Bialgebra module tensor actions and their coherence

For left $H$-modules $V,W$, the four-variable map $(a,b,v,w)\mapsto av\otimes bw$ is $k$-multilinear, so the tensor universal property [F4] defines an action of $H\otimes H$ on $V\otimes W$ by $(a\otimes b)(v\otimes w)=av\otimes bw$. On pure tensors, $(a\otimes b)((c\otimes d)(v\otimes w))=acv\otimes bdw=((a\otimes b)(c\otimes d))(v\otimes w)$ by [F2] and [F3], and $1_H\otimes 1_H$ acts as the identity; pure tensors span, so this is a unital algebra action. Composing its representation map with the algebra map $\mathrm{\Delta }$ gives precisely the candidate action of [F1], $h(v\otimes w)=\sum h_{(1)}v\otimes h_{(2)}w$, as a unital left $H$-module action.

If $f:V\to V^{\mathrm{\prime }}$ and $g:W\to W^{\mathrm{\prime }}$ are $H$-linear, then on each pure tensor $(f\otimes g)(h(v\otimes w))=\sum h_{(1)}f(v)\otimes h_{(2)}g(w)=h((f\otimes g)(v\otimes w))$. Both sides are linear and pure tensors span by [F4], so $f\otimes g$ is $H$-linear. The tensor maps preserve identities and compositions because their values agree on pure tensors.

Since $\epsilon$ is a unital algebra map, $h\cdot c=\epsilon (h)c$ satisfies $1_H\cdot c=c$ and $(ab)\cdot c=a\cdot (b\cdot c)$; hence $k_{\epsilon }$ is a left $H$-module by [F2]. The maps in [F5] are the proposed unit maps.

For $u\in U,v\in V,w\in W$, coassociativity writes the iterated coproduct in either parenthesization as $\sum h_{(1)}\otimes h_{(2)}\otimes h_{(3)}$. Both $\alpha (h\cdot ((u\otimes v)\otimes w))$ and $h\cdot \alpha ((u\otimes v)\otimes w)$ are therefore $\sum h_{(1)}u\otimes (h_{(2)}v\otimes h_{(3)}w)$ after applying the canonical associator; thus $\alpha$ is $H$-linear. For the left unit map, $\lambda (h\cdot (c\otimes v))=\sum \epsilon (h_{(1)})c{h}_{(2)}v=chv=h\cdot \lambda (c\otimes v)$ by the left counit identity. For the right unit map, $\rho (h\cdot (v\otimes c))=\sum h_{(1)}v\epsilon (h_{(2)})c=h\cdot \rho (v\otimes c)$ by the right counit identity. Pure tensors span, so all three maps are $H$-linear on their full domains.

For the converse, use the regular left modules $H,H$. The action formula gives $b\cdot (1_H\otimes 1_H)=\mathrm{\Delta }(b)$ and $(ab)\cdot (1_H\otimes 1_H)=\mathrm{\Delta }(ab)$. By the assumed associativity of this tensor-module action, $\mathrm{\Delta }(ab)=(ab)\cdot (1_H\otimes 1_H)=a\cdot (b\cdot (1_H\otimes 1_H))=a\cdot \mathrm{\Delta }(b)=\sum a_{(1)}{b}_{(1)}\otimes a_{(2)}{b}_{(2)}=\mathrm{\Delta }(a)\mathrm{\Delta }(b)$, where the last equality uses the tensor-product algebra multiplication [F3]. The assumed unital action law gives $1_H\otimes 1_H=1_H\cdot (1_H\otimes 1_H)=\mathrm{\Delta }(1_H)$. Thus $\mathrm{\Delta }$ is a unital algebra map.

The assumed unital $H$-module structure on $k_{\epsilon }$ gives $\epsilon (1_H)=1$ and $\epsilon (ab)=\epsilon (a)\epsilon (b)$ directly from its unit and associativity laws [F2]. Hence $\epsilon$ is a unital algebra map.

Apply $H$-linearity of the canonical associator to the three regular modules at $((1_H\otimes 1_H)\otimes 1_H)$. The action on this left-parenthesized unit tensor is $(\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(h)$, and applying the associator sends it to $\alpha ((\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(h))$ in $H\otimes (H\otimes H)$. On the right-parenthesized unit tensor, the action is $(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })\mathrm{\Delta }(h)$. Since the associator sends the unit tensor to the right-parenthesized unit tensor, its $H$-linearity gives $\alpha ((\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(h))=(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })\mathrm{\Delta }(h)$, precisely coassociativity with the canonical associator [F6].

The naturality squares for $\alpha ,\lambda ,\rho$ commute on pure tensors because each composite applies the same module maps to the same entries; linearity extends equality to all tensors. By [F7], the underlying vector-space associator and unit maps satisfy the pentagon and triangle. Since every arrow in these diagrams is now $H$-linear, the same equalities hold in $H\text{-}\mathrm{M}\mathrm{o}\mathrm{d}$. This proves the forward assertion.

Apply $H$-linearity of $\lambda :k_{\epsilon }\otimes H\to H$ at $1\otimes 1_H$: it gives $\sum \epsilon (h_{(1)})h_{(2)}=h$. Apply $H$-linearity of $\rho :H\otimes k_{\epsilon }\to H$ at $1_H\otimes 1$: it gives $\sum h_{(1)}\epsilon (h_{(2)})=h$. These are the two counit identities, so $\epsilon$ is a counit. Steps 1.5–1.7 prove the other converse axioms.