Modules over the finite double are left-left Yetter–Drinfeld modules

Let $(h_i)$ be a basis of $H$ and $(f_i)$ its dual basis. For a left $D(H)$-module $M$, restrict the action to $H$ and set $\delta (v)={\sum }_i{h}_i\otimes (f_i\otimes 1_H)\cdot v$. By [F1], ${\sum }_i{h}_i\otimes f_i$ is the basis-independent tensor representing ${\mathrm{id}}_H$; hence $\delta$ is basis-independent and $(f\otimes \mathrm{id})\delta (v)=(f\otimes 1_H)\cdot v$ for every $f\in H^{\ast }$.

The coaction counit identity follows from $({\epsilon }_H\otimes \mathrm{id})\delta (v)=({\epsilon }_H\otimes 1_H)\cdot v=v$, since ${\epsilon }_H\otimes 1_H$ is the unit of $D(H)$. To check coassociativity, pair the first two $H$ factors with arbitrary $f,g\in H^{\ast }$. On $(\mathrm{\Delta }\otimes \mathrm{id})\delta (v)$ the result is $(f\ast g\otimes 1_H)\cdot v$ by the definition of convolution $(f\ast g)(x)=\sum f(x_1)g(x_2)$. On $(\mathrm{id}\otimes \delta )\delta (v)$ it is $(g\otimes 1_H)\cdot ((f\otimes 1_H)\cdot v)=(f\ast g\otimes 1_H)\cdot v$ by the $H^{\ast \mathrm{o}\mathrm{p}}$ module law in [L2]. Finite dual separation from [F1] gives coassociativity.

Conversely, let $V$ be a left-left Yetter–Drinfeld module. Give it the $H^{\ast \mathrm{o}\mathrm{p}}$ action of [L2] and define $(f\otimes h)\cdot v:=f\cdot (h\cdot v)$. The $H$ and $H^{\ast \mathrm{o}\mathrm{p}}$ factors act by their algebra structures; it remains to check their crossing relation.

For $h\in H$, $f\in H^{\ast }$ and $v\in V$, the left side of the forward crossing acts as $h\cdot (f\cdot v)=\sum f(v_{-1})h\cdot v_0$. The right side acts as $\sum f_1(S(h_1))f_3(h_3)f_2\cdot (h_2\cdot v)=\sum f(S(h_1)h_2{v}_{-1}S(h_4)h_5)h_3\cdot v_0$ by [L5]. The tensor antipode cancellation $\sum S(h_1)h_2\otimes h_3\otimes S(h_4)h_5=1\otimes h\otimes 1$ follows by applying the two equations [F5] successively and reduces this to $\sum f(v_{-1})h\cdot v_0$. Therefore the action respects the crossing.

By the diagonal-action definition [F2], the $H$ action on $V\otimes W$ is the tensor action, and the $H^{\ast \mathrm{o}\mathrm{p}}$ action is $f\cdot (v\otimes w)=\sum f_1\cdot v\otimes f_2\cdot w$. Pairing with every $f$ shows that its reconstructed coaction is $\delta (v\otimes w)=\sum v_{-1}{w}_{-1}\otimes v_0\otimes w_0$. The associator and unit maps are module maps by [F3] and are the same underlying vector-space maps under the two functors, so the equivalence is monoidal.

Define $c_{V,W}$ by the displayed formula and use the displayed candidate inverse. Applying $S^{-1}$ to the two antipode equations in [F5] and reversing products by [F7] gives $\sum S^{-1}(x_2)x_1=\epsilon (x)1_H=\sum x_2{S}^{-1}(x_1)$. On $w\otimes v$, the composite $c_{V,W}{c}_{V,W}^{-1}$ is $\sum v_{-1,2}{S}^{-1}(v_{-1,1})\cdot w\otimes v_0=w\otimes v$ by the second identity; the other composite is $\sum v_0\otimes S^{-1}(v_{-1,2})v_{-1,1}\cdot w=v\otimes w$ by the first. Thus the maps are inverse.

