The final part is devoted to observation on the adjoint situation between Pos and Quant_{S} . By a *free S -quantale on a poset P* we mean an *S* -quantale *Q *_{S} together with a monotone mapping *ψ* : *P*→*Q *_{S} with the universal property that given any *S* -quantale *A*_{S} and a monotone mapping *f* : *P*→ *A*_{S} , there exists a unique *S* -quantale morphism $\overline{f}:{Q}_{S}\to {A}_{S}$ such that *f* can be factored through.

#### Lemma 5.1

(Th.10) *For a given poset P and a pomonoid S , the free S -poset on P is given by P* × *S , with componentwise order and the action* (*x*,*s*)*t* = (*x*, *st*), *for every x*∈*P*, *s*,*t*∈*S*.

Let (*P*×*S*)_{S} be the free *S* -poset presented in Lemma 5.1. Write

$$Q(P\times S)=\{D\subseteq P\times S\mid D=D\downarrow \},$$

where *D* ↓ is the down-set of *D* for *D* ⊆ *P*×*S* , more precisely,

$$D\downarrow =\{(p,s)\in P\times S\mid (p,s)\le ({p}_{1},{s}_{1})\text{\hspace{0.17em}for some\hspace{0.17em}}({p}_{1},{s}_{1})\in D\}.$$

Note that ( *p* ↓ ×*s* ↓) ↓= *p* ↓ ×*s* ↓ provides that $p\downarrow \times s\downarrow \in Q(P\times S)$ for every element *p*∈*P*, *s*∈*S*. Define an action *∗* on $\text{\hspace{0.17em}}Q(P\times S)$ by

$$D\ast t:=\{(p,s)\in P\times S\mid (p,s)\le ({p}_{1},{s}_{1}t)\text{\hspace{0.17em}for some\hspace{0.17em}}({p}_{1},{s}_{1})\in D\},$$

for *t*∈*S*. Then it is clear that *D*∗*t* = (*Dt*) ↓ . We claim that $(Q(P\times S{)}_{S},\ast ,\subseteq )$ is the free object in Quant_{S}.

#### Proposition 5.2

*Let S be a pomonoid, P be a poset. Then* $(Q(P\times S{)}_{S},\ast ,\subseteq )$ *is an S -quantale*.

*Proof*. Observe first that

$$\begin{array}{rl}(D\ast {t}_{1})\ast {t}_{2}& =\{(p,s)\in P\times S\mid (p,s)\le ({p}_{1},{s}_{1}{t}_{2})\text{\hspace{0.17em}for some\hspace{0.17em}}({p}_{1},{s}_{1})\in D\ast {t}_{1}\}\\ & =\{(p,s)\in P\times S\mid (p,s)\le ({p}_{1},{s}_{1}{t}_{2}),({p}_{1},{s}_{1})\le ({p}_{2},{s}_{2}{t}_{1})\text{\hspace{0.17em}for some\hspace{0.17em}}({p}_{2},{s}_{2})\in D\}\\ & =\{(p,s)\in P\times S\mid (p,s)\le ({p}_{2},{s}_{2}{t}_{1}{t}_{2}),\text{\hspace{0.17em}for some\hspace{0.17em}}({p}_{2},{s}_{2})\in D\}\\ & =D\ast ({t}_{1}{t}_{2})\end{array}$$

for any *t*_{1}, *t*_{2} ∈ *S*, $D\in Q(P\times S),$ and *D*∗1 = (*D*1) ↓= *D*. This shows that $(Q(P\times S),\ast )$ is an *S* -act. Clearly, *D*_{1} ∗ *s* ⊆ *D*_{2} ∗*t*, whenever *D*_{1} ⊆ *D*_{2} in $\text{\hspace{0.17em}}Q(P\times S),$ and *s*⩽*t* in *S* . So $\text{\hspace{0.17em}}Q(P\times S)$ is an *S* -poset. It is straightforward to check that $(\bigcup _{i\in I}{D}_{i})\ast t=\bigcup _{i\in I}({D}_{i}\ast t)$ for every ${D}_{i}\in Q(P\times S),i\in I,t\in S.$

Lemma 5.3 comes true directly by the definition of $\text{\hspace{0.17em}}Q(P\times S)$.

