Double crossing laws and associativity

Pairing the first tensor factor of $\chi (h\otimes f)$ with $x\in H$ gives $\sum f(S(h_1)x{h}_3)h_2$, because ${\mathrm{\Delta }}_{H^{\ast }}^{(2)}f$ evaluates on the three consecutive factors of $S(h_1)x{h}_3$. This determines $\chi$ uniquely by finite tensor separation.

The unit identities are $\chi (1\otimes f)=f\otimes 1$ and $\chi (h\otimes {\epsilon }_H)={\epsilon }_H\otimes h$: the first contracts the first and third dual factors by ${\epsilon }_H$, and the second uses ${\epsilon }_H\circ S={\epsilon }_H$ from [F3] and the two counit identities in [F1].

For $h,k\in H$, the two sides of $\chi (hk\otimes f)=(\mathrm{id}\otimes m_H)(\chi \otimes \mathrm{id})(\mathrm{id}\otimes \chi )(h\otimes k\otimes f)$, paired with $x$, both equal $\sum f(S(k_1)S(h_1)x{h}_3{k}_3)h_2{k}_2$. On the left use $S(h_1{k}_1)=S(k_1)S(h_1)$ from [F3]; on the right, the successive crossings evaluate $f_1(S(k_1))f_2(S(h_1))f_3(x)f_4(h_3)f_5(k_3)$, which is the same dual coproduct evaluation.

For $f,g\in H^{\ast }$, pairing the first tensor factor of the left side of $\chi (h\otimes (f\odot g))=(m_{H^{\ast \mathrm{o}\mathrm{p}}}\otimes \mathrm{id})(\mathrm{id}\otimes \chi )(\chi \otimes \mathrm{id})(h\otimes f\otimes g)$ with $x$ gives the $H$-valued expression $\sum (g\ast f)(S(h_1)x{h}_3)h_2$; the $h_2$ factor remains from $\chi$. Since $\mathrm{\Delta }(S(h_1)x{h}_3)=\sum S(h_2)x_1{h}_4\otimes S(h_1)x_2{h}_5$ by [F3] and multiplicativity of $\mathrm{\Delta }$, this is $\sum g(S(h_2)x_1{h}_4)f(S(h_1)x_2{h}_5)h_3$, which is also the result of the two successive crossings on the right.

Write $\chi (h,b)=\sum b_1\otimes h_1$, $\chi (k,c)=\sum c_1\otimes k_1$, and, for each such term, $\chi (h_1,c_1)=\sum c_2\otimes h_2$. Starting with $(a\otimes h),(b\otimes k),(c\otimes l)$, the left association first crosses $h$ past $b$, then crosses the product $h_{1}k$ past $c$; the $H$ distributive law of step 1.3 expands that second crossing as first crossing $k$ past $c$, then $h_1$ past the resulting $c_1$. It therefore gives $\sum a{b}_1{c}_2\otimes h_2{k}_{1}l$. The right association first crosses $k$ past $c$, then crosses $h$ past the product $b{c}_1$; the $H^{\ast \mathrm{o}\mathrm{p}}$ distributive law of step 1.4 expands that crossing as first crossing $h$ past $b$, then $h_1$ past $c_1$, giving the same sum. Associativity of both factor products follows for $H$ from [F1] and for $H^{\ast \mathrm{o}\mathrm{p}}$ from [F4]. Thus the associations agree on simple tensors and, by multilinearity, everywhere. The unit identities of step 1.2 make ${\epsilon }_H\otimes 1_H$ a two-sided unit.

The maps $f\mapsto f\otimes 1_H$ and $h\mapsto {\epsilon }_H\otimes h$ preserve products by step 1.2 and the defining multiplication. They are injective because applying ${\mathrm{id}}_{H^{\ast }}\otimes {\epsilon }_H$ or ${\epsilon }_{H^{\ast }}\otimes {\mathrm{id}}_H$ recovers the input; ${\epsilon }_H(1_H)=1$.