Let (B, μ, 1_{B}, Δ, ε) be a bialgebra (the ground field will be taken as k), we define the convolution in End(B) for f, g ∈ End(B) and x ∈ B as

f ∗ g(x) = μ(f ⊗ g)Δ(x)  (1)

one can check that (exercise 11.9 (2) of the tutorial) this is an algebra with the linear mapping

convunit : x ↦ ε(x)1_{B}  (2)

as unit. One can check easily that this mapping is the projector on the line k.1_{B} and kernel

B^{+} = ker(ε) = {x ∈ B | ε(x) = 0} .  (3)

We will denote I^{+} = Id_{B} −convunit the complementary projector (in fact, in my previous letter Δ^{+} = (I^{+} ⊗ I^{+}) ∘ Δ and it can be proved that Δ^{+} is coassociative).

The antipode S, when it exists, is defined as the convolutional inverse of Id_{B}. But Id = convunit + I^{+} (of type 1 + X), and then if the sum

S = ∑_{n=0}^{∞}(−1)^{n}(I^{+})^{∗n}  (4)

converges (in some sense), it is the antipode. Here, in the two cases of U and G, the sequence of endomorphisms (I^{+})^{∗n} is locally finite which means that, for all x ∈ B it exists a number N ∈ ℕ from which (I^{+})^{∗n}(x) = 0. Formally

(∀x ∈ B)(∃N ∈ ℕ)(∀n ≥ N)((I^{+})^{∗n}(x) = 0)  (5)

then, it is easy to check that the sum

S_{n}(x) = ∑_{m=0}^{n}(−1)^{m}(I^{+})^{∗m}(x)  (6)

is stationnary from N. Taking as S(x) the common value of the S_{n}(x), one has then the antipode. It is obvious that, if x ∈ B^{+}, S(x) = ∑_{n=1}^{∞}(−1)^{n}(I^{+})^{∗n}(x) as (I^{+})^{∗0}(x) = convunit(x) = 0 and the explains formula (33) of the paper. Now, if you have two bialgebras (B_{i}, μ_{i}, 1_{B_{i}}, Δ_{i}, ε_{i}), i = 1, 2 with a morphism of bialgebras ϕ : B_{1} → B_{2}, this is a morphism for the convolutions in the sense that, if f_{i}, g_{i} ∈ End(B_{i}) with

f_{2} ∘ ϕ = ϕ ∘ f_{1} ; g_{2} ∘ ϕ = ϕ ∘ g_{1}

then (f_{2} ∗_{2} g_{2}) ∘ ϕ = ϕ ∘ (f_{1} ∗_{1} g_{1}) (almost straigthforward). On the other hand,

I_{2}^{+}ϕ = ϕ ∘ I_{1}^{+}

and then,

S_{2}∘ϕ = (∑_{n=0}^{∞}(−1)^{n}(I_{2}^{+})^{∗n})∘ϕ = (∑_{n=0}^{∞}(−1)^{n}ϕ∘(I_{1}^{+})^{∗n}) = ϕ∘(∑_{n=0}^{∞}(−1)^{n}(I_{1}^{+})^{∗n}) = ϕ∘S_{1} (7)