The knotted diagram of the isomorphism $\mathcal{I}^{XYZ}_{12}$ is shown in the figure <ref> on the left, the diagram of the isomorphism $\mathcal{I}^{XYZ}_{23}$ is shown in the figure <ref> on the right.

[scale=0.5, baseline=([yshift=-1.1ex]current bounding box.center)]
(0, -3) rectangle ++(1, 3) node[pos=0.5] $a$;
[knot_diagram] (1, -0.5) – (2, -0.5) node[above=-1.5pt, midway] $X$ circle (0.03);
[knot_diagram] (1, -1.5) – (2, -1.5) node[above=-1.5pt, midway] $Y$ circle (0.03);
[knot_diagram] (1, -2.5) – (2, -2.5) node[above=-1.5pt, midway] $Z$ circle (0.03);
[scale=0.5, baseline=([yshift=-1.1ex]current bounding box.center)]
(0, -3) rectangle ++(1, 3) node[pos=0.5] $a$;

clip width=5,

ignore endpoint intersections=false,

(1, -0.5) .. controls (2.0, -0.5) and (2.0 , -1.5) .. (3, -1.5);
(1, -1.5) .. controls (2.0, -1.5) and (2.0 , -0.5) .. (3, -0.5);
(1, -2.5) – (3, -2.5);

(3, -3) rectangle ++(4, 3) node[pos=0.5] $u^{(Y\otimes X)\otimes Z}_{v_{X}' v_{Y}' (v_{Z}')^{-1}}$;
[knot_diagram] (7, -0.5) – (8, -0.5) node[above=-1.5pt, midway] $Y$ circle (0.03);
[knot_diagram] (7, -1.5) – (8, -1.5) node[above=-1.5pt, midway] $X$ circle (0.03);
[knot_diagram] (7, -2.5) – (8, -2.5) node[above=-1.5pt, midway] $Z$ circle (0.03);

[scale=0.5, baseline=([yshift=-1.1ex]current bounding box.center)]
(0, -3) rectangle ++(1, 3) node[pos=0.5] $a$;
[knot_diagram] (1, -0.5) – (2, -0.5) node[above=-1.5pt, midway] $X$ circle (0.03);
[knot_diagram] (1, -1.5) – (2, -1.5) node[above=-1.5pt, midway] $Y$ circle (0.03);
[knot_diagram] (1, -2.5) – (2, -2.5) node[above=-1.5pt, midway] $Z$ circle (0.03);
[scale=0.5, baseline=([yshift=-1.1ex]current bounding box.center)]
(0, -3) rectangle ++(1, 3) node[pos=0.5] $a$;

clip width=5,

ignore endpoint intersections=false,

(1, -1.5) .. controls (2.0, -1.5) and (2.0 , -2.5) .. (3, -2.5);
(1, -2.5) .. controls (2.0, -2.5) and (2.0 , -1.5) .. (3, -1.5);
(1, -0.5) – (3, -0.5);

(3, -3) rectangle ++(4, 3) node[pos=0.5] $u^{(X\otimes Z)\otimes Y}_{(v_{X}')^{-1} v_{Y}' v_{Z}'}$;
[knot_diagram] (7, -0.5) – (8, -0.5) node[above=-1.5pt, midway] $X$ circle (0.03);
[knot_diagram] (7, -1.5) – (8, -1.5) node[above=-1.5pt, midway] $Z$ circle (0.03);
[knot_diagram] (7, -2.5) – (8, -2.5) node[above=-1.5pt, midway] $Y$ circle (0.03);

Isomorphism $\mathcal{I}^{XYZ}_{12}$ (on the left) and $\mathcal{I}^{XYZ}_{23}$ (on the right)

For any admissible triplet $(X,Y,Z)$ the two isomorphisms $\mathcal{I}^{XYZ}_{12}$ and $\mathcal{I}^{XYZ}_{23}$ are identities.

Consider five different cases with respect to different admissible triplets $(X, Y, Z)$.

$X = Y = Z = \U$. In this case $\mathcal{I}^{\U\U\U}_{12}(a) = a\circ (c_{\U, \U}\otimes id_{\U})\circ u^{\U}_{1} = a$ and $\mathcal{I}^{\U\U\U}_{23}(a) = a\circ \alpha_{\U, \U, \U}\circ (id_{\U}\otimes c_{\U, \U})\circ \alpha_{\U, \U, \U}^{-1}\circ u^{\U}_{1} = a$, because $c_{\U, \U} = id_{\U}$ and $\alpha_{\U, \U, \U} = id_{\U}$.

$X = \U$, $Y = Z = \A$. In this case $\mathcal{I}^{\U\A\A}_{12}(a) = a\circ (c_{\U, \A}\otimes id_{\A})\circ u^{\U + \A}_{1} = a$ and $\mathcal{I}^{\U\A\A}_{23}(a) = a\circ \alpha_{\U, \A, \A}\circ (id_{\U}\otimes c_{\A, \A})\circ \alpha_{\U, \A, \A}^{-1}\circ u^{\U + \A}_{\frac{1}{\beta_{\varepsilon}^2}}$. The diagram of the second composition is shown in the figure <ref>.

[scale=0.5, baseline=([yshift=-0.7ex]current bounding box.center)]
(0, -1) rectangle ++(1, 1) node[pos=0.5] $\U$;

(3, -1) rectangle ++(1, 1) node[pos=0.5] $\U$;
(3, -2) rectangle ++(1, 1) node[pos=0.5] $\A$;
[-latex] (1, -0.5) – (3, -0.5) node[above, midway] $a$;

(6, -1) rectangle ++(1, 1) node[pos=0.5] $\U$;
(6, -2) rectangle ++(1, 1) node[pos=0.5] $\A$;
[-latex] (4, -0.5) – (6, -0.5) node[above, midway] $\beta_{\varepsilon}^2$;
[-latex] (4, -1.5) – (6, -1.5) node[below, midway] $\beta_{\varepsilon}$;

(9, -1) rectangle ++(1, 1) node[pos=0.5] $\U$;
(9, -2) rectangle ++(1, 1) node[pos=0.5] $\A$;
[-latex] (7, -0.5) – (9, -0.5) node[above, midway] $\frac{1}{\beta_{\varepsilon}^2}$;
[-latex] (7, -1.5) – (9, -1.5) node[below, midway] $\frac{1}{\beta_{\varepsilon}^2}$;
[scale=0.5, baseline=([yshift=-1.2ex]current bounding box.center)]
(0, -1) rectangle ++(1, 1) node[pos=0.5] $\U$;

(3, -1) rectangle ++(1, 1) node[pos=0.5] $\U$;
(3, -2) rectangle ++(1, 1) node[pos=0.5] $\A$;
[-latex] (1, -0.5) – (3, -0.5) node[above, midway] $a$;

Diagram of the compositions $a\circ \alpha_{\U, \A, \A}\circ (id_{\U}\otimes c_{\A, \A})\circ \alpha_{\U, \A, \A}^{-1}\circ u^{\U + \A}_{\frac{1}{\beta_{\varepsilon}^2}}$ and $a\circ (c_{\A, \A}\otimes id_{\U})\circ u^{\U + \A}_{\frac{1}{\beta_{\varepsilon}^2}}$

$X = \A$, $Y = \U$, $Z = \A$. In this case $\mathcal{I}^{\A\U\A}_{12}(a) = a\circ (c_{\A, \U}\otimes id_{\A})\circ u^{\U + \A}_{1} = a$ and $\mathcal{I}^{\A\U\A}_{23}(a) = a\circ \alpha_{\A, \U, \A}\circ (id_{\A}\otimes c_{\U, \A})\circ \alpha_{\A, \A, \U}^{-1}\circ u^{\U + \A}_{1} = a$, because $c_{\A, \U} = c_{\U, \A} = id_{\A}$ and $\alpha_{\A, \U, \A} = \alpha_{\A, \A, \U} = id_{\U + \A}$.

$X = Y = \A$, $Z = \U$. In this case $\mathcal{I}^{\A\A\U}_{12}(a) = a\circ (c_{\A, \A}\otimes id_{\U})\circ u^{\U + \A}_{\frac{1}{\beta_{\varepsilon}^2}}$ and $\mathcal{I}^{\A\A\U}_{23}(a) = a\circ \alpha_{\A, \A, \U}\circ (id_{\A}\otimes c_{\A, \U})\circ \alpha_{\A, \U, \A}^{-1}\circ u^{\U + \A}_{1} = a$. The digram of the first composition is shown in the figure <ref>.

$X = Y = Z = \A$. In this case $\mathcal{I}^{\A\A\A}_{12}(a) = a\circ (c_{\A, \A}\otimes id_{\A})\circ u^{\A + \U + \A}_{\frac{1}{\beta_{\varepsilon}}}$ and $\mathcal{I}^{\A\A\A}_{23}(a) = a\circ \alpha_{\A, \A, \A}\circ (id_{\A}\otimes c_{\A, \A})\circ \alpha_{\A, \A, \A}^{-1}\circ u^{\A + \U + \A}_{\frac{1}{\beta_{\varepsilon}}}$. Diagram of the first composition is shown in the figure <ref>. Diagram of the second composition is shown in the figure <ref>.

