For objects V, W of Rep(H) set

V ⊗ W = x ∈ V ⊗_{k} W|x = ∆(1) · x ⊂ V ⊗_{k} W —– (1)

with the obvious action of H via the comultiplication ∆ (here ⊗_{k} denotes the usual tensor product of vector spaces). Note that ∆(1) is an idempotent and therefore V ⊗ W = ∆(1) × (V ⊗_{k} W). The tensor product of morphisms is the restriction of usual tensor product of homomorphisms. The standard associativity isomorphisms (U ⊗ V ) ⊗ W → U ⊗ (V ⊗ W ) are functorial and satisfy the pentagon condition, since ∆ is coassociative. We will suppress these isomorphisms and write simply U ⊗ V ⊗ W.

The target counital subalgebra H_{t} ⊂ H has an H-module structure given by h · z = ε_{t}(hz),where h ∈ H, z ∈ H_{t}.

H_{t} is the unit object of Rep(H).

Define a k-linear homomorphism l_{V} : H_{t} ⊗ V → V by l_{V}(1_{(1)} · z ⊗ 1_{(2)} · v) = z · v, z ∈ H_{t}, v ∈ V.

This map is H-linear, since

l_{V} h · (1_{(1)} · z ⊗ 1_{(2)} · v) = l_{V}(h(1) · z ⊗ h_{(2)} · v) = ε_{t}(h_{(1)}z)h_{(2)} · v = hz · v = h · l_{V} (1_{(1)} · z ⊗ 1_{(2)} · v),

∀ h ∈ H. The inverse map l^{−1}_{V}: → H_{t} ⊗ V is given by V

l^{−1}_{V}(v) = S(1_{(1)}) ⊗ (1_{(2)} · v) = (1_{(1)} · 1) ⊗ (1_{(2)} · v)

The collection {l_{V}}_{V} gives a natural equivalence between the functor H_{t} ⊗ (·) and the identity functor. Indeed, for any H -linear homomorphism f : V → U we have:

l_{U} ◦ (id ⊗ f)(1_{(1)} · z ⊗ 1_{(2)} · v) = l_{U} 1_{(1)} · z ⊗ 1_{(2)} · f(v) = z · f(v) = f(z·v) = f ◦ l_{V}(1_{(1)} · z ⊗ 1_{(2)} · v)

Similarly, the k-linear homomorphism r_{V} : V ⊗ H_{t} → V defined by r_{V}(1_{(1)} · v ⊗ 1_{(2)} · z) = S(z) · v, z ∈ H_{t}, v ∈ V, has the inverse r^{−1}_{V}(v) = 1_{(1)} · v ⊗ 1_{(2)} and satisfies the necessary properties.

Finally, we can check the triangle axiom idV ⊗ l_{W} = r_{V} ⊗ id_{W} : V ⊗ H_{t} ⊗ W → V ⊗ W ∀ objects V, W of Rep(H). For v ∈ V, w ∈ W we have

(id_{V} ⊗ l_{W})(1_{(1)} · v ⊗ 1′_{(1)}1_{(2)} · z ⊗ 1′_{(2)} · w)

= 1_{(1)} · v ⊗ 1′_{(2)}z · w) = 1_{(1)}S(z) · v ⊗ 1_{(2)} · w

=(r_{V} ⊗ id_{W}) (1′_{(1)} · v ⊗ 1′_{(2)} 1_{(1)} · z ⊗ 1_{(2)} · w),

therefore, id_{V} ⊗ l_{W} = r_{V} ⊗ id_{W}

Using the antipode S of H, we can provide Rep(H) with a duality. For any object V of Rep(H), define the action of H on V^{∗} = Hom_{k}(V, k) by

(h · φ)(v) = φ S(h) · v —– (2)

where h ∈ H , v ∈ V , φ ∈ V^{∗}. For any morphism f : V → W , let f^{∗}: W^{∗} → V^{∗} be the morphism dual to f. For any V in Rep(H), we define the duality morphisms d_{V} : V^{∗} ⊗ V → H_{t}, b_{V} : H_{t} → V ⊗ V^{∗ }as follows. For ∑_{j} φ^{j} ⊗ v_{j} ∈ V^{*} ⊗ V, set

d_{V}(∑_{j} φ^{j} ⊗ v_{j}) = ∑_{j} φ^{j} (1_{(1)} · v_{j}) 1_{(2)} —– (3)

