Evaluation, coevaluation, and triangle identities for left duals

Give $H\otimes H$ its tensor-product coalgebra, ${\mathrm{\Delta }}_{H\otimes H}(a\otimes b)=\sum (a_{(1)}\otimes b_{(1)})\otimes (a_{(2)}\otimes b_{(2)})$ and counit $\epsilon \otimes \epsilon$. The tensor universal property [F5] defines these maps from their bilinear formulas, and the coalgebra identities follow on pure tensors from those for $H$; pure tensors span. Thus convolution on ${\mathrm{Hom}}_k(H\otimes H,H)$, $(f\ast g)(z)=\sum f(z_{(1)})g(z_{(2)})$, is associative by coassociativity and associativity of multiplication, with unit $u(\epsilon \otimes \epsilon )$, where $u(c)=c{1}_H$. The pairings $P(a\otimes b)=S(ab)$ and $Q(a\otimes b)=S(b)S(a)$ are bilinear and hence define linear maps by [F5]. The antipode equations applied to $ab$ give $(P\ast m)(a\otimes b)=\epsilon (a)\epsilon (b)1_H=(m\ast P)(a\otimes b)$, where $m(a\otimes b)=ab$. The same equations give $(Q\ast m)(a\otimes b)=\sum S(b_{(1)})S(a_{(1)})a_{(2)}{b}_{(2)}=\epsilon (a)\epsilon (b)1_H$ and $(m\ast Q)(a\otimes b)=\sum a_{(1)}{b}_{(1)}S(b_{(2)})S(a_{(2)})=\epsilon (a)\epsilon (b)1_H$. Hence $P$ and $Q$ are both two-sided convolution inverses of $m$; in any associative unital algebra a two-sided inverse is unique, so $S(ab)=S(b)S(a)$. Substituting $1_H$ into either antipode identity gives $S(1_H)=1_H$.

For $h,g\in H$, $v\in V$ and $f\in V^{\ast }$, $1_H\cdot f=f$ by $S(1_H)=1_H$, and $(h\cdot (g\cdot f))(v)=f(S(g)S(h)v)=f(S(hg)v)=((hg)\cdot f)(v)$ by step 1.1. The action is linear in each variable from linearity of $S$, the module action on $V$, and [F3]. Thus the candidate formula is a unital left $H$-module action.

For $h\in H$, $f\in V^{\ast }$ and $v\in V$, the tensor action and evaluation give $\mathrm{ev}(h\cdot (f\otimes v))=\sum (h_{(1)}\cdot f)(h_{(2)}v)=\sum f(S(h_{(1)})h_{(2)}v)=\epsilon (h)f(v)=h\cdot \mathrm{ev}(f\otimes v)$ by the left antipode identity. Since step 2.1 makes $V^{\ast }$ a left $H$-module, the tensor action is defined; linearity extends this equality from pure tensors to all of $V^{\ast }\otimes V$.

Fix a finite basis $(e_i)$ of $V$ and its dual basis $(e^i)$. Under the canonical isomorphism $V\otimes V^{\ast }\to {\mathrm{End}}_k(V)$, $x\otimes f\mapsto [y\mapsto f(y)x]$, the tensor $h\cdot \mathrm{coev}(1)={\sum }_{(h),i}{h}_{(1)}{e}_i\otimes (h_{(2)}\cdot e^i)$ acts on $v$ as ${\sum }_{(h),i}{e}^i(S(h_{(2)})v)h_{(1)}{e}_i={\sum }_{(h)}{h}_{(1)}S(h_{(2)})v=\epsilon (h)v$ by the dual-basis identity and the right antipode identity. Therefore $h\cdot \mathrm{coev}(1)=\epsilon (h)\mathrm{coev}(1)=\mathrm{coev}(h\cdot 1)$, so coevaluation is $H$-linear.

For $v\in V$, the first triangle composite sends $v$ to ${\sum }_i{e}_i{e}^i(v)=v$. For $f\in V^{\ast }$, the second sends $f$ to ${\sum }_{i}f(e_i)e^i=f$. These are the two identities in [F4], so the triangle identities hold as maps of vector spaces and hence as maps in $H\text{-}\mathrm{M}\mathrm{o}\mathrm{d}$, since evaluation and coevaluation are $H$-linear by steps 3.1 and 3.2.

If $V=0$, its basis is empty, the coevaluation sum is $0$, and both triangle identities are the identity of the zero space. If $V$ is infinite-dimensional, every tensor ${\sum }_{i=1}^r{v}_i\otimes f_i\in V\otimes V^{\ast }$ corresponds to an endomorphism with image in $\mathrm{span}(v_1,\dots ,v_r)$; the identity on $V$ has infinite-dimensional image, so it cannot be such a tensor. This explains why the finite-dimensional hypothesis is essential for the stated coevaluation. No choice principle is used: only a finite basis of the given finite-dimensional $V$ is used.