[scale=0.5, baseline=([yshift=-0.4ex]current bounding box.center)]
(0, -1) rectangle ++(1, 1) node[pos=0.5] $\U$;

(3, -1) rectangle ++(1, 1) node[pos=0.5] $\A$;
(3, -2) rectangle ++(1, 1) node[pos=0.5] $\U$;
(3, -3) rectangle ++(1, 1) node[pos=0.5] $\A$;
[-latex] (1, -0.5) – (3, -1.5) node[above, midway] $a$;

(6, -1) rectangle ++(1, 1) node[pos=0.5] $\A$;
(6, -2) rectangle ++(1, 1) node[pos=0.5] $\U$;
(6, -3) rectangle ++(1, 1) node[pos=0.5] $\A$;
[-latex] (4, -0.5) – (6, -0.5) node[above=-2.0pt, midway] $\beta_{\varepsilon}^2$;
[-latex] (4, -1.5) – (6, -1.5) node[above=-2.0pt, midway] $\beta_{\varepsilon}$;
[-latex] (4, -2.5) – (6, -2.5) node[below, midway] $\beta_{\varepsilon}$;

(9, -1) rectangle ++(1, 1) node[pos=0.5] $\A$;
(9, -2) rectangle ++(1, 1) node[pos=0.5] $\U$;
(9, -3) rectangle ++(1, 1) node[pos=0.5] $\A$;
[-latex] (7, -0.5) – (9, -0.5) node[above=-3.0pt, midway] $\frac{1}{\beta_{\varepsilon}}$;
[-latex] (7, -1.5) – (9, -1.5) node[above=-3.0pt, midway] $\frac{1}{\beta_{\varepsilon}}$;
[-latex] (7, -2.5) – (9, -2.5) node[below, midway] $\frac{1}{\beta_{\varepsilon}}$;
[scale=0.5, baseline=([yshift=-0.75ex]current bounding box.center)]
(0, -1) rectangle ++(1, 1) node[pos=0.5] $\U$;

(3, -1) rectangle ++(1, 1) node[pos=0.5] $\A$;
(3, -2) rectangle ++(1, 1) node[pos=0.5] $\U$;
(3, -3) rectangle ++(1, 1) node[pos=0.5] $\A$;
[-latex] (1, -0.5) – (3, -1.5) node[above, midway] $a$;

Diagram of the composition $a\circ (c_{\A, \A}\otimes id_{\A})\circ u^{\A + \U + \A}_{\frac{1}{\beta_{\varepsilon}}}$

[scale=0.5, baseline=([yshift=-0.4ex]current bounding box.center)]
(0, -1) rectangle ++(1, 1) node[pos=0.5] $\U$;

(3, -1) rectangle ++(1, 1) node[pos=0.5] $\A$;
(3, -2) rectangle ++(1, 1) node[pos=0.5] $\U$;
(3, -3) rectangle ++(1, 1) node[pos=0.5] $\A$;
[-latex] (1, -0.5) – (3, -1.5) node[above, midway] $a$;

(6, -1) rectangle ++(1, 1) node[pos=0.5] $\A$;
(6, -2) rectangle ++(1, 1) node[pos=0.5] $\U$;
(6, -3) rectangle ++(1, 1) node[pos=0.5] $\A$;
[ma_blue] (4, -0.5) – (6, -0.5);
[ma_green] (4, -0.5) – (6, -2.5);
[ma_red] (4, -2.5) – (6, -0.5);
[ma_orange] (4, -2.5) – (6, -2.5);
[ma_black] (4, -1.5) – (6, -1.5);

(9, -1) rectangle ++(1, 1) node[pos=0.5] $\A$;
(9, -2) rectangle ++(1, 1) node[pos=0.5] $\U$;
(9, -3) rectangle ++(1, 1) node[pos=0.5] $\A$;
[-latex] (7, -0.5) – (9, -0.5) node[above=-2.0pt, midway] $\beta_{\varepsilon}^2$;
[-latex] (7, -1.5) – (9, -1.5) node[above=-2.0pt, midway] $\beta_{\varepsilon}$;
[-latex] (7, -2.5) – (9, -2.5) node[below, midway] $\beta_{\varepsilon}$;

(12, -1) rectangle ++(1, 1) node[pos=0.5] $\A$;
(12, -2) rectangle ++(1, 1) node[pos=0.5] $\U$;
(12, -3) rectangle ++(1, 1) node[pos=0.5] $\A$;
[ma_blue] (10, -0.5) – (12, -0.5);
[ma_green] (10, -0.5) – (12, -2.5);
[ma_red] (10, -2.5) – (12, -0.5);
[ma_orange] (10, -2.5) – (12, -2.5);
[ma_black] (10, -1.5) – (12, -1.5);

(15, -1) rectangle ++(1, 1) node[pos=0.5] $\A$;
(15, -2) rectangle ++(1, 1) node[pos=0.5] $\U$;
(15, -3) rectangle ++(1, 1) node[pos=0.5] $\A$;
[-latex] (13, -0.5) – (15, -0.5) node[above=-3.0pt, midway] $\frac{1}{\beta_{\varepsilon}}$;
[-latex] (13, -1.5) – (15, -1.5) node[above=-3.0pt, midway] $\frac{1}{\beta_{\varepsilon}}$;
[-latex] (13, -2.5) – (15, -2.5) node[below, midway] $\frac{1}{\beta_{\varepsilon}}$;
[scale=0.5, baseline=([yshift=-0.75ex]current bounding box.center)]
(0, -1) rectangle ++(1, 1) node[pos=0.5] $\U$;

(3, -1) rectangle ++(1, 1) node[pos=0.5] $\A$;
(3, -2) rectangle ++(1, 1) node[pos=0.5] $\U$;
(3, -3) rectangle ++(1, 1) node[pos=0.5] $\A$;
[-latex] (1, -0.5) – (3, -1.5) node[above, midway] $a$;

Diagram of the composition $a\circ \alpha_{\A, \A, \A}\circ (id_{\A}\otimes c_{\A, \A})\circ \alpha_{\A, \A, \A}^{-1}\circ u^{\A + \U + \A}_{\frac{1}{\beta_{\varepsilon}}}$

For any admissible triplet $(X, Y, Z)$, define the module $H(X, Y, Z)$ obtained by identifying $H^{XYZ}$, $H^{XZY}$, $H^{YXZ}$, $H^{YZX}$, $H^{ZXY}$ and $H^{ZYX}$ by isomorphisms generated by $\mathcal{I}^{XYZ}_{12}$ and $\mathcal{I}^{XYZ}_{23}$. In fact, to get an arbitrary element $a\in H(X, Y, Z)$ we should choose the same element $a$ from all these six modules.

§.§.§ Pairing

Let $w_{\U}\in Hom (\U, \U)$ be a morphism of the category $\E$, defined by the value $1$ (i.e. $w_{\U} = id_{\U}$), and let $w_{\A}\in Hom (\A, \A)$ be a morphism of the category $\E$, defined by the non-zero value $z\in \mathbb{C}$.

For each simple object $X\in I$: $(w_{X}\otimes id_{X})\circ d_X = (id_{X}\otimes w_{X})\circ d_X$.

The lemma statement is obvious.

For any admissible triplet of simple objects $(X, Y, Z)$, define the pairing $(\cdot, \cdot)^{XYZ}\colon H(X, Y, Z)\times H(X, Y, Z)\to\mathbb{C}$ as follows. For any two elements $a, b\in H(X, Y, Z)$, the value $(a, b)^{XYZ}$ is equal to the morphism shown in the figure <ref>. This morphism is a morphism from $\U$ to $\U$, so it is defined by a complex number.

(0, -3) rectangle ++(1, 3) node[pos=0.5] $a$;
(0, -7) rectangle ++(1, 3) node[pos=0.5] $b$;

[knot_diagram] (1, -0.5) – (2.1, -0.5) node[above=-1.5pt, midway] $X$;
[knot_diagram] (1, -1.5) – (2.1, -1.5) node[above=-1.5pt, midway] $Y$;
[knot_diagram] (1, -2.5) – (2.1, -2.5) node[above=-1.5pt, midway] $Z$;

[knot_diagram] (1, -4.5) – (2, -4.5) node[above=-1.5pt, midway] $Z$;
[knot_diagram] (1, -5.5) – (2, -5.5) node[above=-1.5pt, midway] $Y$;
[knot_diagram] (1, -6.5) – (2, -6.5) node[above=-1.5pt, midway] $X$;

(2.1, -0.9) rectangle ++(0.8, 0.8) node[pos=0.5] $w_X$;
(2.1, -1.9) rectangle ++(0.8, 0.8) node[pos=0.5] $w_Y$;
(2.1, -2.9) rectangle ++(0.8, 0.8) node[pos=0.5] $w_Z$;

[knot_diagram] (2.0, -4.5) – (2.9, -4.5) arc (-90:90:1);
[knot_diagram] (2.0, -5.5) – (2.9, -5.5) arc (-90:90:2);
[knot_diagram] (2.0, -6.5) – (2.9, -6.5) arc (-90:90:3);

Knotted diagram of the morphism for the pairing $(a, b)^{XYZ}$

$(a, b)^{\U\U\U} = ab$, $(a, b)^{\U\A\A} = ab y^2 z^2$ and $(a, b)^{\A\A\A} = ab xy^3 z^3$.

For any triplet of simple objects $(X, Y, Z)$ and for any $a, b\in H(X, Y, Z)$, the morphism $(a, b)^{XYZ}$ is a composition $\mu_1\circ\ldots \circ \mu_9$ of nine morphisms, where:

$\mu_1 = a\otimes b$,

$\mu_2 = ((w_X\otimes w_Y)\otimes w_Z)\otimes ((id_Z\otimes id_Y)\otimes id_X)$,

$\mu_3 = \alpha^{-1}_{(X\otimes Y)\otimes Z, Z\otimes Y, X}$,

$\mu_4 = \alpha^{-1}_{(X\otimes Y)\otimes Z,Z, Y}\otimes id_X$,

$\mu_5 = (\alpha_{X\otimes Y, Z, Z}\otimes id_Y)\otimes id_X$,

$\mu_6 = (((id_X\otimes id_Y)\otimes d_Z)\otimes id_Y)\otimes  id_X$,

$\mu_7 = \alpha_{X, Y, Y}\otimes id_X$,

$\mu_8 = (id_X\otimes d_Y)\otimes id_X$,

$\mu_9 = d_X$.

