Proof. We only prove the equivalence of (1) and (2). The proof of equivalence for ( $1^{\prime }$ ) and ( $2^{\prime }$ ) is similar.

First, we show that $f$ is order-preserving under any of the above four given equations. Here, we only prove that $f$ is order-preserving when $f$ satisfies the equation: $\mathord {\downarrow } f(\!\cap _{a\in A}\mathord {\downarrow } a)=\cap _{a\in A}\mathord {\downarrow } f(a)$ . Let $x,y\in P$ and $x\leq y$ . Then, $x\in \mathord {\downarrow } x\cap \mathord {\downarrow } y$ . It implies that $f(x)\in \mathord {\downarrow } f(\mathord {\downarrow } x\cap \mathord {\downarrow } y)=\mathord {\downarrow } f(x)\cap \mathord {\downarrow } f(y)$ . So, $f(x)\leq f(y)$ .

$(1)\Rightarrow (2)$ : For any $A\subseteq P$ , we know that $f(\!\cap _{a\in A}\mathord {\downarrow } a)\subseteq \cap _{a\in A}\mathord {\downarrow } f(a)$ . Conversely, let $x\in \cap _{a\in A}\mathord {\downarrow } f(a)$ . Then, $x\in \mathord {\downarrow } f(\!\cap _{a\in A}\mathord {\downarrow } a)$ by (1). Thus, there is an element $p\in \cap _{a\in A}\mathord {\downarrow } a$ such that $x\leq f(p)$ . Since $f$ is surjective, there exists an element $p_{x}\in P$ such that $f(p_{x})=x$ . Note that $x\in \mathord {\downarrow } f(p_{x})\cap f(p)=\mathord {\downarrow } f(\mathord {\downarrow } p_{x}\cap \mathord {\downarrow } p)$ . It follows that there exists an element $y\in \mathord {\downarrow } p_{x}\cap \mathord {\downarrow } p$ such that $x\leq f(y)$ . So we conclude that $y\in \cap _{a\in A}\mathord {\downarrow } a$ and $x=f(y)$ . This implies $x=f(y)\in f(\!\cap _{a\in A}\mathord {\downarrow } a)$ . Thus, $\cap _{a\in A}\mathord {\downarrow } f(a)\subseteq f(\!\cap _{a\in A}\mathord {\downarrow } a)$ . Therefore, $f(\!\cap _{a\in A}\mathord {\downarrow } a)=\cap _{a\in A}\mathord {\downarrow } f(a)$ .

$(2)\Rightarrow (1)$ : For any $A\subseteq P$ , we have $f(\!\cap _{a\in A}\mathord {\downarrow } a)\subseteq \mathord {\downarrow } f(\!\cap _{a\in A}\mathord {\downarrow } a)$ . By (2), $f(\!\cap _{a\in A}\mathord {\downarrow } a)=\cap _{a\in A}\mathord {\downarrow } f(a)\subseteq \mathord {\downarrow } f(\!\cap _{a\in A}\mathord {\downarrow } a)$ . Conversely, since $f$ is order-preserving, $\mathord {\downarrow } f(\!\cap _{a\in A}\mathord {\downarrow } a)\subseteq \cap _{a\in A}\mathord {\downarrow } f(a)$ . So, $\mathord {\downarrow } f(\!\cap _{a\in A}\mathord {\downarrow } a)=\cap _{a\in A}\mathord {\downarrow } f(a)$ .

Proof. Assume first that $(k\times k)^{-1}(\Delta )$ is a subdcpo of $L\times L$ . Let $D$ be a directed subset of $L$ . As $k$ is a kernel operator, it is order-preserving, $\sup _{L}k(D)\leq k(\!\sup _{L}D)$ . Define $\mathcal{D}=\{(d,k(d)\mid d\in D)\}$ . Then $\mathcal{D}$ is directed in $(k\times k)^{-1}(\Delta )$ . Since $(k\times k)^{-1}(\Delta )$ is a subdcpo of $L\times L$ , $\sup \mathcal{D}=(\!\sup _{L} D,\sup _{L} k(D))\in (k\times k)^{-1}(\Delta )$ . It means that $k(\!\sup _{L} D)=k(\!\sup _{L} k(D))$ . Note that $k(\!\sup _{L} k(D))\leq \sup _{L} k(D)$ , which implies that $k(\!\sup _{L} D)\leq \sup _{L} k(D)$ . So, $\sup _{L}k(D)=k(\!\sup _{L}D)$ , that is, $k$ preserves sups of directed sets. Conversely, assume that $k$ preserves sups of directed sets. Let $\mathcal{D}$ be a directed subset in $(k\times k)^{-1}(\Delta )$ . Set

\begin{align*} D_{1}=\{d_{1}\in L\mid \textrm { there exists an element }d_{2}\in L \textrm { such that }(d_{1},d_{2})\in \mathcal{D}\},\\ D_{2}=\{d_{2}\in L\mid \textrm { there exists an element }d_{1}\in L \textrm { such that }(d_{1},d_{2})\in \mathcal{D}\}. \end{align*}

