Proof. First we prove that $h\left(\nu ,\theta \right)$ is a mapping. Suppose that $\alpha ,{\alpha }^{\prime }\in \mathcal{T}$ and $\alpha \sigma {\alpha }^{\prime }$. Then from (5.5) and the fact that $\theta$ is a bimorphism we obtain

$$e_{\alpha }\theta \stackrel{r}{\approx }{e}_{\alpha \prime }\theta ,f_{\alpha }\theta \stackrel{l}{\approx }{f}_{\alpha \prime }\theta .$$

Further if $h{\omega }^r{e}_{\alpha }\left(h{\omega }^l{f}_{\alpha }\right)$

$$\begin{array}{rcll}{\left(h\wedge \alpha \right)}^{\nu }& =& {\left({\tau }^r\left(h,h{\tau }^r\left(e_{\alpha }\right)\right)\circ \alpha \right)}^{\nu }& \\ & =& {\tau }^r{\left(h,h{\tau }^r\left(e_{\alpha }\right)\right)}^{\nu }\circ {\alpha }^{\nu }& \\ & =& {\tau }^r\left(h\theta ,h\theta {\tau }^r\left(e_{\alpha }\theta \right)\right)\circ {\alpha }^{\nu }& \end{array}$$

by (c2) and (c3). Therefore

$${\left(h\wedge \alpha \right)}^{\nu }=h\theta \wedge {\alpha }^{\nu }\left({\left(\alpha \wedge h\right)}^{\nu }={\alpha }^{\nu \wedge }h\theta \right)$$

(6.4)

Hence by (5.5)

$$\begin{array}{rcll}{e}_{{\alpha }^{\prime }}\theta \wedge {\alpha }^{\nu }& =& {\left(e_{{\alpha }^{\prime }}\wedge \alpha \right)}^{\nu }& \\ & =& {\left({\alpha }^{\prime }\wedge f_{\alpha }\right)}^{\nu }& \\ & =& \left({\alpha }^{\prime \nu }\wedge f_{\alpha }\theta \right),& \end{array}$$

which implies that ${\alpha }^{\nu }\sigma {\alpha }^{\prime v}$. Thus $h\left(\nu ,\theta \right)$ is a mapping. Moreover by (5.8), and (6.2)

$$\begin{array}{rcll}\left(\overline{\alpha }\cdot \overline{\beta }\right)h\left(\nu ,\theta \right)& =& \left(\overline{\alpha \wedge h\circ h\wedge \beta }\right)h\left(\nu ,\theta \right)& \\ & =& {\left(\overline{\alpha \wedge h\circ h\wedge \beta }\right)}^{\nu }& \end{array}$$

where $\alpha ,\beta \in \mathcal{T}$, $h\in s\left(f_{\alpha },e_{\beta }\right)$. Since $h\theta \in s\left(f_{\alpha }\theta ,e_{\beta }\theta \right)$ we have by (c3) and (6.4),

$$\begin{array}{rcll}\left(\overline{\alpha }\cdot \overline{\beta }\right)h\left(\nu ,\theta \right)& =& \overline{{\alpha }^{\nu }\wedge h\theta \circ h\theta \wedge {\beta }^{\nu }}& \\ & =& \overline{{\alpha }^{\nu }}\circ \overline{{\beta }^{\nu }}& \\ & =& \overline{\alpha }h\left(\nu ,\theta \right)\cdot \overline{\beta }h\left(\nu ,\theta \right).& \end{array}$$

Thus $h\left(\nu ,\theta \right)$ is a homomorphism of $\overline{\mathcal{T}}$ into ${\overline{\mathcal{T}}}^{\prime }$. Also for $e\in E$,

$$\begin{array}{rcll}\left(e{\chi }_E\right)h\left(\nu ,\theta \right)& =& \overline{ϵ{\left(e,e\right)}^{\nu }}& \\ & =& \overline{ϵ\left(e\theta ,e\theta \right)}& \\ & =& \left(e\theta \right){\chi }_{E^{\prime }}& \end{array}$$

and therefore $h\left(\nu ,\theta \right)$ extends $\theta$.