The theorem follows from the careful computation of these morphisms for three different cases where $(X, Y, Z)$ coincides with $(\U, \U, \U)$, $(\U, \A, \A)$ or $(\A, \A, \A)$.

§.§.§ 6j-symbols

For any collection of simple objects $X_1, Y_1, Z_1, X_2, Y_2, Z_2\in I$, define 6j-symbol
X_1 & Y_1 & Z_1 \\
X_2 & Y_2 & Z_2
\end{array}\right|\colon H(X_1, Y_1, Z_1)\otimes H(X_1, Y_2, Z_2)\otimes H(Y_1, Z_2, X_2)\otimes H(Z_1, X_2, Y_2)\to\mathbb{C}$$
as follows. If at least one of the triplets $(X_1, Y_1, Z_1)$, $(X_1, Y_2, Z_2)$, $(Y_1, Z_2, X_2)$ or $(Z_1, X_2, Y_2)$ is not admissible, then the corresponding module is trivial and the 6j-symbol is also a trivial map. If all triplets $(X_1, Y_1, Z_1)$, $(X_1, Y_2, Z_2)$, $(Y_1, Z_2, X_2)$ and $(Z_1, X_2, Y_2)$ are admissible, then all four modules $H(X_1, Y_1, Z_1)$, $H(X_1, Y_2, Z_2)$, $H(Y_1, Z_2, X_2)$ and $H(Z_1, X_2, Y_2)$ are isomorphic to $\mathbb{C}$. In this case the 6j-symbol maps $a_1\otimes a_2\otimes a_3\otimes a_4$, for any $a_1\in H(X_1, Y_1, Z_1)$, $a_2\in H(X_1, Y_1, Z_2)$, $a_3\in H(Y_1, Z_2, X_2)$, $a_4\in H(Z_1, X_2, Y_2)$, to the value of the morphism shown in the figure <ref>. This morphism is a morphism from $\U$ to $\U$, so it is defined by a complex value. This complex value defines the image
X_1 & Y_1 & Z_1 \\
X_2 & Y_2 & Z_2
\end{array}\right|(a_1\otimes a_2\otimes a_3\otimes a_4).$$

(0, 0) rectangle ++(1, 3) node[pos=0.5] $a_4$;
(0, 5) rectangle ++(1, 3) node[pos=0.5] $a_1$;
(5, 2) rectangle ++(1, 3) node[pos=0.5] $a_3$;
(9, 4) rectangle ++(1, 3) node[pos=0.5] $a_2$;

(2.1, 1.1) rectangle ++(0.8, 0.8) node[pos=0.5] $w_{X_2}$;
(2.1, 5.1) rectangle ++(0.8, 0.8) node[pos=0.5] $w_{Z_1}$;
(2.1, 6.1) rectangle ++(0.8, 0.8) node[pos=0.5] $w_{Y_1}$;
(2.1, 7.1) rectangle ++(0.8, 0.8) node[pos=0.5] $w_{X_1}$;
(11.1, 4.1) rectangle ++(0.8, 0.8) node[pos=0.5] $w_{Z_2}$;
(11.1, 5.1) rectangle ++(0.8, 0.8) node[pos=0.5] $w_{Y_2}$;

[knot_diagram] (1, 0.5) – (2.1, 0.5) node[above=-1.5pt, midway] $Y_2$ – (11.9, 0.5) arc (-90:90:2.5);
[knot_diagram] (1, 1.5) – (2.1, 1.5) node[above=-1.5pt, midway] $X_2$;
[knot_diagram] (2.9, 1.5) – (7.1, 1.5) arc (-90:90:0.5) – (6, 2.5) node[above=-1.5pt, midway] $X_2$;
[knot_diagram] (1, 2.5) – (2.1, 2.5) node[above=-1.5pt, midway] $Z_1$ – (2.9, 2.5) arc (-90:90:1.5);
[knot_diagram] (1, 5.5) – (2.1, 5.5) node[above=-1.5pt, midway] $Z_1$;
[knot_diagram] (1, 6.5) – (2.1, 6.5) node[above=-1.5pt, midway] $Y_1$;
[knot_diagram] (1, 7.5) – (2.1, 7.5) node[above=-1.5pt, midway] $X_1$;
[knot_diagram] (6, 4.5) – (7.1, 4.5) node[above=-1.5pt, midway] $Y_1$ arc (-90:90:1) – (2.9, 6.5);
[knot_diagram] (6, 3.5) – (7.1, 3.5) node[above=-1.5pt, midway] $Z_2$ – (11.9, 3.5) arc (-90:90:0.5);
[knot_diagram] (10, 4.5) – (11.1, 4.5) node[above=-1.5pt, midway] $Z_2$;
[knot_diagram] (10, 5.5) – (11.1, 5.5) node[above=-1.5pt, midway] $Y_2$;
[knot_diagram] (10, 6.5) – (11.1, 6.5) node[above=-1.5pt, midway] $X_1$ arc (-90:90:0.5) – (2.9, 7.5);

Knotted diagram of the 6j-symbol

All 6j-symbols are symmetrical in the following sense: for any simple objects $X_1, Y_1, Z_1, X_2, Y_2, Z_2\in I$ and $a_1\in H(X_1, Y_1, Z_1)$, $a_2\in H(X_1, Y_2, Z_2)$, $a_3\in H(Y_1, Z_2, X_2)$, $a_4\in H(Z_1, X_2, Y_2)$ we have
X_1 & Y_1 & Z_1 \\
X_2 & Y_2 & Z_2
\end{array}\right|(a_1\otimes a_2\otimes a_3\otimes a_4) = \left|\begin{array}{lll}
X_1 & Z_2 & Y_1 \\
X_2 & Z_1 & Y_1
\end{array}\right|(a_2\otimes a_1\otimes a_3\otimes a_4) = \left|\begin{array}{lll}
Z_1 & Y_2 & X_2 \\
Z_2 & Y_1 & X_1
\end{array}\right|(a_4\otimes a_1\otimes a_2\otimes a_3).$$

Non-trivial 6j-symbols are as follows:

\U & \U & \U \\
\U & \U & \U
\end{array}\right|(a_1\otimes a_2\otimes a_3\otimes a_4) = a_1 a_2 a_3 a_4$,

\U & \U & \U \\
\A & \A & \A
\end{array}\right|(a_1\otimes a_2\otimes a_3\otimes a_4) = a_1 a_2 a_3 a_4 \frac{y^3 z^3}{\sqrt{\varepsilon}}$,

\U & \A & \A \\
\U & \A & \A
\end{array}\right|(a_1\otimes a_2\otimes a_3\otimes a_4) = a_1 a_2 a_3 a_4 \frac{y^4 z^4}{\varepsilon}$,

\U & \A & \A \\
\A & \A & \A
\end{array}\right|(a_1\otimes a_2\otimes a_3\otimes a_4) = a_1 a_2 a_3 a_4 \frac{x y^5 z^5}{\varepsilon}$,

\A & \A & \A \\
\A & \A & \A
\end{array}\right|(a_1\otimes a_2\otimes a_3\otimes a_4) = a_1 a_2 a_3 a_4 \frac{x^2 y^6 z^6}{-\varepsilon^2}$.

For any six simple objects $X_1, Y_1, Z_1, X_2, Y_2, Z_2$ which define non-trivial 6j-symbols, and for any $a_1\in H(X_1, Y_1, Z_1)$, $a_2\in H(X_1, Y_2, Z_2)$, $a_3\in H(Y_1, Z_2, X_2)$, $a_4\in H(Z_1, X_2, Y_2)$ the morphism
X_1 & Y_1 & Z_1 \\
X_2 & Y_2 & Z_2
\end{array}\right|(a_1\otimes a_2\otimes a_3\otimes a_4)
is a composition $\mu_1\circ\ldots\circ \mu_{21}$, where

$\mu_1 = a_1\otimes a_4$,

$\mu_2 = ((w_{X_1}\otimes w_{Y_1})\otimes w_{Z_1})\otimes ((id_{Z_1}\otimes w_{X_2})\otimes id_{Y_2})$,

$\mu_3 = \alpha^{-1}_{(X_1\otimes Y_1)\otimes Z_1, Z_1\otimes X_2, Y_2}$,

$\mu_4 = \alpha^{-1}_{(X_1\otimes Y_1)\otimes Z_1, Z_1, X_2}\otimes id_{Y_2}$,

$\mu_5 = (\alpha_{X_1\otimes Y_1, Z_1, Z_1}\otimes id_{X_2})\otimes id_{Y_2}$,

$\mu_6 = (((id_{X_1}\otimes id_{Y_1})\otimes d_{Z_1})\otimes id_{X_2})\otimes id_{Y_2}$,

$\mu_7 = (((id_{X_1}\otimes id_{Y_1})\otimes a_3)\otimes id_{X_2})\otimes id_{Y_2}$,

$\mu_8 = (\alpha^{-1}_{X_1\otimes Y_1, Y_1\otimes Z_2, X_2}\otimes id_{X_2})\otimes id_{Y_2}$,

$\mu_9 = ((\alpha^{-1}_{X_1\otimes Y_1, Y_1, Z_2}\otimes id_{X_2})\otimes id_{X_2})\otimes id_{Y_2}$,

$\mu_{10} = (((\alpha_{X_1, Y_1, Y_1}\otimes id_{Z_2})\otimes id_{X_2})\otimes id_{X_2})\otimes id_{Y_2}$,

$\mu_{11} = ((((id_{X_1}\otimes d_{Y_1})\otimes id_{Z_2})\otimes id_{X_2})\otimes id_{X_2})\otimes id_{Y_2}$,

$\mu_{12} = \alpha_{X_1\otimes Z_2, X_2, X_2}\otimes id_{Y_2}$,

$\mu_{13} = ((id_{X_1}\otimes id_{Z_2})\otimes d_{X_2})\otimes id_{Y_2}$,

$\mu_{14} = ((id_{X_1}\otimes a_2)\otimes id_{Z_2})\otimes id_{Y_2}$,

$\mu_{15} = ((id_{X_1}\otimes ((id_{X_1}\otimes w_{Y_2})\otimes w_{Z_2}))\otimes id_{Z_2})\otimes id_{Y_2}$,

$\mu_{16} = \alpha_{X_1, (X_1\otimes Y_2)\otimes Z_2, Z_2}\otimes id_{Y_2}$,

$\mu_{17} = (id_{X_1}\otimes \alpha_{X_1\otimes Y_2, Z_2, Z_2})\otimes id_{Y_2}$,

$\mu_{18} = (id_{X_1}\otimes ((id_{X_1}\otimes id_{Y_2})\otimes d_{Z_2}))\otimes id_{Y_2}$,

$\mu_{19} = \alpha^{-1}_{X_1, X_1, Y_2}\otimes id_{Y_2}$,

$\mu_{20} = (d_{X_1}\otimes id_{Y_2})\otimes id_{Y_2}$,

$\mu_{21} = d_{Y_2}$.