Then, $\sup \mathcal{D}=(\!\sup D_{1},\sup D_{2})$ . We claim that $(\!\sup D_{1},\sup D_{2})\in (k\times k)^{-1}(\Delta )$ . It is easy to see that $k(D_{1})=k(D_{2})$ . Now, as $k$ preserves sups of directed sets,

\begin{align*} k(\!\sup D_{1})=\sup \limits _{d_{1}\in D_{1}}k(d_{1})=\sup \limits _{d_{2}\in D_{2}}k(d_{2})=k(\!\sup D_{2}), \end{align*}

which implies that $(\!\sup D_{1},\sup D_{2})\in (k\times k)^{-1}(\Delta )$ . So, $(k\times k)^{-1}(\Delta )$ is a subdcpo of $L\times L$ .

Proof. We only prove the equivalence of (1) and (2).

From Proposition O-3.10 in Gierz et al. (Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003), $k:P\rightarrow P$ is a kernel operator on $P$ if and only if there is an adjunction $(g,d)$ between $P$ and some $T$ where $k=d\circ g$ . Next, we prove that $R=ker(g)$ . For any $(x,y)\in R$ , we have $k(x)=k(y)$ . That is, $dg(x)=dg(y)$ . Thus, $gdg(x)=gdg(y)$ . By Theorem2.13, $g(x)=g(y)$ . So, $(x,y)\in ker(g)$ . Conversely, if $(x,y)\in ker(g)$ , then $g(x)=g(y)$ ; it implies that $dg(x)=dg(y)$ . Thus, $k(x)=k(y)$ . So, $(x,y)\in R$ . Therefore, $R=ker(g)$ .

Proof. $(1)\Rightarrow (2)$ : If $R=P\times P$ , it naturally holds. Otherwise, for any $p\in P$ , from Condition 2 of Definition 3.2, $\inf [p]_{R}$ exists and $\inf [p]_{R}\in [p]_{R}$ . Define a map $k:P\rightarrow P$ by $k(p)=\inf [p]_{R}$ , where $p\in P$ . Then, $k$ is well defined and $k(p)\leq p$ . In fact, $k(p)$ is the smallest element in $[p]_{R}$ . It implies $k(k(p))=k(p)$ , that is, $k$ is idempotent. Next, for any $x,y\in P$ , suppose that $x\leq y$ . Then, $k(y)\mathrel {R}y$ . By Condition 3 of Definition 3.2, there exists an element $u\in P$ with $u\leq k(y)$ and $u\mathrel {R}x$ . Whence, $k(x)=k(u)\leq u\leq k(y)$ . It shows that $k$ is monotone. Thus, $k$ is a kernel operator.

Indeed, if $(x,y)\in R$ , then $k(x)=k(y)$ , and vice versa; therefore, we see that $(k\times k)^{-1}(\Delta )=R$ . From Condition 1 of Definition 3.2, $(k\times k)^{-1}(\Delta )$ is a subdcpo of $P\times P$ . So, $k$ preserves sups of directed sets by Lemma 2.10.

$(2)\Rightarrow (3)$ : If $R=P\times P$ , then $P/R$ is a singleton set. Otherwise, assume that there is a kernel operator $k$ on $P$ preserving sups of directed sets such that $R=(k\times k)^{-1}(\Delta )$ . Since the image of a kernel operator $k$ on $P$ preserving sups of directed sets is a subdcpo of $P$ (see Gierz et al. (Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003), Proposition O-3.12), $R$ is a dcpo congruence (see Mahmoudi et al. (Reference Mahmoudi, Moghbeli and Pióro2019), Corollary 4.7) and $P/R$ with the relation defined by $[x]_{R}\leq _{P/R}[y]_{R}$ if and only if $k(x)\leq k(y)$ is a dcpo and $q:P\rightarrow P/R$ preserves sups of directed sets (see Mahmoudi et al. (Reference Mahmoudi, Moghbeli and Pióro2019), Theorem 4.1). By the first isomorphism theorem for dcpos (see Mahmoudi et al. (Reference Mahmoudi, Moghbeli and Pióro2019), Corollary5.6), $(P/R,\leq _{P/R})\cong k(P)$ and since $k(P)$ is a domain by Lemma 2.7, we obtain that $P/R$ is a domain as well. Applying Decomposition Theorem for dcpos (see Mahmoudi et al. (Reference Mahmoudi, Moghbeli and Pióro2019), Theorem5.3), there is a Scott-continuous map $f:P/R\rightarrow P$ given by $f([p]_{R})=k(p)$ for any $[p]_{R}\in P/R$ , and $k=f\circ q$ . Based on Proposition O-3.10 in Gierz et al. (Reference Gierz, Hofmann, Keimel, Lawson, Mislove and Scott2003), $(q,f)$ is an adjunction. Therefore, $q$ is an upper adjoint preserving sups of directed sets.