The inverse crossing of [L4], acting on $v$, gives $(f\otimes 1_H)\cdot ((1\otimes h)\cdot v)=\sum f_1(h_1)f_3(S(h_3))h_2\cdot (f_2\cdot v)=\sum f(h_1{v}_{-1}S(h_3))h_2\cdot v_0$. By the reconstruction identity in step 1.1, this is $(f\otimes \mathrm{id})\delta (h\cdot v)$; since it holds for every $f$, finite dual separation yields $\delta (h\cdot v)=\sum h_1{v}_{-1}S(h_3)\otimes h_2\cdot v_0$, which is the compatibility condition [L5]. Thus restriction and reconstruction produce a Yetter–Drinfeld module.

A $D(H)$-module map preserves the restricted $H$-action and commutes with every $f\otimes 1_H$ action. The reconstruction identity of step 1.1 then shows it preserves the coaction, so restriction defines a functor to Yetter–Drinfeld modules.

For $f,g\in H^{\ast }$ and $h,k\in H$, first apply the crossing identity of step 1.4 to $h\cdot (g\cdot (k\cdot v))$, obtaining $\sum g_1(S(h_1))g_3(h_3)g_2\cdot (h_2\cdot (k\cdot v))$. Applying $f$ on the left and using the $H^{\ast \mathrm{o}\mathrm{p}}$ action law gives $\sum g_1(S(h_1))g_3(h_3)(g_2\ast f)\cdot ((h_{2}k)\cdot v)$, exactly the action of $(fh)(gk)$ in [L1]. Bilinearity extends this identity to all elements of $D(H)$. Thus every Yetter–Drinfeld module determines a left $D(H)$-module, and maps preserving both structures are exactly the resulting module maps.

Starting with a $D(H)$-module, reconstructing its coaction and then the action of step 1.3 returns the original action, because $f\otimes h=(f\otimes 1_H)({\epsilon }_H\otimes h)$ in normal order. Starting with a Yetter–Drinfeld module, the reconstructed coaction has the same evaluation against every $f$ as the original coaction, so finite dual separation makes the coactions equal. The two functors are inverse.

The map $c_{V,W}$ is $H$-linear: applying the Yetter–Drinfeld equation [L5] to $h_1\cdot v$ gives $c(h\cdot (v\otimes w))=\sum (h_1{v}_{-1}S(h_3)h_4)\cdot w\otimes h_2\cdot v_0=\sum (h_1{v}_{-1})\cdot w\otimes h_2\cdot v_0=h\cdot c(v\otimes w)$, where $S(h_3)h_4$ contracts by [F5].

The map is $H$-colinear as well. The coaction of its output is $\sum v_{-1,1}{w}_{-1}S(v_{-1,3})v_{-1,4}\otimes v_{-1,2}\cdot w_0\otimes v_{0,0}$. The adjacent factors $S(v_{-1,3})v_{-1,4}$ contract to the unit by [F5], leaving $\sum v_{-1,1}{w}_{-1}\otimes v_{-1,2}\cdot w_0\otimes v_{0,0}$, which is $(\mathrm{id}\otimes c)\delta (v\otimes w)$ by coassociativity and the tensor-product coaction. Pairing this identity with every $f\in H^{\ast }$ proves $H^{\ast \mathrm{o}\mathrm{p}}$-linearity; together with step 2.5, $c$ is a $D(H)$-module map.

Naturality follows because Yetter–Drinfeld morphisms preserve both the action and coaction. The two hexagons required by [F4] follow by expanding the tensor-product coaction: $c_{V,W\otimes U}(v\otimes w\otimes u)=\sum v_{-1,1}\cdot w\otimes v_{-1,2}\cdot u\otimes v_0$, and $c_{V\otimes W,U}(v\otimes w\otimes u)=\sum (v_{-1}{w}_{-1})\cdot u\otimes v_0\otimes w_0$; these are exactly the two iterated composites of the displayed $c$. Hence the family is a braiding.