Quasi-Hopf, Weak Hopf, and Relative Tensor Products

For left $H$-modules $V,W$, the formula in the Statement is a unital left action: its unit acts as $\mathrm{\Delta }(1_H)=1_H\otimes 1_H$, and on every elementary tensor the two sides of the module associativity law are respectively $\mathrm{\Delta }(h)\mathrm{\Delta }(g)$ and $\mathrm{\Delta }(hg)$ acting on that tensor. They agree because $\mathrm{\Delta }$ is an algebra map. The tensor maps induced by $H$-linear maps are $H$-linear by the same formula on elementary tensors, and preserve identities and compositions.

Using the canonical vector-space identifications of [F5] and [F8], the map $a_{U,V,W}$ is invertible because $\mathrm{\Phi }$ is invertible; its inverse is the reverse rebracketing followed by ${\mathrm{\Phi }}^{-1}$. It is $H$-linear since, on the underlying triple tensor, $\mathrm{\Phi }(\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(h)=(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })\mathrm{\Delta }(h)\mathrm{\Phi },$ which is the quasi-coassociativity identity in [F1] multiplied on the right by $\mathrm{\Phi }$.

The associator is natural: for module maps $f,g,j$, both composites send an elementary tensor to the same tensor with $f,g,j$ applied to the three components, because these maps commute with each $H$-action. The unit object $k_{\epsilon }$ is a left module since $\epsilon$ is a unital algebra map. The canonical left and right unit maps are natural vector-space isomorphisms by [F8], with inverses $v\mapsto 1\otimes v$ and $v\mapsto v\otimes 1$. They are $H$-linear by the two counit equations: their values on $c\otimes v$ and $v\otimes c$ reduce respectively to $\sum \epsilon (h_{(1)})c{h}_{(2)}v=chv$ and $\sum h_{(1)}v\epsilon (h_{(2)})c=chv$. Their inverses are therefore $H$-linear as well.

Both pentagon paths go from $((U\otimes V)\otimes W)\otimes X$ to $U\otimes (V\otimes (W\otimes X))$. Under the coherent vector-space identifications of [F8], the short composite acts by $(\mathrm{i}\mathrm{d}\otimes \mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })(\mathrm{\Phi })(\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d}\otimes \mathrm{i}\mathrm{d})(\mathrm{\Phi })$. The long composite acts by $(1\otimes \mathrm{\Phi })(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})(\mathrm{\Phi })(\mathrm{\Phi }\otimes 1)$. Each coproduct appears where a component of $\mathrm{\Phi }$ acts on a two-factor module, by step 1.1; later operators multiply on the left. The products are equal by [F1], and the underlying rebracketing paths agree by [F8], so the pentagon commutes as an equality of $H$-linear maps.

In the triangle equation, $a_{U,k_{\epsilon },V}$ acts, after rebracketing, by $(\mathrm{i}\mathrm{d}\otimes \epsilon \otimes \mathrm{i}\mathrm{d})(\mathrm{\Phi })=1\otimes 1$. Both composites from $(U\otimes k)\otimes V$ to $U\otimes V$ therefore send $(u\otimes c)\otimes v$ to $cu\otimes v=u\otimes cv$, by the vector-space unit triangle in [F8]. Elementary tensors span, so the triangle commutes.

Steps 1.1–2.2 provide the bifunctor, unit, natural isomorphisms, pentagon, and triangle required by [F6]; hence $H\text{-}\mathrm{M}\mathrm{o}\mathrm{d}$ is monoidal with the stated convention.

Taking the middle argument to be $1$ in the weak counit identity gives $\epsilon (gh)=\sum \epsilon (g{1}_{(2)})\epsilon (1_{(1)}h)=\epsilon (g{\epsilon }_t(h))$. Substitution into the definition of ${\epsilon }_t$ yields ${\epsilon }_t(gh)={\epsilon }_t(g{\epsilon }_t(h))$. Coalgebra counitality gives ${\epsilon }_t(1)=1$; setting $g=1$ in the displayed identity gives ${\epsilon }_t^2={\epsilon }_t$. The corresponding identities for ${\epsilon }_s$ follow by reversing the tensor factors.