The theorem follows from the careful computation of these morphisms for all necessary combinations of objects $X_1, Y_1, Z_1, X_2, Y_2, Z_2\in I$.

§.§.§ Special spines and definition of $tv_{\varepsilon}$

In the book [17] the construction of the Turaev – Viro type invariant, derived from modular category, is constructed by using triangulations of 3-manifolds. We will use the dual approach and describe the invariant $tv_{\varepsilon}$ using special spines of 3-manifolds.

A two-dimensional polyhedron $P$ is called special, if it satisfies to the following conditions:

* The link of each point $x\in P$ is homeomorphic either to the circle (figure <ref> on the left, these points are called regular points), or to the circle with diameter (figure <ref> in the centre, these points are called triple points), or to the circle with two diameters (figure <ref> on the right, these points are called true vertices);

* The union of all triple points is a disjoint set of intervals called triple lines;

* The union of all regular points is a disjoint union of open discs called 2-components.

[scale=0.5, baseline=([yshift=-0.7ex]current bounding box.center)]
[knot_diagram] (0, 0) – (2, 2) – (6, 2) – (4, 0) – (0, 0);
(3, 1) circle (0.05) node[below] $x$;

[scale=0.5, baseline=([yshift=-2.0ex]current bounding box.center)]
clip width=5,
ignore endpoint intersections=false,
(5, 2.75) – (5, 1);
(2, 2) – (6, 2);

[knot_diagram] (2, 2) – (0, 0) – (4, 0) – (6, 2);
[knot_diagram] (5, 1) – (1, 1) – (1, 2.75) – (5, 2.75);
(3, 1) circle (0.05) node[below] $x$;

[scale=0.5, baseline=([yshift=0.7ex]current bounding box.center)]

clip width=5,
ignore endpoint intersections=false,
flip crossing/.list=3
(5, 2.75) – (5, 1) – (1, 1);
(2, 2) – (6, 2);
(4, 1.75) – (4, 0.25) – (2, -1.75);
(0, 0) – (4, 0);

[knot_diagram] (0, 0) – (2, 2);
[knot_diagram] (1, 1) – (1, 2.75) – (5, 2.75);
[knot_diagram] (4, 0) – (6, 2);
[knot_diagram] (2, -1.75) – (2, 0) – (4, 2) – (4, 1.75);
(3, 1) circle (0.05) node[below right] $x$;

Regular point (on the left), triple point (in the center), true vertex (on the right)

The special polyhedron is called a special spine of the closed 3-manifold $M$, if its complement $M\setminus P$ is homomorphic to an open 3-ball. We will restrict ourselves to closed 3-manifolds, so we don't need a general definition of the spine.

Now we are ready to define the invariant $tv_{\varepsilon}$. Let $P$ be a special spine of the closed 3-manifold $M$. Let $\mathcal{V}(P)$ be the set of all true vertices, $\mathcal{E}(P)$ the set of all triple lines and $\mathcal{C}(P)$ the set of all 2-components of the spine $P$. The colouring of $P$ is a map $\zeta\colon \mathcal{C}(P)\to \{\U, \A\}$. Define the weight of the coloured spine $\{P\}_{\zeta}$ as follows. For each true vertex $v\in \mathcal{V}(P)$, associate the 6j-symbol $\left|\begin{array}{lll}
X_1 & Y_1 & Z_1 \\
X_2 & Y_2 & Z_2
\end{array}\right|$, where $X_1, X_2, X_3$ are colours of three 2-components incident to a triple line in the neighbourhood of $v$, and $X_2, Y_2, Z_2$ are colours of opposite 2-components (figure <ref> on the left). Use the 6j-symbols from the theorem <ref>. Then the weight $\{P\}_{\zeta}$ is a contraction of all tensors associated with all true vertices. The contruction is done by pairings from the theorem <ref>.

From an equivalent point of view we can define $\{P\}_{\zeta}$ as follows. Assign to each triple line $e\in \mathcal{E}(P)$, incident to 2-components with colours $X, Y, Z$, the value opposite to $(1, 1)^{XYZ}$ (figure <ref> on the right). Then the weight $\{P\}_{\zeta}$ is equal to the product of these assigned values and the values of the 6j-symbols associated to the true vertices, calculated with the argument $1\otimes 1\otimes 1\otimes 1$.

[scale=0.5, baseline=([yshift=0.9ex]current bounding box.center)]

clip width=5,
ignore endpoint intersections=false,
flip crossing/.list=3
(5, 2.75) – (5, 1) – (1, 1);
(2, 2) – (6, 2);
(4, 1.75) – (4, 0.25) – (2, -1.75);
(0, 0) node[below right] $Z_1$ – (4, 0);

[knot_diagram] (0, 0) – (2, 2) node[below right=-1.0pt] $Y_1$;
[knot_diagram] (1, 1) – (1, 2.75) node[below left] $X_1$ – (5, 2.75);
[knot_diagram] (4, 0) – (6, 2);
[knot_diagram] (2, -1.75) – (2, 0) node[below right=-1.0pt] $X_2$ – (4, 2) – (4, 1.75);
(4.3, 0.15) node[right] $Y_2$;
(5.3, 1.25) node[right] $Z_2$;
(3, 1) circle (0.05);
$\to \left|\begin{array}{lll}
X_1 & Y_1 & Z_1 \\
X_2 & Y_2 & Z_2
\end{array}
\right|$
[scale=0.5, baseline=([yshift=0.0ex]current bounding box.center)]

clip width=5,
ignore endpoint intersections=false,

(3, 0) – (2.25, -1.25);
(1, -1) – (4, -1);

[knot_diagram] (2.25, -1.25) – (-0.75, -1.25) – (0, 0) – (3, 0);
[knot_diagram] (1, -1) – (0, 0) – (0, 1.25) – (3, 1.25) – (3, 0) – (4, -1);

(-1, -1) node $Y$;
(0.1, 1) node[left] $X$;
(3.9, -0.25) node $Z$;
$\to \dfrac{1}{(1, 1)^{XYZ}}$

6j-symbol corresponds to a true vertex (on the left), opposite value to pairing corresponds to a triple line (on the right)

Let $\zeta$ be a colouring. Denote by $|\zeta|_{\A}$ the number of 2-components coloured by $\A$. Then
$$tv_{\varepsilon}(P) = \sum\limits_{\zeta}\varepsilon^{|\zeta|_{\A}}\cdot \{P\}_{\zeta},$$
where the sum is taken over all colourings of the spine $P$.

There is only one difference in our definition of the Turaev - Viro type invariant with respect to the original ones from <cit.>. We do not use the coefficient $\frac{1}{D^{2V}}$, where $V$ is a number of vertices in the triangulation of the manifold. In our case, the complement of the spine is always an open 3-ball. So we can ignore this coefficient.

If $P_1$ and $P_2$ are special spines of closed 3-manifold $M$ then $tv_{\varepsilon}(P_1) = tv_{\varepsilon}(P_2)$.

The theorem statement follows from <cit.>.

So we can correctly define $tv_{\varepsilon}(M)$ as equal to $tv_{\varepsilon}(P)$ for any special spine $P$ of the closed 3-manifold $M$.

For any closed 3-manifold $M$:
$$|tr_{\varepsilon}(M)|^2 = \frac{tv_{\varepsilon}(M)}{\varepsilon + 2}.$$

The theorem statement is a corollary of <cit.>.

§.§ $\varepsilon$-invariant

There are several equivalent ways to define the $\varepsilon$-invariant for 3-manifolds (see <cit.>). We will use the approach that is very close to the definition of the $tv_{\varepsilon}$ invariant.

Let $P$ be a special spine of the closed 3-manifold $M$, and let $\eta\colon \mathcal{C}(P)\to \{0, 1\}$ be a colouring of $P$ by two colours: 0 and 1. Define the colour weights as follows: $\omega_0 = 1$ and $\omega_{1} = \varepsilon$, where, as before, $\varepsilon^2 = \varepsilon + 1$.