If $h:\overline{\mathcal{T}}\to {\overline{\mathcal{T}}}^{\prime }$ is a homomorphism, Theorem 4.10 and Lemma 5.9 show that ${\theta }_h$ is a bimorphism. Evidently ${\theta }_h$ is a restriction of $h$ to $E$. Now define $v$ by

$${\alpha }^{\nu }={\chi }_{E\prime }\circ \mathfrak{a}\left(\overline{\alpha }h,\overline{{\alpha }^{-1}}h\right)\circ {\chi }_{E^{\prime }}^{-1},$$

(6.5)

for all $\alpha \in \mathcal{T}$. Since $\overline{\alpha }h\in {\overline{\mathcal{T}}}^{\prime }$ and $\overline{{\alpha }^{-1}}h\in i\left(\overline{\alpha }h\right)$ by (5.14), ${\alpha }^{\nu }\in {\mathcal{T}}^{\prime }$. By definition of ${\theta }_h$

$$\begin{array}{rcll}{e}_{{\alpha }^{\nu }}{\chi }_{E^{\prime }}& =& \overline{\alpha }h\cdot \overline{{\alpha }^{-1}}h& \\ & =& \left(e_{\alpha }{\chi }_E\right)h& \\ & =& \left(e_{\alpha }{\theta }_{\alpha }\right){\chi }_{E^{\prime }}& \end{array}$$

and hence $e_{{\alpha }^{\nu }}=e_{\alpha }{\theta }_h$. Similarly $f_{{\alpha }^{\nu }}=f_{\alpha }{\theta }_h$. Hence $v$ satisfies axiom (c1).

Let $\alpha ,\beta \in \mathcal{T}$ and $f_{\alpha }\omega e_{\beta }$. Since $f_{\alpha }\in \mathcal{S}\left(f_{\alpha },e_{\beta }\right)$ we have

$$\overline{\alpha }\cdot \overline{\beta }=\overline{\alpha \circ \beta }$$

and

$$\overline{{\beta }^{-1}}\cdot \overline{{\alpha }^{-1}}=\overline{{\beta }^{-1}\circ {\alpha }^{-1}}.$$

Substituting these in (6.5), we obtain

$${\left(\alpha \circ \beta \right)}^{\nu }={\chi }_{E^{\prime }}\circ \mathfrak{a}\left(\left(\overline{\alpha }\cdot \overline{\beta }\right)h,\left(\overline{{\beta }^{-1}},\overline{{\alpha }^{-1}}\right)h\right)\circ {\chi }_{E^{\prime }}^{-1}.$$

Since $\overline{{\alpha }^{-1}}\cdot \overline{\alpha }=f_{\alpha }{\chi }_E\omega e_{\beta }{\chi }_E=\overline{\beta }\cdot \overline{{\beta }^{-1}}$ we have by (4.16) that

$$\mathfrak{a}\left(\left(\overline{\alpha }\cdot \overline{\beta }\right)h,\left(\overline{{\beta }^{-1}}\overline{{\alpha }^{-1}}\right)h\right)=\mathfrak{a}\left(\overline{\alpha }h,\overline{{\alpha }^{-1}}h\right)\circ \mathfrak{a}\left(\overline{\beta }h,\overline{{\beta }^{-1}}h\right).$$

Therefore

$${\left(\alpha \circ \beta \right)}^{\nu }=\left({\chi }_{E^{\prime }}\circ \mathfrak{a}\left(\overline{\alpha }h,\overline{{\alpha }^{-1}}h\right)\circ {\chi }_{E^{\prime }}^{-1}\right)\circ \left({\chi }_{E^{\prime }}\circ \mathfrak{a}\left(\overline{\beta }h,\overline{{\beta }^{-1}}h\right),\circ {\chi }_{E^{\prime }}^{-1}\right)={\alpha }^{\nu }\circ {\beta }^{\nu }.$$

Similarly it can be seen that the same equality holds when $e_{\beta }\omega f_{\alpha }$. Thus $v$ satisfies (c3) also.