$(3)\Rightarrow (1)$ : We need to check the conditions of I-congruence relation on domains in Definition 3.2. If $P/R$ is a singleton set, then we can directly get that $R=P\times P$ . Otherwise, since the quotient map $q$ preserves sups of directed sets, we have $R$ is a subdcpo of $P\times P$ by Remark 2.11.

Since $q$ is an upper adjoint, there exists an order-preserving map $h:P/R\rightarrow P$ such that $(q,h)$ is an adjunction. Let $p\in P$ . For any element $x\in [p]_{R}$ , we have that $q(x)=q(p)$ , and it follows that $h\circ q(x)=h\circ q(p)$ . Because $(q,h)$ is an adjunction, $h\circ q(p)=h\circ q(x)\leq x$ and $q\circ h\circ q(p)=q(p)$ by Theorem2.13. Thus, $h\circ q(p)$ is the smallest element in $[p]_{R}$ .

Let $v$ , $a$ , $b\in P$ . Assume that $a\leq b$ and $b\mathrel {R}v$ . Then, $q(a)\leq q(b)=q(v)$ . So, $h\circ q(a)\leq h\circ q(v)\leq v$ and $h\circ q(a)\mathrel {R}a$ by Theorem2.13. Thus, $R$ satisfies the dual-link property.

Proof. Obviously, $\leq _{R}$ is reflexive.

Assume that $[x]_{R}\leq _{R} [y]_{R}$ and $[y]_{R}\leq _{R} [x]_{R}$ . Then there exist $x_{1}$ , $x_{2}\in [x]_{R}$ , $y_{1}$ , and $y_{2}\in [y]_{R}$ such that $x_{1}\leq y_{1}$ and $y_{2}\leq x_{2}$ . It means that $x_{1}\leq y_{1}$ and $y_{1}\mathrel {R} y_{2}$ . Since $R$ satisfies the dual-link property, then there exists an element $x_{3}\in [x]_{R}$ such that $x_{3}\leq y_{2}$ . Thus, $x_{3}\leq y_{2}\leq x_{2}$ . Since $R$ is a I-congruence relation on $P$ , there exists an order $\leq _{P/R}$ on $P/R$ such that the quotient map $q:P\rightarrow P/R$ is an upper adjoint preserving sups of directed sets by Theorem3.3. Thus, $[x_{3}]_{R}\leq _{P/R}[y_{2}]_{R}\leq _{P/R}[x_{2}]_{R}$ . Because $[x_{3}]_{R}=[x_{2}]_{R}$ , we have that $[x]_{R}=[x_{3}]_{R}=[y_{2}]_{R}=[y]_{R}$ .

At last, for the transitivity of $\leq _{R}$ , suppose $[x]_{R}\leq _{R} [y]_{R}$ and $[y]_{R}\leq _{R} [z]_{R}$ . Then there exist $x_{1}\in [x]_{R}$ , $y_{1}$ , $y_{2}\in [y]_{R}$ and $z_{2}\in [z]_{R}$ such that $x_{1}\leq y_{1}$ and $y_{2}\leq z_{2}$ . Because $R$ satisfies the dual-link property, there exists $x_{2}\in [x]_{R}$ with $x_{2}\leq y_{2}$ . That is, $x_{2}\leq y_{2}\leq z_{2}$ . Thus, $[x]_{R}\leq _{R} [z]_{R}$ .

Therefore, $\leq _{R}$ is an order on $P/R$ .

Suppose that $\leq _{P/R}$ is an arbitrary order on $P/R$ such that $q$ is an upper adjoint. Then there is an order-preserving map $f:P/R\rightarrow P$ such that $(q,f)$ is an adjunction. For any $x,y\in P$ , if $[x]_{R}\leq _{P/R} [y]_{R}$ , then $f([x]_{R})\leq f([y]_{R})$ , that is, $fq(x)\leq fq(y)$ . Since $qfq(x)=q(x)$ and $qfq(y)=q(y)$ , we have that $[x]_{R}\leq _{R}[y]_{R}$ by the definition of $\leq _{R}$ . It follows that $\leq _{P/R}\subseteq \leq _{R}$ . Conversely, assume that $[x]_{R}\leq _{R} [y]_{R}$ . Then there exist $x_{0}\in [x]_{R}$ and $y_{0}\in [y]_{R}$ such that $x_{0}\leq y_{0}$ . Thus, $q(x)=q(x_{0})\leq _{P/R} q(y_{0})=q(y)$ . It means that $[x]_{R}\leq _{P/R}[y]_{R}$ . Thus, $\leq _{R}\subseteq \leq _{P/R}$ . Therefore, $\leq _{R}=\leq _{P/R}$ .

Proof. It is easy to prove that $(2)\Rightarrow (1)$ and $(3)\Rightarrow (1)$ .

