Weak-bialgebra module tensor products are relative tensor products

Put $E=\mathrm{\Delta }(1)$. Multiplicativity gives $E^2=E$, $\mathrm{\Delta }(h)E=\mathrm{\Delta }(h)$, and $E\mathrm{\Delta }(h)=\mathrm{\Delta }(h)$. Thus the diagonal action on $V\otimes W$ restricts to a unital action on $\mathrm{im}P$: $1_H$ acts there as $P$, and the action law follows from $\mathrm{\Delta }(hg)=\mathrm{\Delta }(h)\mathrm{\Delta }(g)$.

The restriction ${\epsilon }_s^{\mathrm{\prime }}{\mathrm{\mid }}_{H_t}$ is an anti-isomorphism, so $v◃x={\epsilon }_s^{\mathrm{\prime }}(x)v$ is a right $H_t$-module action. For the separability idempotent $e=\sum e^1\otimes e^2$ of [F2], define $P_e(v\otimes w)=\sum (v◃e^1)\otimes e^{2}w$. The identity $\mathrm{\Delta }(1)=\sum {\epsilon }_s^{\mathrm{\prime }}(e^1)\otimes e^2$ gives $P_e=P$. The relation $\sum x{e}^1\otimes e^2=\sum e^1\otimes e^{2}x$ implies $P_e((v◃x)\otimes w)=P_e(v\otimes xw)$, so $P$ kills every balancing relation.

In the balanced quotient, $q{P}_e(v\otimes w)=\sum [v◃e^1\otimes e^{2}w]=\sum [v\otimes e^1{e}^{2}w]=[v\otimes w]$, because $\sum e^1{e}^2=1$. Since $P=P_e$ is idempotent, $q{\mathrm{\mid }}_{\mathrm{im}P}$ is inverse to the map $[v\otimes w]\mapsto P(v\otimes w)$; hence it is an isomorphism. Transporting the image action through this isomorphism gives $h\cdot [v\otimes w]=\sum [h_{(1)}v\otimes h_{(2)}w]$, since $P\mathrm{\Delta }(h)=\mathrm{\Delta }(h)P=\mathrm{\Delta }(h)$. Given $H$-linear maps $f:V\to V^{\mathrm{\prime }}$ and $g:W\to W^{\mathrm{\prime }}$, the bilinear pairing $(v,w)\mapsto [f(v)\otimes g(w)]$ is balanced: for $x\in H_t$, [F8] gives $f(v◃x)=f({\epsilon }_s^{\mathrm{\prime }}(x)v)={\epsilon }_s^{\mathrm{\prime }}(x)f(v)=f(v)◃x$ and $g(xw)=xg(w)$. Thus [F5] induces $T_{f,g}:V{\otimes }_{H_t}W\to V^{\mathrm{\prime }}{\otimes }_{H_t}{W}^{\mathrm{\prime }}$, $T_{f,g}([v\otimes w])=[f(v)\otimes g(w)]$. It is $H$-linear because $T_{f,g}(h\cdot [v\otimes w])=\sum [f(h_{(1)}v)\otimes g(h_{(2)}w)]=\sum [h_{(1)}f(v)\otimes h_{(2)}g(w)]=h\cdot T_{f,g}([v\otimes w])$. These maps preserve identities and composition on elementary classes, which span the quotient, so they define a bifunctor. The quotient-to-image isomorphism is natural because $(f\otimes g)(P(v\otimes w))=\sum 1_{(1)}f(v)\otimes 1_{(2)}g(w)=P(f(v)\otimes g(w))$.

The weak-unit identities give $(\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})E=(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })E=(E\otimes 1)(1\otimes E)=(1\otimes E)(E\otimes 1)$. Thus the two projections used for either parenthesization of three factors commute and have common image; the canonical rebracketing restricts to an isomorphism of these images, and is $H$-linear because $\mathrm{\Delta }$ is coassociative. For four factors, repeated use of this identity shows that each of the five parenthesized tensor products is the image of the same iterated coproduct $D_4=(\mathrm{\Delta }\otimes \mathrm{\Delta })\mathrm{\Delta }(1)$ on $U\otimes V\otimes W\otimes X$; coassociativity makes every iterated coproduct equal to $D_4$. Each associator is the restriction of canonical vector-space rebracketing, so the two pentagon composites send every elementary quadruple tensor to the same rebracketed tensor. Elementary tensors span, hence the pentagon commutes.

The target base $H_t$ is a unital left $H$-module under $h\cdot x={\epsilon }_t(hx)$: ${\epsilon }_t(x)=x$ for $x\in H_t$, and ${\epsilon }_t(h{\epsilon }_t(gx))={\epsilon }_t(hgx)$ by [F2]. Its right $H_t$-action induced by ${\epsilon }_s^{\mathrm{\prime }}$ is ordinary multiplication, because ${\epsilon }_t({\epsilon }_s^{\mathrm{\prime }}(x)z)=zx$ for $x,z\in H_t$. The canonical relative-product unit maps are $H_t{\otimes }_{H_t}V\to V$, $[x\otimes v]\mapsto xv$, and $V{\otimes }_{H_t}{H}_t\to V$, $[v\otimes x]\mapsto v◃x$. They are balanced and inverse to $v\mapsto [1\otimes v]$ and $v\mapsto [v\otimes 1]$, respectively. Write the two maps as $\lambda$ and $\rho$. The first identity in [F2] gives $\lambda (h\cdot [x\otimes v])=\sum {\epsilon }_t(h_{(1)}x)h_{(2)}v=(hx)v=h\lambda ([x\otimes v])$. For the right unit map the order is $\rho (h\cdot [v\otimes x])=\sum {\epsilon }_s^{\mathrm{\prime }}({\epsilon }_t(h_{(2)}x))h_{(1)}v=h{\epsilon }_s^{\mathrm{\prime }}(x)v=h\rho ([v\otimes x])$, using the second identity in [F2]. Thus both unit maps are $H$-linear on elementary balanced classes and hence on their full domains.

The triangle maps agree on every elementary balanced class: both send $[(v\otimes x)\otimes w]$ to $v◃x\otimes w=[v\otimes xw]$. For $H$-linear maps $f,g,j$, both naturality composites for the associator send each elementary triple class to the class with $f,g,j$ applied to its three entries, with only the indicated rebracketing changed. The unit naturality squares commute because $f(xv)=xf(v)$ and $f(v◃x)=f(v)◃x$ by [F8]. These classes span, so the maps are natural. Thus the tensor product, unit, associator, and unit maps satisfy the requirements of [F6], proving the monoidal assertion.