sg.symplectic geometry - Is a Poisson Group a group object in the category of Poisson Manifolds?

I realized that I am very confused about a certain sign in the definition of a Poisson group. I will give some definitions, and then point out my confusion.

Definitions

Group objects

Let $mathcal C$ be a category with Cartesian products. Recall that a group object in $mathcal C$ is an object $G in mathcal C$ along with chosen maps $e: 1to G$ and $m: Gtimes G to G$ (choose an initial object $1$ and a particular instance of the categorical product, and they imply all the others), such that (i) the two maps $G^{times 3} to G$ agree, (ii) the three natural maps $Gto G$ agree, and (iii) the map $p_1 times m: G^{times 2} to G^{times 2}$ is an isomorphism, where $p_1$ is the "project on the first factor" map $G^{times 2}to G$.

You may be used to seeing axiom (iii) presented slightly differently. Namely, if $p_1 times m: G^{times 2} to G^{times 2}$ is an isomorphism, then consider the map $i = p_2 circ (p_1times m)^{-1} circ (etimes text{id}) : G = 1times G to G$. It satisfies the usual axioms of the inverse map. Conversely, if $i: Gtimes G$ satisfies the usual axioms, then $p_1 times (m circ (itimes text{id}))$ is an inverse to $p_1 times m$. I learned this alternate presentation from Chris Schommer-Pries.

Poisson manifolds

A Poisson manifold is a smooth manifold $M$ along with a smooth bivector field, i.e. a section $pi in Gamma(wedge^2 TM)$, satisfying an axiom. Recall that if $vin Gamma(TM)$, then $v$ defines a (linear) map $C^infty(M) to C^infty(M)$ by differentiating in the direction of $v$. Well, if $pi in Gamma(wedge^2 TM)$, then it similarly defines a map $C^infty(M)^{wedge 2} to C^infty(M)$. The axiom states that this map is a Lie brackets, i.e. it satisfies the Jacobi identity.

A morphism of Poisson manifolds is a smooth map of manifolds such that the induced map on $C^infty$ is a Lie algebra homomorphism.

The category of Poisson manifolds has products (Wikipedia).

Poisson groups

A Poisson Group is a manifold $G$ with a Lie group structure $m : Gtimes G to G$ and a Poisson structure $pi in Gamma(wedge^2 TG)$, such that $m$ is a morphism of Poisson manifolds. Recall that a Lie group $G$ is a group object in the category of smooth manifolds.

Recall also that a Lie group is almost entirely controlled by its Lie algebra $mathfrak g = T_eG$. Then it is no surprise that the Poisson structure can be described infinitesimally. Indeed, by left-translating, identify $TG = mathfrak g times G$. Consider the adjoint action of $G$ on the abelian Lie group $mathfrak g$. Then we can define a Lie group structure $mathfrak g^{wedge 2} rtimes G$ on $wedge^2 TG$. Recall that a section $pi in Gamma(wedge^2 TG)$ is just a manifold map $G to wedge^2 TG$ that splits the projection $wedge^2 TG to G$. Then a Poisson manifold $(G,pi)$ is a Poisson group if and only if $pi : G to wedge^2 TG$ is a map of Lie groups.

Thus, a Poisson group structure is precisely the same as a Lie algebra $dpi : mathfrak g to wedge^2 mathfrak g rtimes mathfrak g$ splitting the obvious projection (here $wedge^2 mathfrak g$ is an abelian Lie algebra, and $mathfrak g$ acts on it via the adjoint action), and such that $(dpi)^* : wedge^2mathfrak g^* to mathfrak g^*$ satisfies the Jacobi identity. (Any failure of $G$ to be simply-connected, which might prevent such a map from lifting, also fails in $wedge^2 TG$, so this really is a one-to-one identification of Poisson group structures on $G$ and "Lie bialgebra" structures on $mathfrak g$.)

From this perspective, then, it is more or less clear that the inverse map $i:Gto G$ is not a morphism of Poisson manifolds (Wikipedia). Indeed, infinitesimally, $di = -1: mathfrak g to mathfrak g$, which takes $dpi$ to $-dpi$ (as $dpi$ has one $mathfrak g$ on the left and two on the right). Instead, $i$ is an "anti-Poisson map". The monoid $(mathbb R,times)$ acts on the category of Poisson manifolds by doing nothing to the underlying smooth manifolds and rescaling the Poisson structures; a smooth map is anti-Poisson if it becomes Poisson after twisting by the action of $-1$.

The unit map $e: 1 to G$, on the other hand, is Poisson; it follows from the axioms of a Poisson group that $pi(e) = 0$, and the terminal object in the category of Poisson manifolds is $1 = {text{pt}}$ with the trivial Poisson structure. ($C^infty({text{pt}}) = mathbb R$ can only support this Poisson structure.)

My question

Suppose that $G$ is a Poisson group. Then $p_1 times m: G^{times 2} to G^{times 2}$ is a Poisson map, and an isomorphism of smooth manifolds. Thus, I would expect that it is an isomorphism of Poisson manifolds. On the other hand, in the first section above I constructed the inverse map $i : Gto G$ out of this isomorphism and the other structure maps, all of which are Poisson when $G$ is a Poisson group. And yet $i$ is not a Poisson map. So where am I going wrong?