$(1)\Rightarrow (2)$ : Since $[x]_{R}\leq _{R} [y]_{R}$ , there exist $x_{0}\in [x]_{R}$ and $y_{0}\in [y]_{R}$ such that $x_{0}\leq y_{0}$ . Note that $\inf [y]_{R}\mathrel {R} y_{0}$ and $x_{0}\leq y_{0}$ , then there is an element $c\in [x]_{R}$ such that $c\leq \inf [y]_{R}$ , which implies that $\inf [x]_{R}\leq c\leq \inf [y]_{R}$ ; hence, $(2)$ holds.

$(1)\Rightarrow (3)$ : Since $[x]_{R}\leq _{R} [y]_{R}$ , there exist $x_{0}\in [x]_{R}$ and $y_{0}\in [y]_{R}$ such that $x_{0}\leq y_{0}$ . For any element $b\in [y]_{R}$ , because $b\mathrel {R}y_{0}$ and $x_{0}\leq y_{0}$ , there exists an element $a_{b}\in [x]_{R}$ such that $a_{b}\leq b$ . Thus, $(3)$ holds.

Proof. By Theorem3.3, it is straightforward to check that $(2)\Rightarrow (1)$ , $(3)\Rightarrow (1)$ , and $(4)\Rightarrow (1)$ .

If $R$ is a I-congruence relation on $P$ , then $\leq _{R}$ is an order on $P/R$ by Lemma 3.5. Moreover, from Proposition 3.6, $\leq _{R}=\leq _{R}^{1}=\leq _{R}^{2}$ . Hence, we only need to prove $(1)\Rightarrow (3)$ . Again by Theorem3.3, $R$ is a I-congruence relation if and only if $R=P\times P$ or there is a kernel operator $k$ on $P$ preserving sups of directed sets such that $R=(k\times k)^{-1}(\Delta )$ , where $\Delta$ is the diagonal of $P\times P$ . Recall that the proof of $(2)\Rightarrow (3)$ of Theorem3.3, the order on $P/R$ is defined as follows:

\begin{align*} [x]_{R}\leq _{P/R}[y]_{R}\textrm { if and only if }k(x)\leq k(y)\textrm { for all }x,y\in P. \end{align*}

Now it remains to prove that $\inf [p]_{R}=k(p)$ for any element $p\in P$ . As $R=(k\times k)^{-1}(\Delta )$ and $k$ is a kernel operator, we can see that $k(p)\in [p]_{R}$ . Furthermore, let $x\in [p]_{R}$ . It means that $k(p)=k(x)\leq x$ . So, $k(p)$ is the smallest element in $[p]_{R}$ . Whence, $\inf [p]_{R}=k(p)$ .

Proof. If $\mathrm{card}(\,f(P))=1$ , then $M$ is a singleton set. It implies that $M$ is a continuous lattice. Otherwise, $f$ is an upper adjoint. Then there exists an order-preserving map $g:Q\rightarrow P$ such that $(\,f,g)$ is an adjunction between $Q$ and $P$ . Since for any pair of adjunctions $(\,f,g)$ , the image of $f$ is order-isomorphic with the image of $gf$ which is a kernel operator and, of course, a projection map, by Lemma 2.7, this theorem trivially holds.

Proof. If $\mathrm{card}(\,f(P))=1$ , then $ker(\,f)=P\times P$ , and it is a I-congruence relation on $P$ and $f(P)\cong P/ker(\,f)$ . Otherwise, $f$ is an upper adjoint, based on Theorem2.17, the kernel of an upper adjoint is the kernel of a kernel operator and by Theorem3.3, $ker(\,f)$ is a I-congruence relation on $P$ and $f(P)\cong P/ker(\,f)$ , because the kernels of kernel operators are characterized as I-congruence relations on $P$ .

Proof. Let $B$ be a block of $R$ . If $D\subseteq B$ is directed, note that there is an element $p\in P$ such that $B=[p]_{R}$ . Define $\mathcal{D}=\{(d,p)\mid d\in D\}$ . Then $\mathcal{D}$ is directed in $R$ . Since $R$ is a subdcpo of $P\times P$ , $\sup \mathcal{D}=(\!\sup D,p)\in R$ . It implies that $\sup D\in [p]_{R}=B$ . That is, $B$ is a subdcpo of $P$ . Furthermore, from Condition 2 of Definition 5.1, $B$ is directed in $P$ , we deduce that $\sup B$ exists, and hence, $\sup B\in B$ . It shows that $B$ has a top element.

Proof. $(1)\Rightarrow (2)$ : If $R=P\times P$ , it naturally holds. Otherwise, by Theorem 2.12 in M. Mahmoudi, H. Moghbeli & K. Pióro (Reference Mahmoudi, Moghbeli and Pióro2017), Lemma 2.9 and Proposition 5.2, (2) holds.