#### Lemma 5.3

*Let S be a pomonoid, P be a poset. Then* $D=\bigcup _{(p,s)\in D}(p\downarrow \times s\downarrow )$ *for every* $D\in Q(P\times S)$.

#### Lemma 5.4

*Let S be a pomonoid, P be a poset. Then p* ↓ ×*t* ↓= (*p* ↓ ×1↓)∗*t holds in* $\text{\hspace{0.17em}}Q(P\times S{)}_{S}$ *for every p* ∈ *P*, *t* ∈ *S*.

*Proof*. It is clear that (*q*, *s*) ∈ (*p* ↓ ×1↓)∗*t* for every (*q*, *s*) ∈ *p* ↓ ×*t* ↓ , since (*q*, *s*)⩽(*p*,*t*) . On the other hand, for any $(q,s)\in (p\downarrow \times 1\downarrow )\ast t,(q,s)\le ({p}_{1},{s}_{1}t)=({p}_{1},{s}_{1})t$ for some (*p*_{1}, *s*_{1}) ∈ *p* ↓ ×1↓ , it follows that (*q*, *s*)⩽( *p*,1)*t* = (*p*,*t*). Hence (*q*, *s*) ∈ *p* ↓ ×*t* ↓.

#### Theorem 5.5

*Let S be a pomonoid, P be a poset. Then the free S -quantale on P is given by the S -quantale* $\text{\hspace{0.17em}}Q(P\times S{)}_{S}$.

*Proof*. Define a mapping $\tau :P\to Q(P\times S{)}_{S}$ by *τ*(*p*) = *p* ↓ ×1↓ for every *p*∈*P*. Obviously, *τ* is order-preserving. Let *Q*_{S} be an *S* -quantale, *f* : *P*→*Q*_{S} be any monotone mapping. Define a mapping $\overline{f}:Q(P\times S{)}_{S}\to {Q}_{S}$ by

$$\overline{f}(D)=\bigvee \{f(p)s\mid (p,s)\in D\},$$

for every $D\in Q(P\times S{)}_{S}.$ We claim that *f¯* is the unique *S* -quantale morphism with the property that $\overline{f}\tau =f.$

It is clear that *f¯* preserves *S* -actions. Take ${D}_{i}\in Q(P\times S{)}_{S},i\in I,$ then equalities

$$\overline{f}\left(\bigcup _{i\in I}{D}_{i}\right)=\bigvee \{f(p)s|(p,s)\in \bigcup _{i\in I}{D}_{i}\}=\underset{i\in I}{\bigvee}\{\bigvee \{f(p)s|(p,s)\in {D}_{i}\}\}=\underset{i\in I}{\bigvee}\overline{f}({D}_{i})$$

indicate that$\overline{f}$ preserves arbitrary joins. Evidently, for any *p* ∈ *P*,

$$\overline{f}\tau (p)=\overline{f}(p\downarrow \times 1\downarrow )=\bigvee \{f(q)s\mid (q,s)\in p\downarrow \times 1\downarrow \}\le f(p),$$

while the fact that *f*(*p*) being one of the terms in the sup that defines *fτ*(*p*) guarantees the opposite implication. Suppose that ${f}^{{}^{\prime}}:Q(P\times S{)}_{S}\to {Q}_{S}$ is an *S* -quantale morphism such that ${f}^{{}^{\prime}}\tau =f.$ Then by Lemma 5.3 and Lemma 5.4, we achieve that

$$\begin{array}{rl}{f}^{{}^{\prime}}(D)& ={f}^{{}^{\prime}}\left(\bigcup _{(p,s)\in D}(p\downarrow \times s\downarrow )\right)=\underset{(p,s)\in D}{\bigvee}{f}^{{}^{\prime}}(p\downarrow \times s\downarrow )=\underset{(p,s)\in D}{\bigvee}{f}^{{}^{\prime}}((p\downarrow \times 1\downarrow )\ast s)\\ & =\underset{(p,s)\in D}{\bigvee}{f}^{{}^{\prime}}(p\downarrow \times 1\downarrow )s=\underset{(p,s)\in D}{\bigvee}{f}^{{}^{\prime}}\tau (p)s=\underset{(p,s)\in D}{\bigvee}f(p)s=\overline{f}(D),\end{array}$$

for every $D\in Q(P\times S{)}_{S},$ which finishes our proof.