Let $v\in\mathcal{V}(P)$ be a true vertex of the spine $P$. The neighbourhood of this vertex contains six 2-components. Let $i, j, k\in \{0, 1\}$ be the colours of the components incident to a triple line in the neighbourhood of the vertex $v$, and let $l, m, n$ be the colours of the opposite 2-components. Define the weight $\omega_v$ of the vertex $v$ as follows:
$$\omega_v = \left|\begin{array}{lll}
i & j & k \\
l & m & n
\end{array}\right|',$$

0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right|' = 1$, $\left|\begin{array}{lll}
0 & 0 & 0 \\
1 & 1 & 1
\end{array}\right|' = \frac{1}{\sqrt{\varepsilon}}$, $\left|\begin{array}{lll}
0 & 1 & 1 \\
0 & 1 & 1
\end{array}\right|' = \frac{1}{\varepsilon}$, $\left|\begin{array}{lll}
0 & 1 & 1 \\
1 & 1 & 1
\end{array}\right|' = \frac{1}{\varepsilon}$, $\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right|' = -\frac{1}{\varepsilon^2}$,

and $\omega_v = 0$ in all other cases.

Here we use 6j-symbols with prime to distinguish them from the 6j-symbols used before. The value of the $\varepsilon$-invariant (denoted by $t$) for the manifold $M$ is defined as follows:
$$t(M) = \sum\limits_{\eta}\left(\prod_{c\in \mathcal{C}(p)}\omega_{\eta(c)} \cdot \prod_{v\in \mathcal{V}(P)}\omega_v\right),$$
where the sum is taken over all colourings of $P$.

For any closed 3-manifold $M$:
$$tv_{\varepsilon}(M) = t(M).$$

The main step of the proof is to show that the value $tv_{\varepsilon}(M)$ does not depend on the parameters $x, y$ and $z$ used in the construction.

Let $P$ be a special spine of the manifold $M$, and let $\zeta\colon\mathcal{C}(P)\to \{\U, \A\}$ be a non-trivial colouring (i.e. $\{P\}_{\zeta} \neq 0$). Divide all true vertices of the coloured spine $P$ into five classes with respect to the number of 2-components coloured by $\A$ in the neighbourhood of the vertex. Let $k_0$ be the number of vertices where all incident 2-components are coloured by $\U$, $k_1$ the number of vertices with three incident 2-components coloured by $\A$, $k_2$ the number of vertices with four incident 2-components coloured by $\A$, $k_3$ the number of vertices with five incident 2-components coloured by $\A$, and finally $k_4$ the number of vertices with all incident 2-components coloured by $\A$. Then
$$\{P\}_{\zeta} = \left(\frac{y^3z^3}{\sqrt{\varepsilon}}\right)^{k_1}\cdot \left(\frac{y^4 z^4}{\varepsilon}\right)^{k_2}\cdot \left(\frac{xy^5 z^5}{\varepsilon}\right)^{k_3}\cdot \left(- \frac{x^2 y^6 z^6}{\varepsilon^2}\right)^{k_4}\cdot \left(\frac{1}{xy^3 z^3}\right)^{k_3 + 2k_4}\cdot \left(\frac{1}{z^2 y^2}\right)^{\frac{3}{2}k_1 + 2 k_2 + k_3}.$$
The last two multipliers come from triple lines of $P$. The number of triple lines with only one incident 2-component coloured by $\U$ is equal to $\frac{3}{2}k_1 + 2k_2 + k_3$, and the number of triple lines with all three incident 2-components coloured by $\A$ is equal to $k_3 + 2k_4$.

In the expression $\{P\}_{\zeta}$, the total power of the parameter $x$ is $k_3 + 2k_4 - k_3 - 2k_4 = 0$, and the total powers of the parameters $y$ and $z$ are equal and equal to $3k_1 + 4k_2 + 5k_3 + 6k_4 - 3k_3 - 6k_4 - 3k_1 - 4k_2 - 2k_3 = 0$.

The value $\{P\}_{\zeta}$ does not depend on the parameters $x, y, z$. So we can choose $x = y = z = 1$, and then the formula for $tv_{\varepsilon}(M)$ is exactly the same as the formula for $t(M)$ (after changing $\U$ to $0$ and $\A$ to $1$).

The $tr_{\varepsilon}$ invariant is stronger than $tv_{\varepsilon}$. It's known that there are only four different values of the $\varepsilon$-invariant for lens spaces ([11], <cit.>).

It's easy to calculate $tv_{\varepsilon}(L_{1, 1}) = tv_{\varepsilon}(L_{4, 1}) = 1$, but $tr_{\varepsilon}(L_{1, 1}) = \frac{1}{\sqrt{\varepsilon + 2}}\in \mathbb{R}$ and $tr_{\varepsilon}(L_{4, 1}) = \frac{(\varepsilon + 1)\cdot (1 + \beta_{\varepsilon})^2}{(\varepsilon + 2)^{\frac{3}{2}}}\in\mathbb{C}\setminus\mathbb{R}$. So, $tr_{\varepsilon}(L_{1, 1})\neq tr_{\varepsilon}(L_{4, 1})$.

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$\widehat{h(b_{i},a)}\times(X\setminus\widehat{b_{i}})$.

This shows that $-\colon\mathrm{dom}(-)\to Y$ is lower semicontinuous.
Monotonicity follows from item 1 of Lemma A.5 and the fact that $\kappa$ is
monotone. ∎

We can recover $f$ just from the partial operation $-\colon\mathrm{dom}(-)\to
Y$ as follows:

######  Lemma A.7.

For every $u\in A^{\delta}$ and $v\in B^{\delta}$,

$f(u,v)=\bigwedge\\{z\in Z\mid\exists x\in X\text{ s.t. }u\leq x,\
(z,x)\in\mathrm{dom}(-)\text{ and }\,v\leq z-x\\}.$

###### Proof.

For $(x,y,z)\in X\times Y\times Z$, we have

$f(x,y)\leq z{\quad\text{iff}\quad}\kappa^{-1}(z)\nleq
f(x,y){\quad\text{iff}\quad}f^{\flat}_{1}(x,\kappa^{-1}(z))\nleq y.$

Moreover, $f^{\flat}_{1}(x,\kappa^{-1}(z))\nleq y$ implies $x\leq i(z)$
because the latter inequation is equivalent to
$f^{\flat}_{1}(x,\kappa^{-1}(z))\neq 0$. We obtain that

$f(x,y)\leq z{\quad\text{iff}\quad}[\kappa^{-1}(z-x)\nleq
y{\enspace\text{and}\enspace}x\leq i(z)]{\quad\text{iff}\quad}[y\leq
z-x{\enspace\text{and}\enspace}x\leq i(z)].$

Lastly, since $f(\mkern 1.5mu\text{-}\mkern 1.5mu,\mkern 1.5mu\text{-}\mkern
1.5mu)$ preserves arbitrary meets in both arguments, and every element of
$A^{\delta}$ and $B^{\delta}$ is the meet of the completely meet-irreducible
elements above it, $f(u,v)\leq z$ precisely when $x\leq i(z)$ and $y\leq z-x$
for some $(x,y)\in X\times Y$ satisfying $u\leq x$ and $v\leq y$. ∎

###  A.5. The second partial operation

In the previous section, we presented a partial operation that is mirroring
the role of the pi extension $f$ of $h\colon A\times B\to C$ on the space
side. Although any three-sorted partial operation $Z\times X\rightharpoonup Y$
satisfying the same topological and order-theoretic properties stated in Lemma
A.6 corresponds to a mapping $A^{\delta}\times B^{\delta}\to C^{\delta}$ which
preserves meets in both coordinates, it does not necessarily follow that its
restriction to $A\times B\to C$ is a double operator. This is the reason why
we also need to compute the partial operation corresponding to the sigma
extension $g\colon A^{\delta}\times B^{\delta}\to C^{\delta}$ of $h$.

This time, given an arbitrary $u\in A^{\delta}$, we do not have that
$g(u,\mkern 1.5mu\text{-}\mkern 1.5mu)\colon B^{\delta}\to C^{\delta}$
preserves $0$. On the other hand, since $g$ is monotone, we know that $g(u,0)$
is the smallest element in the image of $g(u,\mkern 1.5mu\text{-}\mkern
1.5mu)$, and so the restriction $g_{1}(u,\mkern 1.5mu\text{-}\mkern
1.5mu)\colon B^{\delta}\to\mathord{\uparrow}\mkern 1.0mug(u,0)$ preserves all
joins and as such it has a right adjoint $g^{\\#}_{1}$:

$g(u,v)\leq w{\quad\text{iff}\quad}g(u,0)\leq w\text{ and }v\leq
g^{\\#}_{1}(u,w)\qquad\forall v\in B^{\delta},\,w\in C^{\delta}.$

######  Lemma A.8.

The following statements hold:

  1. (1)

$g^{\\#}_{1}$ is antitone in the first argument;

  2. (2)

On its domain of definition, $g^{\\#}_{1}$ is a partial map
$\mathcal{J}^{\infty}(A^{\delta})^{\mathrm{op}}\times\mathcal{M}^{\infty}(C^{\delta})\rightharpoonup\mathcal{M}^{\infty}_{1}(B^{\delta})$;

  3. (3)

For $x\in X$ and $z\in Z$, $g(\kappa^{-1}(x),0)\leq z$ iff $i(z)\nleq x$;

  4. (4)

For all $u\in A^{\delta}$ and $z\in Z$, $g^{\\#}_{1}(u,z)\neq 1$.

###### Proof.

Item 1 is easily verified, and item 2 follows from the proof of Theorem 4.4 in
[12].

(3) Because $\kappa^{-1}(x)$ is a closed element,
$g(\kappa^{-1}(x),0)=f(\kappa^{-1}(x),0)=@^{\delta}(\kappa^{-1}(x))$ and so
$@^{\delta}(\kappa^{-1}(x))\leq z$ iff
$\kappa^{-1}(x)\leq@^{\delta}{}^{\\#}(z)=i(z)$ iff $i(z)\nleq x$, where the
penultimate equivalence follows from Lemma A.4 and the definition of $i$.