$(2)\Rightarrow (3)$ : If $R=P\times P$ , then $P/R$ is a singleton set. Otherwise, assume that there is a closure operator $c$ on $P$ preserving sups of directed sets such that $R=(c\times c)^{-1}(\Delta )$ . Since the image of a closure operator $k$ on $P$ preserving sups of directed sets is a subdcpo of $P$ (see Lemma 2.9), $R$ is a dcpo congruence (see Mahmoudi et al. (Reference Mahmoudi, Moghbeli and Pióro2019), Corollary 4.7) and $P/R$ with the partial order $[x]_{R}\leq _{P/R}[y]_{R}$ if and only if $c(x)\leq c(y)$ is a dcpo and $q:P\rightarrow P/R$ preserves sups of directed sets (see Mahmoudi et al. (Reference Mahmoudi, Moghbeli and Pióro2019), Theorem 4.1). By the first isomorphism theorem for dcpos (see Mahmoudi et al. (Reference Mahmoudi, Moghbeli and Pióro2019), Corollary5.6), $(P/R,\leq _{P/R})\cong c(P)$ , we obtain that $P/R$ is a domain as well since $c(P)$ is a domain by Lemma 2.7. Applying Decomposition Theorem for dcpos (see Mahmoudi et al. (Reference Mahmoudi, Moghbeli and Pióro2019), Theorem5.3), there is a Scott-continuous map $f:P/R\rightarrow P$ given by $f([p]_{R})=c(p)$ for any $[p]_{R}\in P/R$ , and $c=f\circ q$ . For $x,y\in P$ , suppose that $x\ll y$ in $P$ . Let $\mathcal{D}\subseteq P/R$ be a directed subset and $[y]_{R}\leq _{P/R}\sup \mathcal{D}$ . Define $D=\{c(d)\mid [d]_{R}\in \mathcal{D}\}$ . Since $f$ is Scott-continuous, we get that $D$ is directed in $c(P)$ and

\begin{align*} f(y)=c(y)\leq f(\!\sup \mathcal{D})=\sup \limits _{[d]_{R}\in \mathcal{D}}f([d]_{R})=\sup \limits _{[d]_{R}\in \mathcal{D}}c(d)=\sup D. \end{align*}

As $x\ll y$ , we have that $c(x)\ll _{c(P)}c(y)\leq \sup D$ by Lemma 2.8. It implies that there exists an element $[d]_{R}\in \mathcal{D}$ such that $c(x)\leq c(d)$ . That is, $[x]_{R}\leq _{P/R}[d]_{R}$ . Thus, $[x]_{R}\ll [y]_{R}$ . This indicates that $q$ preserves the way-below relation $\ll$ .

Finally, we show that for any $x,y\in P$ , $q(\mathord {\uparrow } x\cap \mathord {\uparrow } y))=\mathord {\uparrow } q(x)\cap \mathord {\uparrow } q(y)$ . Since $q$ is Scott-continuous, it is order-preserving, and we get that $q(\mathord {\uparrow } x\cap \mathord {\uparrow } y))\subseteq \mathord {\uparrow } q(x)\cap \mathord {\uparrow } q(y)$ . Conversely, assume that $[z]_{R}\in \mathord {\uparrow } q(x)\cap \mathord {\uparrow } q(y)$ . Then $[x]_{R}\leq _{P/R}[z]_{R}$ and $[y]_{R}\leq _{P/R}[z]_{R}$ . It means that $c(x)\leq c(z)$ and $c(y)\leq c(z)$ . Since $c$ is a closure operator, $x\leq c(x)\leq c(z)$ , $y\leq c(y)\leq c(z)$ , and $c(c(z))=c(z)$ . Therefore, $c(z)\in \mathord {\uparrow } x\cap \mathord {\uparrow } y$ . By (2), $R=(c\times c)^{-1}(\Delta )$ . So, we conclude that $c(z)\in [z]_{R}$ . Hence, $q(c(z))=[z]_{R}$ . It follows that $\mathord {\uparrow } q(x)\cap \mathord {\uparrow } q(y)\subseteq q(\mathord {\uparrow } x\cap \mathord {\uparrow } y))$ . Whence we have that $q(\mathord {\uparrow } x\cap \mathord {\uparrow } y))=\mathord {\uparrow } q(x)\cap \mathord {\uparrow } q(y)$ .

$(3)\Rightarrow (1)$ : We need to check the conditions of Definition 5.1. If $P/R$ is a singleton set, we can directly get that $R=P\times P$ . Otherwise, first, since the quotient map $q$ preserves sups of directed sets, we could show that $R$ is a subdcpo of $P\times P$ by Remark 2.11.

Let $p\in P$ . Then, $p\in [p]_{R}$ . So, $[p]_{R}$ is nonempty. For any elements $x,y\in [p]_{R}$ , we have that $q(x)=q(y)=q(p)$ . It implies that $q(p)\in \mathord {\uparrow } q(x)\cap \mathord {\uparrow } q(y)$ . As $q(\mathord {\uparrow } x\cap \mathord {\uparrow } y)=\mathord {\uparrow } q(x)\cap \mathord {\uparrow } q(y)$ , there exists an element $z\in \mathord {\uparrow } x\cap \mathord {\uparrow } y$ such that $q(z)=q(p)$ . Therefore, $[p]_{R}$ is directed.

