Let $X$ be a Poisson manifold with Poisson bivector field $\Pi.$ Let $x_{0} \in X$ be such that $\Pi (x_{0}) = 0.$ Let $\mathcal O (X)_{x_{0}}$ denote the ring of germs of the smooth functions on $X$ at $x_{0}.$ Consider the maximal ideal $I$ of $\mathcal {O}_{x_{0}}$ consisting of functions vanishing at $x_{0}.$ Consider the Poisson bracket $$\{\cdot, \cdot \} : \mathcal {O}_{x_{0}} \otimes \mathcal {O}_{x_{0}} \longrightarrow \mathcal {O}_{x_{0}}$$ defined by $f \otimes g \mapsto \Pi (df, dg).$ Since $\Pi (x_{0}) = p$ it follows that $\{\cdot, \cdot \} : \mathcal {O}_{x_{0}} \otimes \mathcal {O}_{x_{0}} \longrightarrow I$ and hence we have a Lie bracket $\{\cdot, \cdot \} : I \otimes I \longrightarrow I.$ Moreover if $f \in I$ and $g \in I^{2}$ we have $\{f, g \} = \Pi(df, dg) \in I^{2}$ so that $I^{2}$ is a Lie ideal of $I.$ This induces a Lie algebra structure on the quotient $I/I^{2} \simeq T_{x_{0}}^{\ast} X$ (the cotangent space of $X$ at $x_{0}$).
My question is $:$ How do we have $I/I^{2} \simeq T_{x_{0}}^{\ast} X\ $? Any suggestion in this regard would be warmly appreciated.
Thanks for your time.
Source $:$ Lecture notes on quantum groups by Pavel Etingof (Lecture $2$, Page no. $16$).