(4) We have $1\leq g^{\\#}_{1}(u,z)$ iff $1=g(u,1)\leq z$. But this cannot be
because $z$ is an element of $Z=\mathcal{M}^{\infty}(C^{\delta})$ which does
not contain 1. ∎

As before, we obtain a partial map

$\sim\colon Z\times X\rightharpoonup Y,$

where $\mathrm{dom}(\sim)=\\{(z,x)\mid i(z)\nleq x\\}$, computed as follows:

$z\sim x\enspace\coloneqq\enspace g^{\\#}_{1}(\kappa^{-1}(x),z).$

######  Lemma A.9.

If $\mathrm{dom}(\sim)$ is seen as a subspace of $Z\times X^{\mathrm{op}}$,
the following hold:

  1. (1)

$\mathrm{dom}(\sim)$ is an open up-set of $Z\times X^{\mathrm{op}}$;

  2. (2)

$\sim\colon\mathrm{dom}(\sim)\to Y$ is upper semicontinuous and monotone.

###### Proof.

(1) By the same argument as in Lemma A.6, $\\{(z,x)\mid i(z)\leq x\\}$ is
closed and so its complement $\mathrm{dom}(\sim)$ is open. Further,
$\mathrm{dom}(\sim)$ is an up-set because $i$ is monotone.

(2) The argument is essentially the same as in Lemma A.6. Pick $a\in A$ and
$(z,x)\in\mathrm{dom}(\sim)$, and let $\\{b_{i}\\}_{i}\subseteq A$ be such
that $\kappa^{-1}x=\bigwedge_{i}b_{i}$. Then, $a\leq x\sim z$ iff
$g(\kappa^{-1}x,a)\leq z$ iff $\bigwedge_{i}h(b_{i},a)\leq z$ iff
$h(b_{i},a)\leq z$ for some $i$, where the second equivalence follows from the
fact that $f$ and $g$ agree on closed elements of $A^{\delta}\times
B^{\delta}$. Therefore,

$(\sim)^{-1}(\widehat{a})\enspace=\enspace\mathrm{dom}(\sim)\cap\bigcap_{b\in
B}\left((\widehat{h(b,a)}\times
X)\cup(Z\times(X\setminus\widehat{b}))\right).$

The monotonicity of $\sim$ is a consequence of Lemma A.8 and the monotonicity
of $\kappa$. ∎

Next, we show that $\sim$ is second-order-definable in terms of $-$. The proof
follows the same ideas as the proof of Proposition 3.14 in [6].

######  Lemma A.10.

For all $(z,x)\in\mathrm{dom}(\sim)$,

$z\sim x=\bigvee\\{z-x^{\prime}\mid x^{\prime}\in X\text{ s.t.
}x^{\prime}\nleq x\text{ and }(z,x^{\prime})\in\mathrm{dom}(-)\\}.$

###### Proof.

First, observe that for all $y\in\mathcal{M}^{\infty}(B^{\delta})$,

$\displaystyle\kappa^{-1}(z\sim x)\leq\kappa^{-1}(y)$
$\displaystyle\enspace\Longleftrightarrow\enspace\kappa^{-1}(y)\nleq z\sim x$
$\displaystyle\enspace\Longleftrightarrow\enspace
g(\kappa^{-1}(x),\kappa^{-1}(y))\nleq z$
$\displaystyle\enspace\Longleftrightarrow\enspace
f(\kappa^{-1}(x),\kappa^{-1}(y))\nleq z$
$\displaystyle\enspace\Longleftrightarrow\enspace\kappa^{-1}(z)\leq
f(\kappa^{-1}(x),\kappa^{-1}(y))$
$\displaystyle\enspace\Longleftrightarrow\enspace
f^{\flat}_{1}(\kappa^{-1}(x),\kappa^{-1}(z))\leq\kappa^{-1}(y)$ (12)

where the third equivalence follows because $f$ and $g$ agree on open and
closed elements of the product $A^{\delta}\times B^{\delta}$.

Further, because every element $\kappa^{-1}(x)$ is equal to the meet of
elements of $X=\mathcal{M}^{\infty}(A^{\delta})$ above it, and $f^{\flat}_{1}$
transforms meets into joins in the first argument (item 1 of Lemma A.5),

$\displaystyle f^{\flat}_{1}(\kappa^{-1}(x),\kappa^{-1}(z))$
$\displaystyle=\bigvee\\{f^{\flat}_{1}(x^{\prime},\kappa^{-1}(z))\mid
x^{\prime}\in X\text{ and }\kappa^{-1}(x)\leq x^{\prime}\\}$
$\displaystyle=\bigvee\\{\kappa^{-1}(z-x^{\prime})\mid x^{\prime}\in X,\
x^{\prime}\nleq x\text{ and }(z,x^{\prime})\in\mathrm{dom}(-)\\}$

where the second equality holds because
$f^{\flat}_{1}(x^{\prime},\kappa^{-1}(z))=0$ whenever
$(z,x^{\prime})\notin\mathrm{dom}(-)$. Consequently, by equation (12),

$\kappa^{-1}(z\sim
x)\leq\kappa^{-1}(y)\enspace\Longleftrightarrow\enspace\enspace\kappa^{-1}\enspace(z-x^{\prime})\leq\kappa^{-1}(y)$

for every $x^{\prime}\in X$ such that $x^{\prime}\nleq x$ and
$(z,x^{\prime})\in\mathrm{dom}(-)$. The desired equality then follows because
$\kappa^{-1}$ is an order-isomorphism. ∎

##  Appendix B Extended Priestley duality: the case of ${\mathbf{\Gamma}}$

In this section, we exploit the theory exposed in Appendix A to provide a
detailed justification of the definition of ${\mathbf{\Gamma}}$ and of its
algebraic structure. As explained in Section 3, we wish to obtain
${\mathbf{\Gamma}}$ as a codirected limit of finite chains
${\mathbf{\Gamma}}_{n}=\\{0,\frac{1}{n},\dots,1\\}$. Each such chain, with its
natural order and the discrete topology, is a Priestley space. Its dual
lattice is $(L_{n})^{\mathrm{op}}$, where

$L_{n}\enspace=\enspace(0<\frac{1}{n}<\frac{2}{n}<\dots<1<\top).$

Note that in Section 3, we denoted by $L_{n}$ the order-dual of the lattice
(of the same name) just defined. We make this change here to be able to view
${\mathbf{\Gamma}}_{n}$ as order-embedded into $L_{n}$ (cf. Remark A.1).
Observe that $L_{n}$ is isomorphic to its canonical extension because it is
finite and hence complete. The image of the order-embedding
${\mathbf{\Gamma}}_{n}\hookrightarrow L_{n}$ coincides precisely with the set
$\mathcal{M}^{\infty}(L_{n})$ of (completely) meet-irreducible elements of
$L_{n}$.

Next, we define two binary operators, with the first $\oplus\colon L_{n}\times
L_{n}\to L_{n}$ given as

$\top\oplus u=u\oplus\top=\top\quad\forall u\in
L_{n},{\quad\text{and}\quad}\frac{a}{n}\oplus\frac{b}{n}=\begin{cases}\top&\text{
if }a+b>n\\\ \frac{a+b}{n}&\text{ otherwise}\end{cases}$

and the second $\ominus\colon L_{n}\times L_{n}\to L_{n}$ as

$\displaystyle\top\ominus 0$ $\displaystyle=\top$
$\displaystyle\top\ominus\frac{b}{n}$
$\displaystyle=\frac{n-b+1}{n}\quad\text{ (for $b\geq 1$)}$
$\displaystyle\frac{a}{n}\ominus\frac{b}{n}$
$\displaystyle=\begin{cases}0&\text{ if }b>a\\\ \frac{a-b}{n}&\text{
otherwise}\end{cases}$ $\displaystyle u\ominus\top$
$\displaystyle=0\quad\forall u\in L_{n}.$

In fact, these two operators are adjoint to each other: for all $u,v,w\in
L_{n}$, $u\ominus v\leq w$ if, and only if, $u\leq v\oplus w$. Since $L_{n}$
is isomorphic to its canonical extension,
$\oplus^{\pi}=\oplus^{\sigma}=\oplus$ and
$\ominus^{\pi}=\ominus^{\sigma}=\ominus$.

###  B.1. Dualising $\oplus$ on $L_{n}$.

Observe that $\oplus$ is a three-sorted operator of type $L_{n}\times L_{n}\to
L_{n}$. Because adjoints are uniquely determined, the left adjoint
$f^{\flat}_{1}(u,\mkern 1.5mu\text{-}\mkern 1.5mu)$ to $f(u,\mkern
1.5mu\text{-}\mkern 1.5mu)=u\oplus(\mkern 1.5mu\text{-}\mkern 1.5mu)$ is given
by $f^{\flat}_{1}(u,v)=v\ominus u$. Then, by Lemma A.5, the corresponding
partial operation

$\displaystyle
z-x=\kappa(f^{\flat}_{1}(x,\kappa^{-1}z)),{\quad\text{for}\quad}x\leq z,$

is well defined. The order-isomorphism
$\kappa\colon\mathcal{J}^{\infty}(L_{n})\to\mathcal{M}^{\infty}(L_{n})$ is
given by $\top\mapsto 1$ and $\frac{a}{n}\mapsto\frac{a-1}{n}$, for $a>0$.
Consequently, it is not difficult to see that
$-\colon\mathrm{dom}(-)\to{\mathbf{\Gamma}}_{n}$ is precisely the partial
subtraction on ${\mathbf{\Gamma}}_{n}$.