Let $u$ , $a$ , $b\in P$ . Assume that $a\leq b$ and $u\mathrel {R}a$ . Then, $q(u)=q(a)\leq q(b)$ . Note that $q(b)\in \mathord {\uparrow } q(u)\cap \mathord {\uparrow } q(b)$ . Again since $q(\mathord {\uparrow } u\cap \mathord {\uparrow } b)=\mathord {\uparrow } q(u)\cap \mathord {\uparrow } q(b)$ , there is an element $v\in \mathord {\uparrow } u\cap \mathord {\uparrow } b$ such that $q(v)=q(b)$ , which means that $u\leq v$ and $b\mathrel {R}v$ .

From the above proof, we can see that for all $p\in P$ , $\sup [p]_{R}$ exists. Finally, we need to prove that $\{\sup [p]_{R}\mid p\in P\}$ is a subdcpo of $P$ . Let $D\subseteq \{\sup [p]_{R}\mid p\in P\}$ be a directed subset. It is also directed in $P$ . So, $\sup D$ exists. We claim that $\sup D=\sup [\!\sup D]_{R}$ . Clearly, $\sup D\in [\!\sup D]_{R}$ . It is enough to show that $x\leq \sup D$ for all $x\in [\!\sup D]_{R}$ . Suppose not, there is an element $x_{0}\in [\!\sup D]_{R}$ such that $x_{0}\not \leq \sup D$ . Since $P$ is a domain, there exists an element $y\ll x_{0}$ but $y\not \leq \sup D$ . By (3), as $q$ preserves the way-below relation $\ll$ , we infer that $q(y)\ll q(x_{0})=q(\!\sup D)$ . Because $q$ is Scott-continuous and $D$ is directed,

\begin{align*} q(y)\ll q(x_{0})=q(\!\sup D)=\sup \limits _{d\in D}q(d). \end{align*}

Thus, there exists $d\in D$ such that $q(y)\leq q(d)$ . Note that $q(d)\in \mathord {\uparrow } q(y)\cap \mathord {\uparrow } q(d))=q(\mathord {\uparrow } y\cap \mathord {\uparrow } d))$ , it implies that there is an element $z\in \mathord {\uparrow } y\cap \mathord {\uparrow } d$ such that $q(z)=q(d)$ . That is, $y\leq z$ , $d\leq z$ , and $d\mathrel {R}z$ . Since $d$ is the largest element in $[d]_{R}$ , $y\leq z=d$ . Hence, $y\leq d\leq \sup D$ , which is a contradiction. Whence $x\leq \sup D$ for all $x\in [\!\sup D]_{R}$ . Therefore, $\sup D=\sup [\!\sup D]_{R}$ . It follows that $\sup D\in \{\sup [p]_{R}\mid p\in P\}$ . So, $\{\sup [p]_{R}\mid p\in P\}$ is a subdcpo of $P$ .

Proof. Take elements $a,b,p\in P$ such that $a\leq p\leq b$ and $a\mathrel {R}b$ . By Theorem5.3, there exists an order $\leq _{P/R}$ on the quotient set $P/R$ such that the quotient map $q:P\rightarrow P/R$ is Scott-continuous, so it is also order-preserving. Then $q(a)=q(b)$ and $q(a)\leq q(p)\leq q(b)$ . It implies that $q(a)=q(p)$ . Thus, $a\mathrel {R}p$ . It also follows from the analogous result for posets (see Blyth and Janowitz (Reference Blyth and Janowitz1972)), and dcpos (see Mahmoudi et al. (Reference Mahmoudi, Moghbeli and Pióro2017)), because $R$ is, particularly, a poset congruence, respectively a dcpo congruence.

Proof. Obviously, $\leq _{R}$ is reflexive.

Assume that $[x]_{R}\leq _{R} [y]_{R}$ and $[y]_{R}\leq _{R} [x]_{R}$ . Then there exist $x_{1}$ , $x_{2}\in [x]_{R}$ , $y_{1}$ , $y_{2}\in [y]_{R}$ such that $x_{1}\leq y_{1}$ and $y_{2}\leq x_{2}$ . It means that $x_{1}\leq y_{1}$ and $x_{1}\mathrel {R} x_{2}$ . Since $R$ satisfies the link property, there exists an element $y_{3}\in [y]_{R}$ such that $x_{2}\leq y_{3}$ . Thus $y_{2}\leq x_{2}\leq y_{3}$ . By Proposition 5.5, $[y]_{R}$ is convex, then we have that $x_{2}\in [y]_{R}$ . As equivalence classes are mutually disjoint, we get that $[x]_{R}=[y]_{R}$ . The antisymmetry of $\leq _{R}$ is proved.

At last, for the transitivity of $\leq _{R}$ , suppose $[x]_{R}\leq _{R} [y]_{R}$ and $[y]_{R}\leq _{R} [z]_{R}$ . Then there exist $x_{1}\in [x]_{R}$ , $y_{1}$ , $y_{2}\in [y]_{R}$ , and $z_{2}\in [z]_{R}$ such that $x_{1}\leq y_{1}$ and $y_{2}\leq z_{2}$ . Because $R$ satisfies the link property, there exists $z_{1}\in [z]_{R}$ with $y_{1}\leq z_{1}$ . That is, $x_{1}\leq y_{1}\leq z_{1}$ . Thus, $[x]_{R}\leq _{R} [z]_{R}$ .