Let {f_{i}}_{i} and {ξ^{i}}_{i} be bases of V and V^{∗}, respectively, dual to each other. The element ∑_{i} f_{i} ⊗ ξ^{i} does not depend on choice of these bases; moreover, ∀ v ∈ V, φ ∈ V^{∗} one has φ = ∑_{i} φ(f_{i}) ξ^{i} and v = ∑_{i} f_{i} ξ^{i} (v). Set

b_{V}(z) = z · (∑_{i} f_{i} ⊗ ξ^{i}) —– (4)

The category Rep(H) is a monoidal category with duality. We know already that Rep(H) is monoidal, it remains to prove that d_{V} and b_{V} are H-linear and satisfy the identities

(id_{V} ⊗ d_{V})(b_{V} ⊗ id_{V}) = id_{V}, (d_{V} ⊗ id_{V∗})(id_{V∗} ⊗ b_{V}) = id_{V∗}.

Take ∑_{j} φ^{j} ⊗ v_{j} ∈ V^{∗} ⊗ V, z ∈ H_{t}, h ∈ H. Using the axioms of a quantum groupoid, we have

h · d_{V}(∑_{j} φ^{j} ⊗ v_{j}) = ((∑_{j} φ^{j} (1_{(1)} · v_{j}) ε_{t}(h1_{(2)})

= (∑_{j} φ^{j} ⊗ ε_{s}(1_{(1)}h) · v_{j} 1_{(2)} ∑_{j} φ^{j} S(h_{(1)})1_{(1)}h_{(2)} · v_{j} 1_{(2)}

= (∑_{j} h_{(1) }· φ^{j} )(1_{(1)} · (h_{(2)} · v_{j}))1_{(2)}

= (∑_{j} d_{V}(h_{(1) }· φ^{j} ⊗ h_{(2)} · v_{j}) = d_{V}(h · ∑_{j} φ^{j} ⊗ v_{j})

therefore, d_{V} is H-linear. To check the H-linearity of b_{V} we have to show that h · b_{V}(z) =

b_{V} (h · z), i.e., that

∑_{i} h_{(1)}z · f_{i} ⊗ h_{(2)} · ξ^{i} = ∑_{i} 1_{(1)} ε_{t}(h_{z}) · f_{i} ⊗ 1_{(2)} · ξ^{i}

Since both sides of the above equality are elements of V ⊗_{k} V^{∗}, evaluating the second factor on v ∈ V, we get the equivalent condition

h_{(1)}zS(h_{(2)}) · v = 1_{(1)}ε^{t} (hz)S(1_{(2)}) · v, which is easy to check. Thus, b_{V} is H-linear.

Using the isomorphisms l_{V} and r_{V} identifying H_{t} ⊗ V, V ⊗ H_{t}, and V, ∀ v ∈ V and φ ∈ V^{∗} we have:

(id_{V} ⊗ d_{V})(b_{V} ⊗ id_{V})(v)

=(id_{V} ⊗ d_{V})b_{V}(1_{(1)} · 1) ⊗ 1_{(2)} · v

= (id_{V} ⊗ d_{V})b_{V}(1_{(2)}) ⊗ S^{−1}(1_{(1)}) · v

= ∑_{i} (id_{V} ⊗ d_{V}) 1_{(2)} · f_{i} ⊗ 1_{(3)} · ξ^{i} ⊗ S^{−1} (1_{(1)}) · v

= ∑_{i }1_{(2)} · f_{i} ⊗ 1_{(3)} · ξ^{i} (1′_{(1)}S^{-1} (1_{(1)}) · v) 1′_{(2)}

= 1_{(2) }S(1_{(3)}) 1′_{(1) S-1 }(1_{(1)}) · v ⊗ 1′_{(2)} = v

(d_{V} ⊗ id_{V∗})(id_{V∗} ⊗ b_{V})(φ)

= (d_{V} ⊗ id_{V∗}) 1_{(1)} · φ ⊗ b_{V}(1_{(2)})

= ∑_{i} (d_{V} ⊗ id_{V∗})(1_{(1)} · φ ⊗ 1_{(2) }· 1_{(2) }· 1_{(3)} · ξ^{i} )

= ∑_{i} (1_{(1)} · φ (1′_{(1)}1_{(2) }· f_{i})1′_{(2)} ⊗ 1_{(3)} · ξ^{i}

= 1′_{(2)} ⊗ 1_{(3)}1_{(1)} S(1′_{(1)}1_{(2)}) · φ = φ,