###  B.2. Dualising $\ominus$ on $L_{n}$.

The assignment $f(u,v)=v\ominus u$ is a three-sorted operator of type
$L_{n}\times L_{n}^{\mathrm{op}}\to L_{n}^{\mathrm{op}}$. Just observe that
$u\ominus\top=0=0\ominus u$ and $\top\ominus 0=\top$. Using Lemma A.5, we see
that the right adjoint $f^{\\#}_{1}(u,\mkern 1.5mu\text{-}\mkern 1.5mu)$ to
$f(u,\mkern 1.5mu\text{-}\mkern 1.5mu)$ restricts to
$\mathcal{M}^{\infty}(L_{n})^{\mathrm{op}}\times\mathcal{J}^{\infty}(L_{n}^{\mathrm{op}})\to\mathcal{J}^{\infty}_{0}(L_{n}^{\mathrm{op}})$,
which is simply a monotone map
$\mathcal{M}^{\infty}(L_{n})\times\mathcal{M}^{\infty}(L_{n})\to L_{n}$.
Because right adjoints are unique, $f^{\\#}_{1}(u,v)=u\oplus v$. Hence, the
dual of $\ominus$ on ${\mathbf{\Gamma}}_{n}$ is the partial plus $x+z$,
defined whenever $x\leq 1-z$.

In this specific case, the formula in Lemma A.10 to recover $\ominus$ from the
partial plus looks as follows:

$\displaystyle f(u,v)=\bigvee\\{j\in\mathcal{J}^{\infty}(L_{n})$
$\displaystyle\mid\exists x\in\mathcal{M}^{\infty}(L_{n})\text{ s.t. }u\leq
x,\ (x,\kappa(j))\in\mathrm{dom}(+),$ $\displaystyle\quad\text{ and
}\kappa^{-1}(x+\kappa(j))\leq v\\}.$

###  B.3. Duals to floor and ceiling functions.

Let
$\varepsilon\colon{\mathbf{\Gamma}}_{n}\hookrightarrow{\mathbf{\Gamma}}_{nm}$
be the obvious inclusion. Floor and ceiling functions are right and left
adjoint, respectively, to $\varepsilon$. Just observe that, for all
$\frac{a}{n}\in{\mathbf{\Gamma}}_{n}$ and
$\frac{b}{nm}\in{\mathbf{\Gamma}}_{nm}$, we have

$\varepsilon\left(\tfrac{a}{n}\right)\leq\tfrac{b}{nm}\enspace\iff\enspace\tfrac{a}{n}\leq\lfloor\tfrac{b}{nm}\rfloor{\quad\text{and}\quad}\lceil\tfrac{b}{nm}\rceil\leq\tfrac{a}{n}\enspace\iff\enspace\tfrac{b}{nm}\leq\varepsilon\left(\tfrac{a}{n}\right).$

Therefore, the floor function uniquely extends to the right adjoint of the
lattice embedding $L_{n}\hookrightarrow L_{nm}$. On the other hand, the
ceiling function naturally extends to the left adjoint of the lattice
embedding $L^{\bot}_{n}\hookrightarrow L^{\bot}_{nm}$, with

$L^{\bot}_{n}=(\bot<0<\frac{1}{n}<\dots<\frac{n-1}{n}<1)$

and $L^{\bot}_{nm}$ defined similarly. Therefore, as a matter of convenience,
we prefer to view ceiling functions as functions of the type
$\mathcal{J}^{\infty}(L^{\bot}_{nm})\to\mathcal{J}^{\infty}(L^{\bot}_{n})$
(cf. Section A.2). To summarise, the duals of floor functions are the lattice
embeddings

$i_{n,nm}\colon L_{n}\hookrightarrow L_{nm},$

and the duals of ceiling functions are the lattice embeddings

$i^{\prime}_{n,nm}\colon L_{n}\xrightarrow{\leavevmode\nobreak\
t\leavevmode\nobreak\ }L^{\bot}_{n}\hookrightarrow
L^{\bot}_{nm}\xrightarrow{\leavevmode\nobreak\ t^{-1}\leavevmode\nobreak\
}L_{nm},$

where $t\colon L_{n}\to L^{\bot}_{n}$ is the renaming $\top\mapsto 1$,
$0\mapsto\bot$ and $\frac{a}{n}\mapsto\frac{a-1}{n}$. The duals of floor and
ceiling functions can also be computed in terms of plain Priestley duality,
cf. Section 2.1.

Observe that $i_{n,m}\colon L_{n}\hookrightarrow L_{nm}$ preserves $\oplus$
but not $\ominus$, because

$i_{n,nm}(\top\ominus\tfrac{b}{n})=i_{n,nm}(\tfrac{n-b+1}{n})=\tfrac{n-b+1}{n}$

whereas

$i_{n,nm}(\top)\ominus i_{n,nm}(\tfrac{b}{n})=\tfrac{nm-b+1}{nm},$

for $1\leq b\leq n$. A similar argument shows that $i^{\prime}_{n,nm}$ does
not preserve $\oplus$ nor $\ominus$.

The previous discussion justifies why we take the floor functions
$\lfloor\mkern 1.5mu\text{-}\mkern
1.5mu\rfloor\colon{\mathbf{\Gamma}}_{nm}\twoheadrightarrow{\mathbf{\Gamma}}_{n}$
as the comparison maps in the codirected diagram that defines
${\mathbf{\Gamma}}$ and, moreover, why we equip the finite spaces
${\mathbf{\Gamma}}_{n}$ with partial minus, rather than partial plus.

###  B.4. The partial operations on ${\mathbf{\Gamma}}$.

Saying that ${\mathbf{\Gamma}}$ is the codirected limit of the system of floor
functions $\lfloor\mkern 1.5mu\text{-}\mkern
1.5mu\rfloor\colon{\mathbf{\Gamma}}_{nm}\to{\mathbf{\Gamma}}_{n}$ is
equivalent to saying that ${\mathbf{\Gamma}}$ is the Priestley space dual to
the directed colimit of the lattice embeddings $L_{n}\hookrightarrow L_{nm}$.
It is readily seen that this colimit can be identified with the lattice
$\mathbf{L}=(\mathbb{Q}\cap[0,1])<\\{\top\\}$. Its canonical extension has the
following properties.

######  Lemma B.1.

The following statements hold:

  1. (1)

$\mathbf{L}^{\delta}$ can be identified with the lattice having underlying set

$\\{0^{\circ},0^{+}\\}\cup\\{r^{-},r^{+}\mid
r\in(0,1)\setminus\mathbb{Q}\\}\cup\\{q^{-},q^{\circ},q^{+}\mid
q\in(0,1)\cap\mathbb{Q}\\}\cup\\{1^{-},1^{\circ},\top\\}$

and the linear order defined by the following conditions:

(a) $r^{-}\prec r^{+}$ for all $r\in(0,1)\setminus\mathbb{Q}$ |  (b) $q^{-}\prec q^{\circ}\prec q^{+}$ for all $q\in(0,1)\cap\mathbb{Q}$
---|---
(c) $0^{\circ}\prec 0^{+}$ and $1^{-}\prec 1^{\circ}\prec\top$ |  (d) $x^{+}<y^{-}$ if $x<y$ in $[0,1]$

where $\prec$ is the covering relation, i.e. $\alpha\prec\beta$ if
$\alpha<\beta$ and, whenever $\alpha\leq\gamma\leq\beta$, either
$\alpha=\gamma$ or $\beta=\gamma$.

  2. (2)

$\mathcal{M}^{\infty}(\mathbf{L}^{\delta})=\mathbf{L}^{\delta}\setminus(\\{\top\\}\cup\\{r^{+}\mid
r\in[0,1)\\})$ and
$\mathcal{J}^{\infty}(\mathbf{L}^{\delta})=\mathbf{L}^{\delta}\setminus(\\{0^{\circ}\\}\cup\\{r^{-}\mid
r\in(0,1]\\})$

  3. (3)

The order-isomorphism
$\kappa\colon\mathcal{J}^{\infty}(\mathbf{L}^{\delta})\to\mathcal{M}^{\infty}(\mathbf{L}^{\delta})$
is given by

$\top\mapsto 1^{\circ}\qquad\qquad x^{+}\mapsto\begin{cases}x^{\circ}&\text{
if }x\in\mathbb{Q}\\\ x^{-}&\text{ otherwise}\end{cases}\qquad\qquad
q^{\circ}\mapsto q^{-}.$

###### Proof.

(1) It is easy to see that $\mathbf{L}^{\delta}$ is a complete lattice and
that the embedding $\mathbf{L}\hookrightarrow\mathbf{L}^{\delta}$, defined as
$a\mapsto a^{\circ}$ and $\top\mapsto\top$, satisfies the density and
compactness conditions. Items (2) and (3) are easily verified and left to the
reader. ∎

Because the lattice embeddings $L_{n}\hookrightarrow L_{nm}$ preserve the
operators $\oplus\colon L_{n}\times L_{n}\to L_{n}$, also $\mathbf{L}$ comes
equipped with an operator
$\oplus\colon\mathbf{L}\times\mathbf{L}\to\mathbf{L}$ obtained as the colimit
of the corresponding operators on the finite sublattices $L_{n}$. This can be
expressed as follows:

$\top\oplus u=u\oplus\top=\top,\quad\forall
u\in\mathbf{L},\quad{\quad\text{and}\quad}\quad\frac{a}{n}\oplus\frac{b}{m}=\begin{cases}\top&\text{
if }am+bn>nm\\\ \frac{am+bn}{nm}&\text{ otherwise.}\end{cases}$

This time, though, $\oplus$ does not have a left adjoint because
$\top\ominus\frac{a}{n}$ cannot be defined on $\mathbf{L}$. This mirrors the
fact that the inclusion maps $L_{n}\hookrightarrow L_{mn}$ do not preserve
$\ominus$ defined on finite lattices. Following the recipe of Section A.4, we
compute the partial operation $-\colon\mathrm{dom}(-)\to{\mathbf{\Gamma}}$
from the left adjoint $a\mathbin{\oplus^{\pi\mkern 2.0mu\flat}}(\mkern
1.5mu\text{-}\mkern 1.5mu)$ of $a\oplus^{\pi}(\mkern 1.5mu\text{-}\mkern
1.5mu)$. The pi extension $\oplus^{\pi}$ is commutative because so is
$\oplus$. Therefore, the following lemma provides us with a full description
of $\oplus^{\pi}\colon\mathbf{L}^{\delta}\to\mathbf{L}^{\delta}$.

