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We apply this to the two elements in B
We compute:
(ρ ∗ μ)(x ⊗ y) = ∑ x⊗y
H ⊗ H → H
ν : x ⊗ y ↦→ S(y) ∙ S(x)
ρ : x ⊗ y ↦→ S(x ∙ y)
ρ((x ⊗ y) (1) ) ∙ μ((x ⊗ y) (2) ) [defn. of the convolution ∗]
= ∑ ρ(x (1) ⊗ y (1) )μ(x (2) ⊗ y (2) ) [defn. of the coproduct of H ⊗ H]
= ∑ S(x (1) y (1) )x (2) y (2) [defn. of ρ and μ]
= ∑ (xy)
S((xy) (1) )(xy) (2) [Δ is multiplicative]
= ηɛ(xy) [defn. of the antipode]
= 1 B (x ⊗ y)
It is instructive to do such a calculation graphically:
S
S
= =
The first equality is the multiplicativity of the coproduct in a bialgebra, the second is the
definition of the antipode.
On the other hand, we compute μ ∗ ν:
S
S
= = = =
S
S
S
ηɛ(x) ∙ ɛ(y) = ηɛ(x ∙ y)
where in the last step we used that the counit ɛ is a map of algebras.
Finally, the equality defining the antipode
can be applied to 1 H and then yields
id ∗ S = ηɛ ,
1 H ∙ S(1 H ) = id ∗ S(1 H ) = ηɛ(1 H ) = 1 H ,
where the first equality is unitality of the coproduct Δ and the last identity is the unitality of
the counit ɛ. This identity in H implies S(1 H ) = 1 H . The assertions about the coproduct are
proven in an analogous way: use the identity
Δ ◦ S = Δ −1 = (S ⊗ S) ◦ Δ copp in Hom(H, H ⊗ H) .
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