To verify (c2), let $e\stackrel{r}{\approx }f$. Then

$$\begin{array}{rcll}{\tau }^r\left(e,f\right)\circ {\chi }_E& =& {\chi }_E\circ {\tau }^r\left(e\chi ,f\chi \right)& \\ & =& {\chi }_E\circ \mathfrak{a}\left(f{\chi }_E,e{\chi }_E\right).& \end{array}$$

Hence

$$\begin{array}{rcll}{\tau }^r{\left(e,f\right)}^{\nu }& =& {\chi }_{E\prime }\circ \mathfrak{a}\left(\left(f{\chi }_E\right)h,\left(e{\chi }_E\right)h\right)\circ {\chi }_{E^{\prime }}^{-1}& \\ & =& {\chi }_{E^{\prime }}\circ \mathfrak{a}\left(\left(f{\theta }_h\right){\chi }_{E^{\prime }}\right)\left(e{\theta }_h,{\chi }_{E^{\prime }}\right)\circ {\chi }_{E^{\prime }}^{-1}.& \end{array}$$

Since

$${\tau }^r\left(e{\theta }_h,f{\theta }_h\right)\circ {\chi }_{E^{\prime }}={\chi }_{E^{\prime }}\circ \mathfrak{a}\left(\left(f{\theta }_h\right){\chi }_{E^{\prime }},\left(e{\theta }_h,{\chi }_{E^{\prime }}\right)\right),$$

we have

$$\begin{array}{rcll}{\tau }^r{\left(e,f\right)}^{\nu }& =& {\tau }^r\left(e{\theta }_h,f{\theta }_h\right)\circ {\chi }_{E^{\prime }}\circ {\chi }_{E^{\prime }}^{-1}& \\ & =& {\tau }^r\left(e{\theta }_h,f{\theta }_h\right).& \end{array}$$

Similarly if $e\stackrel{l}{\approx }f$ we obtain

$${\tau }^l{\left(e,f\right)}^{\nu }={\tau }^l\left(e{\theta }_h,f{\theta }_h\right).$$

This together with axiom (c3) implies axiom (c2). Hence the theorem. □

Proof. If $h\left(\nu ,\theta \right)$ is injective, since

$$\theta ={\chi }_E\circ h\left(\nu ,\theta \right)\circ {\chi }_{E^{\prime }}^{-1}$$

$\theta$ is also injective. On the other hand, if $\theta$ is injective

$$h\left(\nu ,\theta \right)|E\left(\overline{\mathcal{T}}\right)=I_{E\left(\overline{\mathcal{T}}\right)}\circ h\left(\nu ,\theta \right)={\chi }_E^{-1}\circ \theta \circ {\chi }_{E^{\prime }},$$

is also injective and so $h\left(\nu ,\theta \right)$ is an idempotent separating homomorphism. Since $\overline{\mathcal{T}}$ is fundamental, this implies that $h\left(\nu ,\theta \right)$ is injective. Thus (a) $\iff$ (c).

Let $\alpha ,\beta \in \mathcal{T}$ and ${\alpha }^{\nu }={\beta }^{\nu }$. Then by Definition 6.1

$$\begin{array}{rcll}{e}_{\alpha }\theta & =& e_{\beta }\theta & \\ f_{\alpha }\theta & =& f_{\beta }\theta & \end{array}$$

and since $\theta$ is injective we have $e_{\alpha }=e_{\beta },f_{\alpha }=f_{\beta }$. Further, since ${\alpha }^{\nu }={\beta }^{\nu }$,

$$\overline{\alpha }h\left(\nu ,\theta \right)=\overline{\beta }h\left(\nu ,\theta \right)$$

which implies that $\alpha \sigma \beta$. Hence $\alpha =\beta$ i.e., $\nu$ is injective and thus (a) $\Rightarrow$ (b).

Let $e\theta =f\theta$. Then

$$\begin{array}{rcll}ϵ{\left(e,e\right)}^{\nu }& =& ϵ\left(e\theta ,e\theta \right)& \\ & =& ϵ\left(f\theta ,f\theta \right)& \\ & =& ϵ{\left(f,f\right)}^{\nu }.& \end{array}$$

Since $v$ is injective we have $ϵ\left(e,e\right)=ϵ\left(f,f\right)$ and hence $e=f$. Thus (b) $\Rightarrow$ (c). □