Let $\mathrm{\Delta }(1)=1_{(1)}\otimes 1_{(2)}=1_{(1)}^{\mathrm{\prime }}\otimes 1_{(2)}^{\mathrm{\prime }}$ denote two copies. The weak-unit identity and multiplicativity give $\mathrm{\Delta }({\epsilon }_t(h))=\epsilon (1_{(1)}h)\mathrm{\Delta }(1_{(2)})=\epsilon (1_{(1)}h)1_{(2)}\otimes 1_{(3)}={\epsilon }_t(h)1_{(1)}\otimes 1_{(2)}=1_{(1)}{\epsilon }_t(h)\otimes 1_{(2)}$. Reversing the tensor factors gives $\mathrm{\Delta }({\epsilon }_s(h))=1_{(1)}\otimes {\epsilon }_s(h)1_{(2)}=1_{(1)}\otimes 1_{(2)}{\epsilon }_s(h)$. The weak-unit and weak-counit equations also give ${\epsilon }_t(1_{(1)}h)1_{(2)}=\epsilon (1_{(1)}^{\mathrm{\prime }}{1}_{(1)}h)1_{(2)}^{\mathrm{\prime }}{1}_{(2)}={\epsilon }_t(h)$ and $1_{(1)}\otimes {\epsilon }_t(1_{(2)})=1_{(1)}\otimes \epsilon (1_{(1)}^{\mathrm{\prime }}{1}_{(2)})1_{(2)}^{\mathrm{\prime }}=1_{(1)}\otimes \epsilon (1_{(2)})1_{(3)}=1_{(1)}\otimes 1_{(2)}$. Hence $\mathrm{\Delta }(1)={\epsilon }_s(1_{(1)})\otimes {\epsilon }_t(1_{(2)})\in H_s\otimes H_t$. The same formulas with the primed counital maps show that they restrict to the identity on their respective images. Thus $\mathrm{\Delta }(x)=x{1}_{(1)}\otimes 1_{(2)}=1_{(1)}x\otimes 1_{(2)}$ for $x\in H_t$, and the factor-reversed formulas hold for $x\in H_s$. Finally, the counital identity $x{\epsilon }_t(h)={\epsilon }_t(xh)$ for $x\in H_t$ gives $xy={\epsilon }_t(xb)\in H_t$ when $x={\epsilon }_t(a)$ and $y={\epsilon }_t(b)$; also $1={\epsilon }_t(1)$. The factor-reversed calculation proves that $H_s$ is a unital subalgebra.

Expanding the counital maps and using the two equal weak-unit products gives ${\epsilon }_t(g){\epsilon }_s(h)=\epsilon (1_{(1)}g)1_{(2)}{1}_{(1)}^{\mathrm{\prime }}\epsilon (h{1}_{(2)}^{\mathrm{\prime }})=\epsilon (1_{(1)}g)1_{(1)}^{\mathrm{\prime }}{1}_{(2)}\epsilon (h{1}_{(2)}^{\mathrm{\prime }})={\epsilon }_s(h){\epsilon }_t(g)$, so the two images commute. For $y\in H_s$, $h\in H$, and $x\in H_t$, the product formula in step 1.1 and the identity $x{\epsilon }_t(h)={\epsilon }_t(xh)$ from step 2.1 give ${\epsilon }_t(yh)={\epsilon }_t(y{\epsilon }_t(h))={\epsilon }_t({\epsilon }_t(h)y)={\epsilon }_t(h){\epsilon }_t(y)$; hence ${\epsilon }_t{\mathrm{\mid }}_{H_s}$ reverses products. Direct substitution into the definitions gives ${\epsilon }_t{\epsilon }_s^{\mathrm{\prime }}(h)=\epsilon (1_{(1)}{1}_{(1)}^{\mathrm{\prime }}\epsilon (1_{(2)}^{\mathrm{\prime }}h))1_{(2)}=\epsilon (1_{(1)}h)1_{(2)}={\epsilon }_t(h)$ by the weak-counit identity. Applying the same calculation with multiplication and coproduct reversed gives ${\epsilon }_s^{\mathrm{\prime }}{\epsilon }_t(h)={\epsilon }_s^{\mathrm{\prime }}(h)$. Since both counital maps are the identity on their images, these equalities make ${\epsilon }_t{\mathrm{\mid }}_{H_s}$ and ${\epsilon }_s^{\mathrm{\prime }}{\mathrm{\mid }}_{H_t}$ mutually inverse. Thus ${\epsilon }_s^{\mathrm{\prime }}(x)\in H_s$; it commutes with $z\in H_t$, and ${\epsilon }_t({\epsilon }_s^{\mathrm{\prime }}(x)z)={\epsilon }_t(z{\epsilon }_s^{\mathrm{\prime }}(x))=z{\epsilon }_t({\epsilon }_s^{\mathrm{\prime }}(x))=zx$.

Put $E=\mathrm{\Delta }(1)$. For every $y\in H$, multiplicativity gives $E\mathrm{\Delta }(y)=\mathrm{\Delta }(y)$; hence $\sum {\epsilon }_t(y_{(1)})y_{(2)}=\sum \epsilon (1_{(1)}{y}_{(1)})1_{(2)}{y}_{(2)}=(\epsilon \otimes \mathrm{i}\mathrm{d})(E\mathrm{\Delta }(y))=y$ by coalgebra counitality. For $x\in H_t$, step 2.1 gives $\mathrm{\Delta }(x)=(x\otimes 1)E=E(x\otimes 1)$, so $\mathrm{\Delta }(hx)=\mathrm{\Delta }(h)(x\otimes 1)E=\mathrm{\Delta }(h)E(x\otimes 1)=\sum h_{(1)}x\otimes h_{(2)}$. Applying the preceding identity to $y=hx$ proves $\sum {\epsilon }_t(h_{(1)}x)h_{(2)}=hx$.