######  Lemma B.2.

Suppose $x,y\in[0,1]$. Then

$x+y\leq 1\enspace\Rightarrow\begin{split}&\leavevmode\nobreak\
x^{\circ}\oplus^{\pi}y^{\circ}\enspace=\enspace(x+y)^{\circ}\\\
&\begin{rcases*}x^{\circ}\oplus^{\pi}y^{-}\\\
x^{-}\oplus^{\pi}y^{-}\end{rcases*}\enspace=\enspace(x+y)^{-}\end{split}\quad\text{and}\quad
x+y<1\ \Rightarrow\begin{split}&\begin{rcases*}x^{+}\oplus^{\pi}y^{+}\\\
x^{+}\oplus^{\pi}y^{\circ}\\\
x^{+}\oplus^{\pi}y^{-}\end{rcases*}\enspace=\enspace(x+y)^{+},\end{split}$

whenever the expressions make sense. In all the remaining cases, we get
$\top$.

###### Proof.

Suppose $x+y\leq 1$. It is not difficult to see that the set of open elements
of $\mathbf{L}^{\delta}$ is
$\mathcal{I}(\mathbf{L}^{\delta})=\mathcal{M}^{\infty}_{1}(\mathbf{L}^{\delta})=\mathcal{M}^{\infty}(\mathbf{L}^{\delta})\cup\\{\top\\}$.
Just observe that every proper ideal in a linearly ordered lattice is prime.
Hence, by item 2 in Lemma B.1,

$\mathcal{I}(\mathbf{L}^{\delta})=\\{0^{\circ}\\}\cup\\{r^{-}\mid
r\in(0,1)\setminus\mathbb{Q}\\}\cup\\{q^{-},q^{\circ}\mid
q\in(0,1)\cap\mathbb{Q}\\}\cup\\{1^{-},1^{\circ},\top\\}.$

By the definition of pi extensions on open elements (cf. Section A.2), we get

$x^{\circ}\oplus^{\pi}y^{\circ}=\bigvee{\\{a\oplus b\mid a,b\in\mathbf{L},\
a\leq x^{\circ}\text{ and }b\leq y^{\circ}\\}}=(x+y)^{\circ}.$

The same equation also tells us the results of $x^{\circ}\oplus^{\pi}y^{-}$
and $x^{-}\oplus^{\pi}y^{-}$, since all the involved elements are open:

$\displaystyle x^{\circ}\oplus^{\pi}y^{-}$ $\displaystyle=\bigvee{\\{a\oplus
b\mid a,b\in\mathbf{L},\ a\leq x^{\circ}\text{ and }b\leq y^{-}\\}}$
$\displaystyle=\bigvee{\\{x\oplus b\mid b\in\mathbb{Q}\cap[0,1],\
b<y\\}}=(x+y)^{-},$

and

$x^{-}\oplus^{\pi}y^{-}=\bigvee{\\{a\oplus b\mid a,b\in\mathbb{Q}\cap[0,1],\
a<x\text{ and }b<y\\}}=(x+y)^{-}.$

Suppose now that $x+y<1$. The definition of pi extensions on possibly non-open
elements, along with the cases covered in the first part of the proof, yield

$\displaystyle x^{+}\oplus^{\pi}y^{+}$
$\displaystyle=\bigwedge{\\{w\oplus^{\pi}z\mid
w,z\in\mathcal{I}(\mathbf{L}^{\delta}),\ x^{+}\leq w\text{ and }y^{+}\leq
z\\}}$ $\displaystyle=\bigwedge{\\{w^{-}\oplus^{\pi}z^{-}\mid w,z\in(0,1],\
x<w\text{ and }y<z\\}}$ $\displaystyle=\bigwedge{\\{(w+z)^{-}\mid
w,z\in(0,1],\ x<w\text{ and }y<z\\}}=(x+y)^{+}.$

All the remaining cases are computed similarly. ∎

Since the assignment $a\mapsto a\oplus 0$ is equal to the identity, its dual
$i\colon{\mathbf{\Gamma}}\to{\mathbf{\Gamma}}$ is also the identity.
Therefore, $\mathrm{dom}(-)$ is equal to $\\{(z,x)\mid x\leq z\\}$. Moreover,
it follows from Lemma B.1, along with Lemma B.3 below, that the partial
operation $z-x=\kappa(x\mathbin{\oplus^{\pi\mkern 2.0mu\flat}}\kappa^{-1}z)$
is exactly the one described in Section 3.2.

######  Lemma B.3.

Let $u,v\in\mathbf{L}^{\delta}$. If $u\leq v$, then
$u\mathbin{\oplus^{\pi\mkern 2.0mu\flat}}v=0^{\circ}$. Otherwise, for
$x,y\in[0,1]$,

$\begin{array}[]{|c|c|c|c|}\hline\cr\mathbin{\oplus^{\pi\mkern
2.0mu\flat}}&y^{\circ}&y^{-}&y^{+}\\\ \hline\cr
x^{\circ}&(x-y)^{\circ}&(x-y)^{+}&(x-y)^{-}\\\ \hline\cr
x^{-}&(x-y)^{-}&(x-y)^{-}&(x-y)^{-}\\\ \hline\cr
x^{+}&(x-y)^{+}&(x-y)^{+}&(x-y)^{-}\\\ \hline\cr\end{array}$

whenever the expressions make sense (read as:
$(\text{row})\mathbin{\oplus^{\pi\mkern 2.0mu\flat}}(\text{colon})$). Finally,

$\top\mathbin{\oplus^{\pi\mkern
2.0mu\flat}}x^{\circ}=\top\mathbin{\oplus^{\pi\mkern
2.0mu\flat}}x^{-}=(1-x)^{+}\ \text{ and }\ \top\mathbin{\oplus^{\pi\mkern
2.0mu\flat}}x^{+}=(1-x)^{-}.$

###### Proof.

Follows by a routine computation using the usual description of left adjoints
in terms of infima, i.e. $u\mathbin{\oplus^{\pi\mkern
2.0mu\flat}}v=\bigwedge{\\{w\in\mathbf{L}^{\delta}\mid u\leq
v\oplus^{\pi}w\\}}$ for every $u,v\in\mathbf{L}^{\delta}$. ∎

Lastly, we compute the expression for the second partial operation
$\sim\colon\mathrm{dom}(\sim)\to{\mathbf{\Gamma}}$. However, instead of
computing the sigma extension of $\oplus$ and its (partial) right adjoint, as
in Section A.5, we use the formula from Lemma A.10. Because
$i\colon{\mathbf{\Gamma}}\to{\mathbf{\Gamma}}$ is the identity mapping,
$\mbox{$\mathrm{dom}$($\sim$)}=\\{(z,x)\mid z\nleq x\\}=\\{(z,x)\mid x<z\\}$.
Furthermore, by Lemma A.10, for every $(z,x)\in\mathrm{dom}(\sim)$ we have

$z\sim x=\bigvee\\{z-x^{\prime}\mid x^{\prime}\in{\mathbf{\Gamma}}\text{ s.t.
}x<x^{\prime}\leq z\\}.$

Now, since $x<z$, there are two possible cases. Either $x=q^{-}<q^{\circ}=z$
for some rational $q\in(0,1]$, in which case the unique $x^{\prime}$ such that
$x<x^{\prime}\leq z$ is $q^{\circ}$, or $x$ is not an immediate predecessor of
$z$. In the latter case, by a similar argument, for every $x<x^{\prime}\leq z$
there is a rational $q\in(0,1]$ such that $x<q^{\circ}\leq x^{\prime}$. Thus,
because $-$ is antitone in the second coordinate, we obtain exactly the same
expression as used in Section 3.2 to define $\sim$:

$z\sim x=\bigvee\\{z-x^{\prime}\mid q\in(0,1]\cap\mathbb{Q}\text{ s.t.
}x<q^{\circ}\leq z\\}.$

The only difference is that $z\sim x$, as described in Section 3.2, is also
defined for $z=x$.

To conclude, we offer a proof of Lemma 5.1, which relies on the theory exposed
above.

###### Proof of Lemma 5.1.

Let $f$ be the pi extension $\oplus^{\pi}$ of $\oplus$. Then,

$f(x,y)\leq z\iff\kappa^{-1}z\nleq f(x,y)\iff
f^{\flat}_{1}(x,\kappa^{-1}z)\nleq y\iff y\leq z-x$

for every $y\in{\mathbf{\Gamma}}$ and $(z,x)\in\mathrm{dom}(-)$. By Lemma B.2,
the restriction of $f$ to

$(x^{\circ},y^{\circ}),(x^{\circ},y^{-}),(x^{-},y^{\circ}),\text{ and
}(x^{-},y^{-})$

where $x+y\leq 1$ (whenever the expressions make sense), is a well defined
partial operation on ${\mathbf{\Gamma}}$. Observe that this domain of
definition is precisely $\\{(x,y)\mid x\leq 1^{\circ}-y\\}$ because $x\leq
1^{\circ}-y$ iff $f(x,y)\leq 1^{\circ}$ iff $f(x,y)\neq\top$. ∎

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