Therefore, $\leq _{R}$ is an order on $P/R$ .

Suppose that $\leq _{P/R}$ is an arbitrary order on $P/R$ such that for any $x,y\in P$ , $q$ satisfies the equation $q(\mathord {\uparrow } x\cap \mathord {\uparrow } y)=\mathord {\uparrow } q(x)\cap \mathord {\uparrow } q(y)$ . For any $x,y\in P$ , if $[x]_{R}\leq _{P/R} [y]_{R}$ , then $[y]_{R}\in \mathord {\uparrow } q(x)\cap \mathord {\uparrow } q(y)=q(\mathord {\uparrow } x\cap \mathord {\uparrow } y)$ . It implies that there exists $z\in \mathord {\uparrow } x\cap \mathord {\uparrow } y$ such that $z\in [y]_{R}$ . Note that $x\leq z$ . So $[x]_{R}\leq _{R}[y]_{R}$ by the definition of $\leq _{R}$ . Thus, $\leq _{P/R}\subseteq \leq _{R}$ . Conversely, assume that $[x]_{R}\leq _{R} [y]_{R}$ . Then there exist $x_{0}\in [x]_{R}$ and $y_{0}\in [y]_{R}$ such that $x_{0}\leq y_{0}$ . Thus, $[y]_{R}\in q(\mathord {\uparrow } x_{0}\cap \mathord {\uparrow } y_{0})=\mathord {\uparrow } q(x_{0})\cap \mathord {\uparrow } q(y_{0})$ . It follows that $[x]_{R}\leq _{P/R}[y]_{R}$ . So $\leq _{R}\subseteq \leq _{P/R}$ . Therefore, $\leq _{R}=\leq _{P/R}$ .

Proof. It is easy to prove that $(2)\Rightarrow (3)$ and $(3)\Rightarrow (1)$ .

$(1)\Rightarrow (2)$ : Since $[x]_{R}\leq _{R} [y]_{R}$ , there exist $x_{0}\in [x]_{R}$ and $y_{0}\in [y]_{R}$ such that $x_{0}\leq y_{0}$ . Note that $\sup [x]_{R}\mathrel {R} x_{0}$ and $x_{0}\leq y_{0}$ , then there is an element $c\in [y]_{R}$ such that $\sup [x]_{R}\leq c$ , which implies that $\sup [x]_{R}\leq c\leq \sup [y]_{R}$ ; hence, $(2)$ holds.

Proof. By Theorem5.3, it is straightforward to check that $(2)\Rightarrow (1)$ , $(3)\Rightarrow (1)$ , and $(4)\Rightarrow (1)$ .

If $R$ is a II-congruence relation on $P$ , then $\leq _{R}$ is an order on $P/R$ by Lemma 5.7. Moreover, from Proposition 5.8, $\leq _{R}=\leq _{R}^{\prime }=\leq _{R}^{\prime \prime }$ . Hence, we only need to prove $(1)\Rightarrow (3)$ . Again by Theorem5.3, $R$ is a II-congruence relation if and only if $R=P\times P$ or there is a closure operator $c$ on $P$ preserving sups of directed sets such that $R=(c\times c)^{-1}(\Delta )$ , where $\Delta$ is the diagonal of $P\times P$ . Recall that the proof of $(2)\Rightarrow (3)$ of Theorem5.3, we define the order on $P/R$ below:

\begin{align*} [x]_{R}\leq _{P/R}[y]_{R}\textrm { if and only if }c(x)\leq c(y)\textrm { for all }x,y\in P. \end{align*}

Now it remains to prove that $\sup [p]_{R}=c(p)$ for any element $p\in P$ . As $R=(c\times c)^{-1}(\Delta )$ and $c$ is a closure operator, we can see that $c(p)\in [p]_{R}$ . Furthermore, let $x\in [p]_{R}$ . It follows that $x\leq c(x)=c(p)$ . So, $c(p)$ is the largest element in $[p]_{R}$ . Whence $\sup [p]_{R}=c(p)$ .

Proof. $(1)\Rightarrow (2)$ : Let $x,y\in P$ and $x\leq y$ . Then, $y\in \mathord {\uparrow } x\cap \mathord {\uparrow } y$ . Whence

\begin{align*} f(y)\in \mathord {\uparrow } f(\mathord {\uparrow } x\cap \mathord {\uparrow } y)=\mathord {\uparrow } f(x)\cap \mathord {\uparrow } f(y). \end{align*}

It implies that $f(x)\leq f(y)$ . Thus, $f$ is order-preserving. The same argument gives that $f(x\vee y)$ is an upper bound of $f(x)$ and $f(y)$ . Let $z$ be another upper bound of $f(x)$ and $f(y)$ . Then, $z\in \mathord {\uparrow } f(x)\cap \mathord {\uparrow } f(y)=\mathord {\uparrow } f(\mathord {\uparrow } x\cap \mathord {\uparrow } y)$ . It means that there is an element $p\in \mathord {\uparrow } x\cap \mathord {\uparrow } y$ such that $f(p)\leq z$ . Note that $x,y\leq p$ . Since $P$ is a sup semilattice, $x\vee y$ exists and $x\vee y\leq p$ . We get that $f(x\vee y)\leq f(p)\leq z$ . Hence, $f(x\vee y)=f(x)\vee f(y)$ . Thus, $f$ preserves sups of finite subsets.