Put $e=({\epsilon }_t\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(1)=\sum e^1\otimes e^2$. Step 2.1 puts both tensor legs in $H_t$, and $e^1=\epsilon (1_{(1)}^{\mathrm{\prime }}{1}_{(1)})1_{(2)}^{\mathrm{\prime }}$, $e^2=1_{(2)}$. For $x\in H_t$, substituting these expressions and using the weak-unit identity yields $\sum \epsilon (x{e}^1)e^2=\epsilon (1_{(1)}^{\mathrm{\prime }}{1}_{(1)})\epsilon (x{1}_{(2)}^{\mathrm{\prime }})1_{(2)}=\epsilon (1_{(1)}x)1_{(2)}=x$. The other contraction is $\sum e^1\epsilon (e^{2}x)=\epsilon (1_{(1)}^{\mathrm{\prime }}{1}_{(1)})1_{(2)}^{\mathrm{\prime }}\epsilon (1_{(2)}x)=\epsilon (1_{(1)}^{\mathrm{\prime }}x)1_{(2)}^{\mathrm{\prime }}={\epsilon }_t(x)=x$, again by the weak-unit identity and $x\in H_t$. Taking $x=1$ in the first contraction gives $\sum \epsilon (e^1)e^2=1$; the weak-unit identity also gives $\sum e^1{e}^2=\sum {\epsilon }_t(1_{(1)})1_{(2)}=1$. The pairing $(y,x)\mapsto \epsilon (yx)$ on $H_t$ is nondegenerate: if $\epsilon (yx)=0$ for every $y\in H_t$, then the second contraction gives $x=\sum e^1\epsilon (e^{2}x)=0$. For any $x,y\in H_t$, both contractions $(\epsilon (y\cdot )\otimes \mathrm{i}\mathrm{d})(xe)$ and $(\epsilon (y\cdot )\otimes \mathrm{i}\mathrm{d})(ex)$ equal $yx$ by the first reconstruction identity; nondegeneracy therefore gives $xe=ex$. Finally, $\mathrm{\Delta }(1)\in H_s\otimes H_t$ and ${\epsilon }_s^{\mathrm{\prime }}{\epsilon }_t{\mathrm{\mid }}_{H_s}={\mathrm{i}\mathrm{d}}_{H_s}$ imply $\mathrm{\Delta }(1)=\sum {\epsilon }_s^{\mathrm{\prime }}(e^1)\otimes e^2$.

For $q\in H_s$, step 2.1 gives $\mathrm{\Delta }(q)=E(1\otimes q)$, so $\mathrm{\Delta }(hq)=\mathrm{\Delta }(h)(1\otimes q)$ and coalgebra counitality yields $\sum h_{(1)}\epsilon (h_{(2)}q)=hq$. The weak counit equation with middle argument $1$ gives $\epsilon (h_{(2)}x)=\sum \epsilon (h_{(2)}{1}_{(1)})\epsilon (1_{(2)}x)$. Since $E\in H_s\otimes H_t$ by step 2.1, applying the preceding contraction to its first leg shows $\sum h_{(1)}\epsilon (h_{(2)}x)=\sum h{1}_{(1)}\epsilon (1_{(2)}x)=h{\epsilon }_s^{\mathrm{\prime }}(x)$. Finally step 3.1 proves ${\epsilon }_s^{\mathrm{\prime }}{\epsilon }_t={\epsilon }_s^{\mathrm{\prime }}$ on all of $H$, and $E\mathrm{\Delta }(h)=\mathrm{\Delta }(h)$ gives $\sum {\epsilon }_s^{\mathrm{\prime }}({\epsilon }_t(h_{(2)}x))h_{(1)}=\sum {\epsilon }_s^{\mathrm{\prime }}(h_{(2)}x)h_{(1)}=\sum 1_{(1)}{h}_{(1)}\epsilon (1_{(2)}{h}_{(2)}x)=\sum h_{(1)}\epsilon (h_{(2)}x)=h{\epsilon }_s^{\mathrm{\prime }}(x)$. The order of the factors is essential: the ${\epsilon }_s^{\mathrm{\prime }}$ factor precedes $h_{(1)}$, as required for the right unit map of left modules. No antipode, finite-dimensionality assumption, or commutativity of $H_t$ has been used.

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.