[proofplan] The Kuranishi theorem realizes the local deformation space of $X$ as the germ at $0$ of the zero set of an analytic obstruction map whose domain is $H^1(X,T_X)$ and whose target is $H^2(X,T_X)$. If $H^2(X,T_X)=0$, this target is the zero [vector space](/page/Vector%20Space), so the obstruction map is identically zero. Hence its zero set is the whole neighbourhood in $H^1(X,T_X)$, which is a smooth complex manifold of dimension $\dim_{\mathbb C}H^1(X,T_X)$. [/proofplan] [step:Realize the Kuranishi base as the zero set of the obstruction map] Let \begin{align*} V:=H^1(X,T_X) \end{align*} and \begin{align*} W:=H^2(X,T_X). \end{align*} Since $X$ is compact, these Dolbeault cohomology groups are finite-dimensional complex vector spaces. By the Kuranishi theorem [citetheorem:9119], there exists an open neighbourhood $U\subset V$ of $0$ and a holomorphic map \begin{align*} \kappa:U\to W \end{align*} such that the Kuranishi base germ of $X$ at $0$ is the analytic germ of \begin{align*} \kappa^{-1}(0_W)\subset U, \end{align*} where $0_W$ denotes the zero vector of $W$. [guided] The Kuranishi theorem [citetheorem:9119] is the structural input. It says that the local deformation space is not presented abstractly: after choosing the Kuranishi model, its base is cut out inside the finite-dimensional tangent space \begin{align*} V:=H^1(X,T_X) \end{align*} by an analytic obstruction map with values in \begin{align*} W:=H^2(X,T_X). \end{align*} Thus there is an open neighbourhood $U\subset V$ of $0$ and a holomorphic map \begin{align*} \kappa:U\to W \end{align*} such that the Kuranishi base germ is the germ at $0$ of the analytic set \begin{align*} \kappa^{-1}(0_W)\subset U. \end{align*} Here $0_W$ is the zero vector of $W$. This formulation is exactly where the obstruction space $H^2(X,T_X)$ enters: it is the target of the equations defining the Kuranishi base. [/guided] [/step] [step:Use $H^2(X,T_X)=0$ to eliminate all Kuranishi equations] By hypothesis, \begin{align*} W=H^2(X,T_X)=0. \end{align*} Therefore $W$ is the zero vector space. A map from $U$ to the zero vector space is necessarily the constant zero map, so \begin{align*} \kappa(u)=0_W \end{align*} for every $u\in U$. Hence \begin{align*} \kappa^{-1}(0_W)=U. \end{align*} [/step] [step:Identify the local deformation space with a smooth neighbourhood of $H^1(X,T_X)$] The [open set](/page/Open%20Set) $U\subset V=H^1(X,T_X)$ is a complex manifold because it is an open subset of a finite-dimensional complex vector space. Since the Kuranishi base germ is the germ of $U$ at $0$, it is smooth at $0$. Its dimension is \begin{align*} \dim_{\mathbb C}U=\dim_{\mathbb C}V=\dim_{\mathbb C}H^1(X,T_X)=h^1(X,T_X). \end{align*} This proves that the Kuranishi base of $X$ is smooth near $0$ and has dimension $h^1(X,T_X)$ at $0$. [/step]