Proof. If $\alpha \in \mathcal{T}$ then ${\alpha }^{\nu }\in \mathcal{T}\prime \left(e_{\alpha }\theta ,f_{\alpha }\theta \right)$ and hence ${\left({\alpha }^{\nu }\right)}^{-1}\in \mathcal{T}\prime \left(f_{\alpha }\theta ,e_{\alpha }\theta \right)$. But since ${\alpha }^{-1}\in \mathcal{T}\left(f,e\right),{\left({\alpha }^{-1}\right)}^{\nu }$ also belongs to $\mathcal{T}\prime \left(f_{\alpha }\theta ,e_{\alpha }\theta \right)$. Further $\overline{{\alpha }^{\nu -1}}$ and $\overline{{\left({\alpha }^{-1}\right)}^{\nu }}$ are inverse of ${\alpha }^{\nu }$ in the $\mathcal{\mathscr{H}}$-class $H_{\left(f\alpha \theta \right){\chi }_{E\prime },\left(e_{\alpha }\theta \right){\chi }_{E\prime }}$ of $\overline{\mathcal{T}}\prime$ and therefore ${\overline{\left({\alpha }^{\nu }\right)}}^{-1}={\overline{\left({\alpha }^{-1}\right)}}^{\nu }$, which implies that ${\left({\alpha }^{\nu }\right)}^{-1}\sigma {\left({\alpha }^{-1}\right)}^{\nu }$. Thus ${\left({\alpha }^{\nu }\right)}^{-1}={\left({\alpha }^{-1}\right)}^{\nu }$.

To prove (b), let $x=\bar{\alpha }$ and $x\prime =\overline{{\alpha }^{-1}}$. Then for $g\in \omega \left(e_{\alpha }\right)$

$$\begin{array}{rcll}& & g\left({\chi }_E\right)\left(\mathfrak{a}\left(x,x\prime \right)\circ h\left(\nu ,\theta \right)\right)& \\ & & =\left(x\prime h\left(\nu ,\theta \right)\right)\left(g{\chi }_{E}h\left(\nu ,\theta \right)\right)\left(xh\left(\nu ,\theta \right)\right)& \\ & & =\left(\left(g{\chi }_E\right)h\left(\nu ,\theta \right)\right)\mathfrak{a}\left(xh\left(\nu ,\theta \right),x\prime h\left(\nu ,\theta \right)\right).& \end{array}$$

Thus

$$\mathfrak{a}\left(x,x\prime \right)\circ h\left(\nu ,\theta \right)=h\left(\nu ,\theta \right)\circ \mathfrak{a}\left(xh\left(\nu ,\theta \right),x\prime h\left(\nu ,\theta \right)\right).$$

(6.6)

For $g\in \omega \left(e_{\alpha }\right)$ we have by (5.14), (6.1), (6.2) and (6.6)

$$\begin{array}{rcll}\left(g\right)\left(\alpha \circ \theta \circ {\chi }_{E\prime }\right)& =& \left(g_{\alpha }\right)\left({\chi }_E\circ h\left(\nu ,\theta \right)\right)& \\ & =& \left(g{\chi }_E\right)\left(\mathfrak{a}\left(x,x\prime \right)\circ h\left(\nu ,\theta \right)\right)& \\ & =& \left(g{\chi }_E\right)\left(h\left(\nu ,\theta \right)\circ \mathfrak{a}\left(xh\left(\nu ,\theta \right),x\prime h\left(\nu ,\theta \right)\right)\right)& \\ & =& \left(g\theta \right)\left({\chi }_{E\prime }\circ \mathfrak{a}\left(xh\left(\nu ,\theta \right),x\prime h\left(\nu ,\theta \right)\right)\right)& \\ & =& \left(\left(g\theta \right){\chi }_{E\prime }\right)\mathfrak{a}\left({\overline{\alpha }}^{\nu },\overline{{\alpha }^{\nu -1}}\right)& \\ & =& \left(\left(g\theta \right){\alpha }^{\nu }\right)x_{E\prime }& \\ & =& \left(g\right)\left(\theta \circ {\alpha }^{\nu }\circ {\chi }_{E\prime }\right).& \end{array}$$

Since ${\chi }_{E\prime }$ is an isomorphism, we have for all $g\in \omega \left(e_{\alpha }\right)$

□

Proof. Since $\theta$ is surjective, there exist $e,f\in E$ such that

Let $e_2\in \mathcal{S}\left(e,e\alpha \right)$ and $e_1=e_2{\tau }^r\left(e\alpha \right)$. Then $e_2\theta \in \mathcal{S}\left(e\theta ,e_2\theta \right)=\mathcal{S}\left(e_{\alpha \prime },e\alpha \theta \right)=\left\{e_{\alpha \prime }\right\}$ since $e_{\alpha \prime }\stackrel{r}{\approx }{e}_{\alpha }\theta$; and

Thus

and $e_2\stackrel{r}{\approx }{e}_1\omega e_{\alpha }$.