#### Corollary 5.6

*The category* Quant_{S} has a separator.

*Proof*. Let *f* , *g* : *A*_{S} → *B*_{S} be a pair of morphisms in Quant_{S} with *f* ≠ *g* . Then there exists *a*∈ *A*_{S} such that *f*(*a*) ≠ *g*(*a*). Let *P* be a poset. Define a mapping *k* : *P* → *A*_{S} by *k*(*p*) = *a*,∀*p*∈*P* . We are aware that *k* is a morphism in Pos . Hence there is a unique *S* -quantale morphism $\overline{k}:Q(P\times S{)}_{S}\to {A}_{S}$ with $\overline{k}\tau =k$, where $\tau :P\to Q(P\times S{)}_{S}$ is defined as in Theorem 5.5. This yields that $f\overline{k}\ne g\overline{k}$, and consequently gives that $Q(P\times S{)}_{S}$ is a separator.

We thereby obtain a free functor from the category of posets into the category of *S* -quantales, which is shown to be left adjoint to the forgetful functor.

#### Proposition 5.7

*There is a free functor F* : Pos → Quant_{S} given by

where $FP=Q(P\times S{)}_{S}$, and

$$Ff(D)=\{(x,y)\in Q\times S\mid (x,y)\le (f(p),s)\mathrm{f}\mathrm{o}\mathrm{r}\phantom{\rule{thinmathspace}{0ex}}\mathrm{s}\mathrm{o}\mathrm{m}\mathrm{e}(p,s)\in D\},$$

for any monotone mapping *f* : *P*→*Q* and *D* ∈ *FP*.

#### Theorem 5.8

*The free functor F* : Pos → Quant_{S} is left adjoint to the forgetful functor $\u230a\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\u230b:\text{Quan}{\text{t}}_{S}\to \text{Pos}.$

*Proof*. Let us prove that $\eta :\text{i}{\text{d}}_{\text{pos}}\to \u230a\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\u230bF$ with ${\eta}_{P}:P\to \lfloor Q(P\times S{)}_{S}\rfloor $, where *P* is a Pos -object, *η*_{P}(*p*) = *p* ↓ × 1 ↓ , ∀*p*∈*P* , is a natural transformation. Suppose that *f* : *P*→ *P*^{'} is a morphism in Pos . Then

$$\begin{array}{rl}Ff\circ {\eta}_{P}(p)& =Ff(p\downarrow \times 1\downarrow )=\{(x,y)\in {P}^{{}^{\prime}}\times S\mid (x,y)\le (f(\stackrel{~}{p}),s)\phantom{\rule{thinmathspace}{0ex}}\mathrm{f}\mathrm{o}\mathrm{r}\phantom{\rule{thinmathspace}{0ex}}\mathrm{s}\mathrm{o}\mathrm{m}\mathrm{e}\phantom{\rule{thinmathspace}{0ex}}(\stackrel{~}{p},s)\in p\downarrow \times 1\downarrow \}\\ & =\{(x,y)\in {P}^{{}^{\prime}}\times S\mid (x,y)\le (f(\stackrel{~}{p}),s)\le (f(p),1),(\stackrel{~}{p},s)\in p\downarrow \times 1\downarrow \},\end{array}$$

for *p* ∈ *P* , and

$$({\eta}_{{P}^{{}^{\prime}}}\circ f)(p)={\eta}_{{P}^{{}^{\prime}}}(f(p))=f(p)\downarrow \times 1\downarrow .$$

It results in $Ff\circ {\eta}_{P}={\eta}_{{P}^{{}^{\prime}}}\circ f$ as needed. Now, by Theorem 5.5 and [12] 19.4(2), we obtain that *F* is left adjoint to$\lfloor \rfloor $.

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