Let $G$ be a Poisson-Lie group and the inversion map $i:G\to G$ defined by $$ g\mapsto g^{-1}. $$ How to show that the inversion map is anti-Poisson, i.e., $\{f\circ i, g\circ i\}(x) = -\{f, g\}(x^{-1})$?
Also, I wonder what's the codomain of the Poisson bracket $\{\cdot, \cdot\}$ here? I understand that it takes elements $f,\;g\in C^\infty(G)$. Also, what does it mean to say $\{f,g\}(x)$?
The answer is a bit long because it requires some intermediate results from Poisson geometry. Although these results are not very complicated, I decided to recall all the necessary background and give an answer to this post so that it does not stay incomplete. In case one has some basic background in differential and Poisson geometry, they could jump directly to the third part.
To answer the simpler of your questions, the codomain of the Poisson bracket is the space of smooth functions of the manifold. More precisely, a Poisson bracket $\{\cdot\,,\cdot\}$ on a manifold $M$ is a map $\{\cdot\,,\cdot\}:C^\infty(M)\times C^\infty(M)\to C^\infty(M)$ satisfying the following three conditions:
for all $f,g,h\in C^\infty(M)$. Equivalently, a Poisson structure on $M$ is given by a bivector field $\pi\in\Gamma(\wedge^2TM)$ that satisfies the equation $[\pi,\pi]=0$; here, $[\cdot\,,\cdot]$ denotes the Schouten bracket on the space of multivector fields $\Gamma(\wedge TM)$. This description will be needed in the next part. Note that in this setting, a Poisson morphism is a smooth map between two Poisson manifolds $\phi:(M_1,\pi_1)\to(M_2,\pi_2)$ such that $\pi_1$ and $\pi_2$ are $\phi$-related, i.e. $\phi_*\pi_1=\pi_2$.
Given two Poisson manifolds $(M_1,\pi_1)$ and $(M_2,\pi_2)$ the product manifold $M_1\times M_2$ inherits a Poisson bivector $\pi$ given point-wise by $\pi_{(p,q)} = (\wedge^2T_pi^1_q)(\pi_{1,p}) + (\wedge^2T_qi^2_p)(\pi_{2,q})$ for $(p,q)\in M_1\times M_2$, where the maps $i^1,i^2$ are the following collections of inclusions $$ i^1_q:M_1\to M_1\times M_2;\ p\mapsto (p,q)\ \text{for a } \textbf{fixed } q\in M_2 $$ $$ i^2_p:M_2\to M_1\times M_2;\ q\mapsto (p,q)\ \text{for a } \textbf{fixed } p\in M_1. $$
A Poisson-Lie group is a Lie group $G$ equipped with a bivector field $\pi\in\Gamma(\wedge^2TG)$ such that the multiplication map $\mu:G\times G\to G$ is a Poisson map, where $G\times G$ is endowed with the product Poisson structure $\Pi\in\Gamma(\wedge^2T(G\times G))$ induced by $\pi$ on $G$. According to the first part, this translates to $\mu_*\Pi = \pi$.
Proposition 1. The compatibility of $\pi$ with the Lie group structure is equivalent to $$ \pi_{gh} = (\wedge^2T_gR_h)(\pi_g) + (\wedge^2T_hL_g)(\pi_h),\ \text{for all } g,h\in G, $$ where $R_h$ and $L_g$ denote the right and left multiplication by $h$ and $g$, respectively.
Proof. Using the identities $\mu\circ i^1_h=R_h$ and $\mu\circ i^2_g=L_g$ we compute $$ \wedge^2T_{(g,h)}\mu (\Pi_{(g,h)}) = \wedge^2T_{(g,h)}\mu(\wedge^2T_gi^1_h(\pi_g)) + \wedge^2T_{(g,h)}\mu(\wedge^2T_hi^2_g(\pi_h)) = \wedge^2T_gR_h(\pi_g) + \wedge^2T_hL_g(\pi_h). $$
Lemma 2. The Poisson structure of a Poisson-Lie group vanishes at the identity.
Proof. Substituting $e$ for $g$ and $h$ in the equation of Proposition 2 yields $\pi_e = \pi_e + \pi_e$ and hence $\pi_e = 0$.
We prove the equivalent statement that the inverse map is a Poisson map between $(G,\pi)$ and $(G,-\pi)$. Let $g\in G$ and substitute $h$ with $g^{-1}$ in the equation of Proposition 1 above. So we have $$ 0 = \pi_e = \wedge^2T_gR_{g^{-1}}(\pi_g) + \wedge^2T_{g^{-1}}L_g(\pi_{g^{-1}}). $$ Using now that $T_g \iota = -T_eL_{g^{-1}}\circ T_gR_{g^{-1}}$, we compute $$ \wedge^2T_g\iota(\pi_g) = \wedge^2T_eL_{g^{-1}}(\wedge^2T_gR_{g^{-1}}(\pi_g)) = - \wedge^2T_eL_{g^{-1}}(\wedge^2T_{g^{-1}}L_g(\pi_{g^{-1}})) = - \wedge^2T_{g^{-1}}\operatorname{id}_G(\pi_{g^{-1}}) = - \pi_{g^{-1}}$$ which proves the claim.