Dually we can prove that there exist $f_1$ and $f_2$ such that

and

$$f_2\stackrel{l}{\approx }{f}_1\omega f_2.$$

Now let $h_1\in \left(f_1{\alpha }^{-1},e_1\right)$. Then $h_1\in \omega \left(e_{\alpha }\right)$ and by Lemma 6.3.

Thus

and if $h_1^{\prime }=h_1^{\alpha }$

Since $h_1\in {\omega }^r\left(e_2\right)$ and $h_1^{\prime }\in {\omega }^l\left(f_2\right)$, for

we have

and

By definitions of $h_2$ and $h_2^{\prime }$

Therefore by (6.4)

Now if

then ${\alpha }_1\sigma h_1\wedge \alpha$ and ${\alpha }_1\in \mathcal{T}\left(h_2,h\prime \right)$. Thus

and

Consequently ${\alpha }_1^{\nu },{\alpha }^{\prime }\in \mathcal{T}\prime \left(e_{\alpha \prime },f_{\alpha \prime }\right)$ and ${\alpha }_1^{\nu },\sigma \alpha \prime$, which imply that ${\alpha }_1^{\nu }=\alpha \prime$. □

Proof. (a) If $h\left(\nu ,\theta \right)$ is surjective, by (4.6) and (6.1), $\theta$ is also surjective. For ${\alpha }^{\prime }\in {\mathcal{T}}^{\prime }$, there exists $\alpha \in \mathcal{T}$ such that ${\alpha }^{\nu }\sigma {\alpha }^{\prime }$. Also by Lemma 6.4 there exists ${\alpha }_1\in \mathcal{T}$ such that ${\alpha }_1^{\nu }={\alpha }^{\prime }$ and thus $\nu$ is surjective.

If $\nu$ is surjective, by definition, $h\left(\nu ,\theta \right)$ is also surjective.

(b) If $\theta$ is surjective, we have by (4.6), and (6.1)

$$\left(E\left(\bar{\mathcal{T}}\right)\right)h\left(\nu ,\theta \right)=E\left({\bar{\mathcal{T}}}^{\prime }\right).$$

Denoting $\left(\bar{\mathcal{T}}\right)h\left(\nu ,\theta \right)$ by ${\bar{\mathcal{T}}}^{\prime \prime }$ we see that ${\bar{\mathcal{T}}}^{\prime \prime }$ is a regular subsemigroup of ${\bar{\mathcal{T}}}^{\prime }$ such that $E\left({\bar{\mathcal{T}}}^{\prime \prime }\right)=E\left({\bar{\mathcal{T}}}^{\prime }\right)$ since $h\left(\nu ,\theta \right)$ is a homomorphism of $\bar{\mathcal{T}}$ onto ${\bar{\mathcal{T}}}^{\prime \prime }$ and ${\bar{\mathcal{T}}}^{\prime \prime }$ is fundamental, we have

$${\mathcal{T}}^{\prime \prime }=\left\{\alpha |\bar{\alpha }\in {\bar{\mathcal{T}}}^{\prime \prime }\right\}\in \mathcal{P}\mathcal{T}\left(E^{\prime }\right).$$

It now follows from (a) that $\nu$ maps $\mathcal{T}$ into ${\mathcal{T}}^{\prime \prime }$. □

Proof. Let $c$ be a ${\delta }_0$-chain in $E$ and define $\tilde{\theta }$ by

$$ϵ{\left(c\right)}^{\tilde{\theta }}=ϵ\left(c\theta \right).$$

(6.8)

Then clearly

$$\overline{ϵ\left(c\right)}\bar{\theta }=\overline{ϵ{\left(c\right)}^{\tilde{\theta }}}.$$

(6.9)

We now prove that $\tilde{\theta }$ is compatible with $\theta$ which will imply that $\bar{\theta }=h\left(\tilde{\theta },\theta \right)$.

