Admissible cuts give a connected graded Hopf algebra and recursive antipode

A rooted-tree isomorphism bijects admissible edge subsets and preserves the isomorphism classes of their pruned forests and root components, so the cut sum is well-defined on $\mathcal{T}$. Each cut partitions the vertices between $P_C$ and $R_C$, hence every coproduct term has the same total degree as $t$; multiplicative extension makes $\mathrm{\Delta }$ a graded algebra map. For nonempty $t$, only the empty cut contributes after applying $\epsilon$ to the first factor, and only the virtual full cut $t\otimes 1$ contributes after applying $\epsilon$ to the second factor. Thus both counit identities hold on tree generators and, since both maps are algebra maps, on all forests.

Adjoin the empty cut $(P,R)=(1,t)$ and the virtual root cut $(P,R)=(t,1)$. A term of either iterated coproduct is equivalently a labeling of every vertex by its tensor position $1$, $2$, or $3$, with labels nonincreasing along every parent-to-child edge: the root-side trunk has label $3$, any middle components have label $2$, and the detached crown has label $1$. In $(\mathrm{\Delta }\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(t)$, the first cut separates the label-$3$ trunk and the second cuts the pruned forest into labels $1$ and $2$; in $(\mathrm{i}\mathrm{d}\otimes \mathrm{\Delta })\mathrm{\Delta }(t)$, the first cut separates the label-$1$ crown and the second cuts the root component into labels $2$ and $3$. Conversely, any such labeling determines these cuts uniquely: cut each edge where the label drops, and use the empty or virtual root cut when a layer is absent. Jumps from $3$ directly to $1$ are represented by the same edge cut in both iterates, with the missing middle factor supplied by the corresponding extreme cut. Thus both iterates sum over exactly the same finite vertex labelings, and each labeling gives the same three forests with the same multiplicity even when distinct labelings yield isomorphic forests. Coassociativity holds on $t$. Both iterated maps are algebra maps, so the identity holds on all forests.

The degree-zero part is exactly $\mathbb{Q}1$, and the counit and grading checked in steps 1.1 and 2.1 make $H$ a connected graded bialgebra. For linear maps $f,g:H\to H$, define convolution $f\ast g=m(f\otimes g)\mathrm{\Delta }$; coassociativity of $\mathrm{\Delta }$ and associativity of multiplication make this product associative with unit $u\epsilon$, where $u$ is the algebra unit map. Put $\pi =\mathrm{i}\mathrm{d}-u\epsilon$.

If $h$ is homogeneous of degree $n$, then ${\pi }^{\ast k}(h)=0$ for $k>n$: the iterated coproduct has $k$ nonnegative homogeneous tensor degrees summing to $n$, so for $k>n$ at least one factor has degree zero and is killed by $\pi$. Therefore $S={\sum }_{k\ge 0}(-1{)}^k{\pi }^{\ast k}$ is a finite sum on each homogeneous input. Multiplying its finite partial sum on either side by $\mathrm{i}\mathrm{d}=u\epsilon +\pi$ telescopes to $u\epsilon$, because the remaining power of $\pi$ vanishes above degree $n$. Thus $S$ is a two-sided convolution inverse of $\mathrm{i}\mathrm{d}$; uniqueness follows by multiplying any two-sided inverse expressions in the associative convolution algebra.

For a positive-degree tree $t$, expand $(S\ast \mathrm{i}\mathrm{d})(t)=0$ using the coproduct in [F1]. The two extreme cuts contribute $S(t)$ and $t$, and each nonempty ordinary cut contributes $S(P_C(t))R_C(t)$. Rearranging gives the displayed recursion. For every such cut both the root component and each tree in the pruned forest have fewer vertices than $t$, so induction on degree terminates.

Let $m:H\otimes H\to H$ be multiplication and define $g(a\otimes b)=S(b)S(a)$. In the convolution algebra $\mathrm{Hom}(H\otimes H,H)$, both $S\circ m$ and $g$ are two-sided inverses of $m$: for example, $(g\ast m)(a\otimes b)=\sum S(b_{(1)})(S(a_{(1)})a_{(2)})b_{(2)}=\epsilon (a)\epsilon (b)1$, and $(m\ast g)(a\otimes b)=\sum a_{(1)}(b_{(1)}S(b_{(2)}))S(a_{(2)})=\epsilon (a)\epsilon (b)1$; the analogous two products for $S\circ m$ are $(S\ast \mathrm{i}\mathrm{d})(ab)$ and $(\mathrm{i}\mathrm{d}\ast S)(ab)$. Uniqueness of a two-sided inverse gives $S(ab)=S(b)S(a)$. Since $H$ is commutative, $S(b)S(a)=S(a)S(b)$, so the antipode is multiplicative.