$(2)\Rightarrow (1)$ : Since $f$ preserves sups of finite sets, then $f$ is order-preserving. So, $\mathord {\uparrow } f(\mathord {\uparrow } x\cap \mathord {\uparrow } y)\subseteq \mathord {\uparrow } f(x)\cap \mathord {\uparrow } f(y)$ . Conversely, let $z\in \mathord {\uparrow } f(x)\cap \mathord {\uparrow } f(y)$ . Then, $f(x)\leq z$ and $f(y)\leq z$ . It implies that $f(x\vee y)=f(x)\vee f(y)\leq z$ . Note that $x\vee y\in \mathord {\uparrow } x\cap \mathord {\uparrow } y$ . It follows that $f(x\vee y)\in f(\mathord {\uparrow } x\cap \mathord {\uparrow } y)$ . Thus, $z\in \mathord {\uparrow } f(\mathord {\uparrow } x\cap \mathord {\uparrow } y)$ . Hence, $\mathord {\uparrow } f(x)\cap \mathord {\uparrow } f(y)\subseteq \mathord {\uparrow } f(\mathord {\uparrow } x\cap \mathord {\uparrow } y)$ . Therefore, $\mathord {\uparrow } f(x)\cap \mathord {\uparrow } f(y)=\mathord {\uparrow } f(\mathord {\uparrow } x\cap \mathord {\uparrow } y)$ .

Proof. Since $P$ is a complete lattice, the map $f$ preserving sups of arbitrary subsets implies it is Scott-continuous and preserves sups of finite sets. Thus, $(2)\Rightarrow (1)$ by Proposition 6.2. When $f$ is surjective, the whole proof will be complete if we prove the following.

$(1)\Rightarrow (2)$ : If $\mathrm{card}(\,f(P))=1$ , then it is easy to verify that $f$ preserves sups of arbitrary subsets and the way-below relation $\ll$ by the assumption that $f$ is surjective. Otherwise, since $f$ is Scott-continuous and preserves sups of finite sets, we have that $f$ preserves sups of nonempty subsets. It suffices to prove that $f$ preserves sups of the empty subset. We know that $\sup \emptyset =\bot$ . So, $f(\!\sup \emptyset )=f(\bot )$ . Note that $f(\bot )$ is a lower bound of $f(P)$ and $f(\bot )\in f(P)$ . Therefore, $f(\bot )$ is the smallest element in $f(P)$ . As $f$ is a surjective map, $f(P)=Q$ . This implies that $f(\bot )$ is the smallest element in $Q$ . Thus, $\sup f(\emptyset )=f(\bot )$ . It follows that $f(\!\sup \emptyset )=\sup f(\emptyset )$ . Whence we get that $f$ preserves sups of arbitrary subsets.

Proof. Since $f$ has an upper adjoint, there is an order-preserving map $g:Q\rightarrow P$ such that $(g,f)$ is an adjunction. By Lemma 2.15, $f$ preserves existing sups. So it is Scott-continuous. Thus, for all $x,y\in P$ , $\mathord {\uparrow } f(\mathord {\uparrow } x\cap \mathord {\uparrow } y)\subseteq \mathord {\uparrow } f(x)\cap \mathord {\uparrow } f(y)$ . Conversely , let $z\in \mathord {\uparrow } f(x)\cap \mathord {\uparrow } f(y)$ . Then $f(x)\leq z$ and $f(y)\leq z$ . So, $x\leq g(z)$ and $y\leq g(z)$ . It follows that $g(z)\in \mathord {\uparrow } x\cap \mathord {\uparrow } y$ . Hence, $fg(z)\in f(\mathord {\uparrow } x\cap \mathord {\uparrow } y)$ . Note that $fg(z)\leq z$ by Theorem2.13. Therefore, $z\in \mathord {\uparrow } f(\mathord {\uparrow } x\cap \mathord {\uparrow } y)$ . This implies that $\mathord {\uparrow } f(x)\cap \mathord {\uparrow } f(y)\subseteq \mathord {\uparrow } f(\mathord {\uparrow } x\cap \mathord {\uparrow } y)$ . Whence we get that for all $x,y\in P$ , $f$ satisfies the equation: $\mathord {\uparrow } f(\mathord {\uparrow } x\cap \mathord {\uparrow } y)=\mathord {\uparrow } f(x)\cap \mathord {\uparrow } f(y)$ .