Since $ϵ\left(c\right)\in {Ẽ}_0\left(e_0,e_n\right)$ and $ϵ\left(c\theta \right)\in {Ẽ}_0^{\prime }\left(e_0^{\theta },e_n^{\theta }\right)$, axiom (c1) is clearly satisfied. Axiom (c2) follows from the definition of $\tilde{\theta }$ itself.

Let $c=\left(e_0\dots e_n\right)$, $c^{\prime }=\left(e_0^{\prime }\dots e_m^{\prime }\right)$ and suppose that $e_n\omega e_0^{\prime }$. Then by 3.2 we have

$$ϵ\left(e_n{c}^{\prime }\right)={\tau }^r\left(e_n,e_n\right)\circ ϵ\left(c^{\prime }\right).$$

Consequently

$$\begin{array}{rcll}ϵ\left(c\cdot e_n{c}^{\prime }\right)& =& ϵ\left(c\right)\circ {\tau }^r\left(e_n,e_n\right)\circ ϵ\left(c^{\prime }\right)& \\ & =& ϵ\left(c\right)\circ ϵ\left(c^{\prime }\right).& \end{array}$$

But by definition of $\left(c\cdot e_n{c}^{\prime }\right)\theta$

$$\left(c\cdot e_n{c}^{\prime }\right)\theta =c\theta \cdot e_n\theta c^{\prime }\theta$$

and $e_n\theta \omega e_0^{\prime }\theta$. Thus

$$ϵ\left(c\cdot e_n{c}^{\prime }\theta \right)=ϵ\left(c\theta \right)\circ ϵ\left(c^{\prime }\theta \right)$$

and

$${\left(ϵ\left(c\right)\circ ϵ\left(c^{\prime }\right)\right)}^{\tilde{\theta }}=ϵ{\left(c\right)}^{\tilde{\theta }}\circ ϵ{\left(c^{\prime }\right)}^{\tilde{\theta }}.$$

The same equality can be proved when $e_0^{\prime }\omega e_n$. Hence $\tilde{\theta }$ satisfies (c3) and we conclude that $\tilde{\theta }$ is compatible with $\theta$.

By Theorem 6.2, $\bar{\theta }$ is injective if and only if $\theta$ is injective. If $\theta$ is surjective, clearly $\bar{\theta }$ is also surjective. On the other hand, if $\bar{\theta }$ is surjective, by Theorem 6.5(b),

$${Ẽ}_0\tilde{\theta }\in \mathcal{P}\mathcal{T}\left(E^{\prime }\right),$$

and hence ${Ẽ}_0^{\prime }\subseteq {Ẽ}_0\tilde{\theta }$. But clearly $\tilde{\theta }$ maps ${Ẽ}_0$ into ${Ẽ}_0^{\prime }$ and therefore ${Ẽ}_0\tilde{\theta }\subseteq {Ẽ}_0^{\prime }$. Thus $\tilde{\theta }$ is surjective and hence by Theorem 6.5(a), $\theta$ is surjective. □

Proof. By Theorem 6.6, $\bar{\theta }$ is a homomorphism of ${\overline{Ẽ}}_0$ into ${\overline{Ẽ}}_0^{\prime }$ and hence ${\overline{Ẽ}}_0\bar{\theta }$ is a regular subsemigroup of ${\overline{Ẽ}}_0^{\prime }$. $E\left(\overline{Ẽ}\bar{\theta }\right)$ is therefore a biordered subset of $E\left({\overline{Ẽ}}_0^{\prime }\right)$. But by (4.6),

$$E\left({\overline{Ẽ}}_0\bar{\theta }\right)=E\left({\overline{Ẽ}}_0\right)\bar{\theta },$$

and hence $E\left({\overline{Ẽ}}_0\bar{\theta }\right){\chi }_{E^{\prime }}^{-1}$ is a biordered subset of $E^{\prime }$. Since

$$\theta ={\chi }_E\circ \theta \circ {\chi }_{E^{\prime }}^{-1}$$

and $E\left({\overline{Ẽ}}_0\right)=E{\chi }_E$, we notice that

$$E\theta =\left(E{\chi }_E\right)\left(\bar{\theta }\circ {\chi }_{E^{\prime }}^{-1}\right)$$

is a biordered subset of $E^{\prime }$ □