and $dx\le dy$, then $dx\in F$ and thus $dy\in F$. Then, $dy = d(dy)\in d(F)$. Therefore, d(F) is a filter of d(L), that is, $d(F)\in {\mathcal {F}}(d(L))$.

• (2) Let G be a filter of d(L). For all $x,y\in d^{-1}(G)$, since $dx, dy\in G$, we get $dx\odot dy = d(x\odot y) \in G$ and $x\odot y\in d^{-1}(G)$. Next, suppose that $x\in d^{-1}(G)$ and $x\le y$. It follows from $dx\in G$ and $dy\le dy$ that $dy\in G$ and hence that $y\in d^{-1}(G)$. Lastly, if $x\in d^{-1}(G)$, then $dx\in G$ and $d(dx) = dx\in G$. This implies $dx\in d^{-1}(G)$. Therefore, $d^{-1}(G)$ is the d-filter, that is, $d^{-1}(G)\in {\mathcal {F}}_d(L)$.$\square$

For a filter F, we define a quotient structure L/F, where $x/F=y/F$ is defined by $x\rightarrow y, y\rightarrow x\in F$ for all $x,y\in L$. Since the set of all residuated lattices forms a variety, the quotient algebra L/F by a filter F is also a residuated lattice.

Proposition 6

Let d be an ideal derivation and F be a d-filter of L. If we define a map $d/F : L/F \rightarrow L/F$ by $(d/F)(x/F) = dx/F$ for all $x/F\in L/F$, then d/F is an ideal derivation of L/F.

Proof

At first, we show that d/F is well defined. Suppose that $x/F=y/F$. Since $x\rightarrow y, y\rightarrow x\in F(\in {\mathcal {F}}_d(L))$, we have $d(x\rightarrow y), d(y\rightarrow x)\in F$. It follows from $d(x\rightarrow y) \le dx\rightarrow dy, d(y\rightarrow x)\le dy\rightarrow dx$ that $dx\rightarrow dy, dy\rightarrow dx\in F$. This means that $(d/F)(x/F)=dx/F=dy/F=(d/F)(y/F)$ and hence that d/F is well defined. It is easy to show that d/F is a good derivation of the residuated lattice L/F. It is sufficient to show that d/F is monotone. Suppose that $x/F\le y/F$. Since $x\rightarrow y\in F$ and F is the d-filter, we have $d(x\rightarrow y)\in F$ and $dx\rightarrow dy\in F$ because of $d(x\rightarrow y)\le dx\rightarrow dy$. This implies $dx/F \le dy/F$ and hence $(d/F)(x/F)=dx/F \le dy/F = (d/F)(y/F)$, namely d/F is monotone. $\square$

We note that $d1/F=1/F$, because, since F is the d-filter, the fact $1\in F$ implies $d1\in F$, that is, $d1/F=1/F$. It follows from the above that the quotient structure $(d/F)(L/F) =\hbox {Fix}_{d/F}(L/F)= (\hbox {Fix}_{d/F}(L/F), \wedge , \vee , \odot , \mapsto , 0/F, 1/F)$ is also a residuated lattice for any ideal derivation d and d-filter F. Moreover, since F is the d-filter, d(F) is the filter of d(L) and thus the quotient structure d(L)/d(F) forms a residuated lattice. It is natural to ask what the relation between two residuated lattices (d/F)(L/F) and d(L)/d(F) is. In order to answer the question, we need the following result.

Lemma 2

If $F\in {\mathcal {F}}_d(L)$, then we have $F \cap d(L) = d(F)$.

Proof

For $x\in F \cap d(L)$, since $x\in d(L)= \hbox {Fix}_d(L)$, we have $x=dx\in d(F)$ and $F \cap d(L) \subseteq d(F)$.

Conversely, suppose $x\in d(F)$. There exists $y\in F$ such that $x=dy$. Since F is the d-filter, we get $x=dy\in F$. On the other hand, it follows from $dx = d(dy) = dy = x$ that $x\in \hbox {Fix}_d(L)=d(L)$ and $d(F)\subseteq F \cap d(L)$. Therefore, $F \cap d(L) = d(F)$. $\square$

From the above, we have a following result which answers the question.

Theorem 7

Let d be an ideal derivation and F be a d-filter of L. Then, we have $(d/F)(L/F)= \hbox {Fix}_{d/F}(L/F)$ is isomorphic to d(L)/d(F), that is,

$$\begin{aligned} (d/F)(L/F)= \hbox {Fix}_{d/F}(L/F) \cong d(L)/d(F). \end{aligned}$$

Proof

We define a map $\Phi : (d/F)(L/F) \rightarrow d(L)/d(F)$ by $\Phi ((d/F)(x/F))=dx/d(F)$. We show that $\Phi$ is an isomorphism. We only show that the map $\Phi$ is well defined and injective. At first, $\Phi$ is well defined, because we have

$$\begin{aligned} (d/F)(x/F)&=(d/F)(y/F) \\&\Rightarrow dx/F=dy/F \\&\Rightarrow dx\rightarrow dy, dy\rightarrow dx\in F \\&\Rightarrow d(dx\rightarrow dy), d(dy\rightarrow dx) \in d(F) \\&\Rightarrow dx\mapsto dy, dy\mapsto dx\in d(F) \\&\Rightarrow dx/d(F)=dy/d(F). \end{aligned}$$

Hence, $\Phi$ is well defined.

Suppose $\Phi ((d/F)(x/F))=\Phi ((d/F)(y/F))$. Since $dx/d(F)=dy/d(F)$, we have $dx\mapsto dy, dy\mapsto dx\in d(F)$ and thus $d(dx\rightarrow dy), d(dy\rightarrow dx)\in d(F)= F\cap d(L)$. This implies $d(dx\rightarrow dy), d(dy\rightarrow dx)\in F$. Since d is contractive, we also get $d(dx\rightarrow dy)\le dx\rightarrow dy, d(dy\rightarrow dx)\le dy\rightarrow dx$ and $dx\rightarrow dy, dy\rightarrow dx\in F$. Therefore, $dx/F=dy/F$ and $(d/F)(x/F)= dx/F =dy/F = (d/F)(y/F)$, namely $\Phi$ is injective. $\square$