A quasitriangular structure gives and is recovered from a module braiding

Write $R={\sum }_i{a}_i\otimes b_i$. The left action of $H\otimes H$ on $V\otimes W$ sends $(a\otimes b)(v\otimes w)$ to $av\otimes bw$; therefore the finite tensor $R$ defines the linear map in the Statement, independently of its chosen sum representation. Its inverse is $({\rho }_V\otimes {\rho }_W)(R^{-1})\circ {\tau }_{W,V}$, so $c_{V,W}$ is an isomorphism.

Now suppose a braiding is given on all left $H$-modules. For $v\in V$, the map $f_v:H\to V$ from the regular module of [F5], $x\mapsto xv$, is $H$-linear because module associativity gives $f_v(hx)=(hx)v=h(xv)$, and module unitality in [F3] gives $f_v(1_H)=v$. Set $R={\tau }_{H,H}{c}_{H,H}(1_H\otimes 1_H)$. Naturality for $f_v$ and $f_w$ then gives $c_{V,W}(v\otimes w)={\sum }_i{b}_{i}w\otimes a_{i}v$ when $R={\sum }_i{a}_i\otimes b_i$.

For $h\in H$, the diagonal tensor action and the quasitriangular relation give $c_{V,W}{\rho }_{V\otimes W}(h)={\tau }_{V,W}({\rho }_V\otimes {\rho }_W)(R\mathrm{\Delta }(h))={\tau }_{V,W}({\rho }_V\otimes {\rho }_W)({\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}(h)R)={\rho }_{W\otimes V}(h)c_{V,W},$ where the last equality is the elementary flip identity $\tau {\rho }_{V,W}({\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}(h))={\rho }_{W,V}(h)\tau$. Hence $c_{V,W}$ is $H$-linear.

If $f:V\to V^{\mathrm{\prime }}$ and $g:W\to W^{\mathrm{\prime }}$ are $H$-linear, then $(g\otimes f)c_{V,W}$ and $c_{V^{\mathrm{\prime }},W^{\mathrm{\prime }}}(f\otimes g)$ both send $v\otimes w$ to ${\sum }_i{b}_{i}g(w)\otimes a_{i}f(v)$, using the $H$-linearity of $f,g$. Thus the family is natural.

On $u\otimes v\otimes w$, the map $c_{U,V\otimes W}$ gives ${\sum }_i(b_i{)}_{(1)}v\otimes (b_i{)}_{(2)}w\otimes a_{i}u$. The second hexagon identity rewrites this coefficient tensor as $R_{13}{R}_{12}={\sum }_{j,k}{a}_j{a}_k\otimes b_k\otimes b_j$, so the result is ${\sum }_{j,k}{b}_{k}v\otimes b_{j}w\otimes a_j{a}_{k}u$, exactly $({\mathrm{id}}_V\otimes c_{U,W})(c_{U,V}\otimes {\mathrm{id}}_W)(u\otimes v\otimes w)$. Equality on spanning pure tensors proves the first hexagon.

On $u\otimes v\otimes w$, the map $c_{U\otimes V,W}$ gives ${\sum }_i{b}_{i}w\otimes (a_i{)}_{(1)}u\otimes (a_i{)}_{(2)}v$. The first hexagon identity rewrites the coefficient tensor as $R_{13}{R}_{23}={\sum }_{j,k}{a}_j\otimes a_k\otimes b_j{b}_k$, so the result is ${\sum }_{j,k}{b}_j{b}_{k}w\otimes a_{j}u\otimes a_{k}v$, exactly $(c_{U,W}\otimes {\mathrm{id}}_V)({\mathrm{id}}_U\otimes c_{V,W})(u\otimes v\otimes w)$. Equality on pure tensors proves the second hexagon.

For $V^{\otimes 3}$, write $c_1={\tau }_{12}{R}_{12}$ and $c_2={\tau }_{23}{R}_{23}$ for their induced operators. Moving the flips past the leg operators by conjugating the affected tensor factors gives $c_1{c}_2{c}_1={\tau }_{12}{\tau }_{23}{\tau }_{12}{R}_{23}{R}_{13}{R}_{12}$ and $c_2{c}_1{c}_2={\tau }_{23}{\tau }_{12}{\tau }_{23}{R}_{12}{R}_{13}{R}_{23}$. The two flip products agree on every triple pure tensor, and [F6] makes the two R-products equal, so ${\check{R}}_V$ satisfies the Yang–Baxter braid relation.

In the composite $c_{W,V}{c}_{V,W}$, conjugating the action of $R$ across the flip changes $R$ to $R_{21}$, so the composite acts on $V\otimes W$ by $R_{21}R$. If $R_{21}R=1\otimes 1$, the composite is the identity, proving symmetry.

Applying the formula of step 1.2 to $V=W=H$ yields $c_{H,H}={\tau }_{H,H}\circ L_R$, where $L_R$ is left multiplication by $R$ on the componentwise algebra $H\otimes H$. Since $c_{H,H}$ and $\tau$ are bijective, $L_R$ is bijective. Surjectivity supplies $X$ with $RX=1$; then $R(XR-1)=0$, and injectivity of $L_R$ gives $XR=1$. Thus $R$ is invertible.

The $H$-linearity equation for $c_{H,H}=\tau L_R$ is $\tau L_R{L}_{\mathrm{\Delta }(h)}=L_{\mathrm{\Delta }(h)}\tau L_R$. Conjugating the right-hand action by the vector-space flip gives $L_R{L}_{\mathrm{\Delta }(h)}=L_{{\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}(h)}{L}_R$, hence $L_{R\mathrm{\Delta }(h)}=L_{{\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}(h)R}$. The conjugation identity holds on elementary tensors by the definition of the flip and extends linearly. [F7] evaluates the equal left-multiplication maps at $1_H\otimes 1_H$, yielding $R\mathrm{\Delta }(h)={\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}(h)R$.

Apply the two hexagons of the given braiding to the regular modules $H,H,H$. By the formula in step 1.2, the first hexagon's two sides are the same output flip composed with left multiplication on $H^{\otimes 3}$ by ${\alpha }^{-1}(\mathrm{id}\otimes \mathrm{\Delta })R$ and by $R_{13}{R}_{12}$, respectively. Cancel the common flip. Equality of the resulting maps and [F7] give the second quasitriangular hexagon. The other hexagon gives, in the same way, left multiplication by $(\mathrm{\Delta }\otimes \mathrm{id})R$ and by $R_{13}{R}_{23}$; [F7] gives the first hexagon.

The formula in step 1.2 uniquely determines every $c_{V,W}$ from $R$. The preceding converse steps prove that the recovered $R$ is invertible and satisfies every quasitriangular axiom. If the braiding is symmetric, then $c_{H,H}{c}_{H,H}=\mathrm{id}$. Using $c_{H,H}=\tau L_R$ from step 2.7, the composite is left multiplication by $R_{21}R$; [F7] therefore gives $R_{21}R=1\otimes 1$. Together with step 2.6 this proves the triangular/symmetric equivalence.