[proofplan] Choose a finite Stein Leray cover of $X$ and present a first-order deformation over the dual-number base by local products. On overlaps, the transition maps reduce to the identity on $X$, so their first-order parts are holomorphic derivations, equivalently local sections of $T_X$. The cocycle condition for transition maps gives a Cech $1$-cocycle, and changing local product charts changes it by a Cech coboundary. Conversely, any tangent-sheaf cocycle glues the local products $U_i \times D$ into a first-order deformation, and these two constructions are inverse modulo isomorphism. [/proofplan] [step:Choose a Stein Leray cover and local product charts] Let $\mathcal U=(U_i)_{i\in I}$ be a finite Stein cover of $X$ such that every nonempty finite intersection $U_{i_0}\cap\cdots\cap U_{i_p}$ is Stein. Since $X$ is compact, such a finite Stein Leray cover exists by taking a sufficiently fine finite cover by coordinate polydiscs. Because $\mathcal U$ is Leray for the coherent sheaf $T_X$, the natural Cech cohomology group $\check H^1(\mathcal U,T_X)$ identifies with $H^1(X,T_X)$. Let $D$ denote the analytic germ with structure algebra $A=\mathbb C[\varepsilon]/(\varepsilon^2)$, and let $\rho:\mathcal Y\to D$ be a first-order analytic deformation of $X$. Thus $\rho$ is flat, its special fibre is identified with $X$, and its ideal of the special fibre is square-zero and generated by $\varepsilon$. We use the standard local product theorem for first-order analytic deformations of complex manifolds: if $V\subset X$ is a Stein coordinate domain and $\rho:\mathcal Y\to D$ is a flat analytic deformation with special fibre $X$, then after restricting to $V$ there is an isomorphism of $D$-spaces \begin{align*} \tau_i:\rho^{-1}(U_i)\to U_i\times D \end{align*} whose restriction to the special fibre is the identity on $U_i$. Its hypotheses hold because each $U_i$ is Stein and a coordinate domain, the fibre $X$ is nonsingular, and the base is the square-zero analytic germ $D$. For each $i\in I$, fix such a product chart $\tau_i$. [/step] [step:Extract a tangent-sheaf cocycle from the transition maps] For every nonempty overlap $U_{ij}=U_i\cap U_j$, define the transition automorphism \begin{align*} g_{ij}:U_{ij}\times D\to U_{ij}\times D \end{align*} by \begin{align*} g_{ij}=\tau_i\circ\tau_j^{-1}. \end{align*} It is an automorphism over $D$ and reduces to the identity modulo $\varepsilon$. Let $\mathcal O_X(U_{ij})$ denote the [algebra of holomorphic functions](/theorems/4949) on $U_{ij}$. For every $f\in\mathcal O_X(U_{ij})$, there is a unique [holomorphic function](/page/Holomorphic%20Function) $\theta_{ij}(f)\in\mathcal O_X(U_{ij})$ such that \begin{align*} g_{ij}^*(f)=f+\varepsilon\theta_{ij}(f). \end{align*} The map \begin{align*} \theta_{ij}:\mathcal O_X(U_{ij})\to\mathcal O_X(U_{ij}) \end{align*} is $\mathbb C$-linear. Since $g_{ij}^*$ is a homomorphism of algebras, for $f,h\in\mathcal O_X(U_{ij})$ we have \begin{align*} f h+\varepsilon\theta_{ij}(f h)=(f+\varepsilon\theta_{ij}(f))(h+\varepsilon\theta_{ij}(h)). \end{align*} Using $\varepsilon^2=0$, this gives \begin{align*} \theta_{ij}(f h)=f\theta_{ij}(h)+h\theta_{ij}(f). \end{align*} Thus $\theta_{ij}$ is a holomorphic derivation, hence a section of $T_X(U_{ij})$. On every nonempty triple intersection $U_{ijk}=U_i\cap U_j\cap U_k$, the transition maps satisfy \begin{align*} g_{ij}\circ g_{jk}\circ g_{ki}=\operatorname{id}_{U_{ijk}\times D}. \end{align*} Pulling back $f\in\mathcal O_X(U_{ijk})$ and comparing the coefficient of $\varepsilon$ gives \begin{align*} \theta_{ij}+\theta_{jk}+\theta_{ki}=0. \end{align*} Therefore $\theta=(\theta_{ij})$ is a Cech $1$-cocycle in $\check Z^1(\mathcal U,T_X)$. [guided] The transition maps compare two ways of identifying the deformation with the product over the same [open set](/page/Open%20Set). Since both product charts restrict to the identity on the special fibre, the comparison map is the identity modulo $\varepsilon$. This is exactly why its first nonzero term is a derivation. For an overlap $U_{ij}=U_i\cap U_j$, set \begin{align*} g_{ij}=\tau_i\circ\tau_j^{-1}:U_{ij}\times D\to U_{ij}\times D. \end{align*} The map $g_{ij}$ is over $D$, and its reduction modulo $\varepsilon$ is the identity on $U_{ij}$. Hence for each holomorphic function $f\in\mathcal O_X(U_{ij})$ there is a unique holomorphic function $\theta_{ij}(f)$ on $U_{ij}$ with \begin{align*} g_{ij}^*(f)=f+\varepsilon\theta_{ij}(f). \end{align*} Uniqueness follows because every function on $U_{ij}\times D$ has a unique form $a+\varepsilon b$ with $a,b\in\mathcal O_X(U_{ij})$. We now verify that $\theta_{ij}$ is a derivation. The pullback $g_{ij}^*$ is an algebra homomorphism, so for $f,h\in\mathcal O_X(U_{ij})$, \begin{align*} g_{ij}^*(f h)=g_{ij}^*(f)g_{ij}^*(h). \end{align*} Substituting the first-order expressions gives \begin{align*} f h+\varepsilon\theta_{ij}(f h)=(f+\varepsilon\theta_{ij}(f))(h+\varepsilon\theta_{ij}(h)). \end{align*} Because $\varepsilon^2=0$, the right-hand side equals \begin{align*} f h+\varepsilon f\theta_{ij}(h)+\varepsilon h\theta_{ij}(f). \end{align*} Comparing the coefficients of $\varepsilon$ gives \begin{align*} \theta_{ij}(f h)=f\theta_{ij}(h)+h\theta_{ij}(f). \end{align*} Thus $\theta_{ij}$ is a holomorphic derivation of $\mathcal O_X(U_{ij})$, and holomorphic derivations are precisely holomorphic vector fields, that is, sections of $T_X$ over $U_{ij}$. Finally, the transition maps of a glued analytic space satisfy the triple-overlap identity \begin{align*} g_{ij}\circ g_{jk}\circ g_{ki}=\operatorname{id}_{U_{ijk}\times D}. \end{align*} Applying pullback to $f\in\mathcal O_X(U_{ijk})$ and keeping only first-order terms gives \begin{align*} f+\varepsilon(\theta_{ij}(f)+\theta_{jk}(f)+\theta_{ki}(f))=f. \end{align*} Therefore \begin{align*} \theta_{ij}+\theta_{jk}+\theta_{ki}=0. \end{align*} This is exactly the Cech cocycle condition for the tangent sheaf on the cover $\mathcal U$. [/guided] [/step] [step:Show that changing product charts changes the cocycle by a coboundary] Choose another system of product charts $\tau_i'$. Define \begin{align*} h_i=\tau_i'\circ\tau_i^{-1}:U_i\times D\to U_i\times D. \end{align*} Each $h_i$ is the identity modulo $\varepsilon$, so there is a unique section $\xi_i\in T_X(U_i)$ such that \begin{align*} h_i^*(f)=f+\varepsilon\xi_i(f) \end{align*} for all $f\in\mathcal O_X(U_i)$. Let $\theta'_{ij}$ be the cocycle obtained from $\tau_i'$. The transition maps satisfy \begin{align*} g'_{ij}=h_i\circ g_{ij}\circ h_j^{-1}. \end{align*} Taking pullbacks and using $(h_j^{-1})^*(f)=f-\varepsilon\xi_j(f)$, we obtain \begin{align*} \theta'_{ij}=\theta_{ij}+\xi_i-\xi_j. \end{align*} Hence $\theta'$ and $\theta$ differ by the Cech coboundary of the $0$-cochain $(\xi_i)$. Therefore the cohomology class $[\theta]\in\check H^1(\mathcal U,T_X)$ is independent of the chosen local product charts. [/step] [step:Glue a first-order deformation from a tangent-sheaf cocycle] Conversely, let $\theta=(\theta_{ij})\in\check Z^1(\mathcal U,T_X)$ be a Cech $1$-cocycle. For each $i\in I$, take the product deformation $U_i\times D\to D$. On $U_{ij}\times D$, define an automorphism \begin{align*} g_{ij}:U_{ij}\times D\to U_{ij}\times D \end{align*} by its pullback on functions: \begin{align*} g_{ij}^*(f+\varepsilon h)=f+\varepsilon h+\varepsilon\theta_{ij}(f) \end{align*} for $f,h\in\mathcal O_X(U_{ij})$. Since $\theta_{ij}$ is a derivation and $\varepsilon^2=0$, this formula defines a homomorphism of local analytic algebras over $A$. Its inverse is given by replacing $\theta_{ij}$ by $-\theta_{ij}$. The cocycle identity $\theta_{ij}+\theta_{jk}+\theta_{ki}=0$ implies the transition identity \begin{align*} g_{ij}\circ g_{jk}\circ g_{ki}=\operatorname{id} \end{align*} on triple overlaps. Therefore the standard analytic gluing theorem applies to the local analytic spaces $U_i\times D$ and produces a complex analytic space $\mathcal Y_\theta$ with a morphism \begin{align*} \rho_\theta:\mathcal Y_\theta\to D. \end{align*} The morphism is flat because flatness is local on the source and each local model is the projection $U_i\times D\to D$, whose structure sheaf is $\mathcal O_{U_i}\otimes_{\mathbb C}A$ and is flat over $A$. The special fibre is obtained by setting $\varepsilon=0$, where all transition maps become the identity, so the special fibre is canonically $X$. [/step] [step:Identify cohomologous cocycles with isomorphic deformations] Suppose $\theta'_{ij}=\theta_{ij}+\xi_i-\xi_j$ for a $0$-cochain $\xi=(\xi_i)$ with $\xi_i\in T_X(U_i)$. Define \begin{align*} h_i:U_i\times D\to U_i\times D \end{align*} by \begin{align*} h_i^*(f+\varepsilon h)=f+\varepsilon h+\varepsilon\xi_i(f). \end{align*} For $f,h,p,q\in\mathcal O_X(U_i)$, the product rule for the derivation $\xi_i$ gives \begin{align*} h_i^*((f+\varepsilon h)(p+\varepsilon q))=fp+\varepsilon(fq+hp)+\varepsilon(f\xi_i(p)+p\xi_i(f)). \end{align*} On the other hand, \begin{align*} h_i^*(f+\varepsilon h)h_i^*(p+\varepsilon q)=(f+\varepsilon h+\varepsilon\xi_i(f))(p+\varepsilon q+\varepsilon\xi_i(p)), \end{align*} which is the same expression because $\varepsilon^2=0$. Thus $h_i^*$ is a homomorphism of $A$-algebras, and the formula with $-\xi_i$ gives its inverse. Hence $h_i$ is an automorphism over $D$. The equality $\theta'_{ij}=\theta_{ij}+\xi_i-\xi_j$ is precisely the compatibility condition saying that the $h_i$ intertwine the gluing maps for $\theta$ and $\theta'$. Hence the $h_i$ glue to an isomorphism of deformations \begin{align*} \mathcal Y_\theta\cong\mathcal Y_{\theta'} \end{align*} over $D$. Thus the deformation constructed from a cocycle depends only on its class in $\check H^1(\mathcal U,T_X)$. [/step] [step:Verify that the two constructions are inverse] Starting with a deformation $\rho:\mathcal Y\to D$, choosing local product charts, extracting the transition cocycle, and then gluing the products by those same transition maps reconstructs $\mathcal Y$ by the uniqueness part of analytic gluing. Thus the deformation maps to the same isomorphism class after the two constructions are composed in this order. Starting with a cocycle $\theta$, the glued deformation $\mathcal Y_\theta$ has local product charts by construction. The transition maps for those charts are exactly the maps whose first-order parts are $\theta_{ij}$, so extracting the associated cocycle returns $\theta$. Passing to cohomology classes and using the Leray identification \begin{align*} \check H^1(\mathcal U,T_X)\cong H^1(X,T_X) \end{align*} gives a natural bijection between isomorphism classes of first-order analytic deformations of $X$ over $D$ and $H^1(X,T_X)$. For the product deformation $X\times D\to D$, all transition maps may be chosen to be the identity, so all $\theta_{ij}$ are zero and the associated cohomology class is $0$. This proves the theorem. [/step]