[proofplan] The Dolbeault cohomology group in bidegree $(0,0)$ is the kernel of $\bar\partial:\Omega^{0,0}(M)\to\Omega^{0,1}(M)$ modulo the image of the nonexistent incoming map from $\Omega^{0,-1}(M)$. Since $(0,0)$-forms are exactly smooth complex-valued functions, the only point is to identify the condition $\bar\partial f=0$ with holomorphicity. This is checked in holomorphic coordinates, where $\bar\partial f$ is the collection of the anti-holomorphic partial derivatives of $f$. [/proofplan] [step:Identify the cochains in bidegree $(0,0)$ with smooth complex-valued functions] By definition of forms of type $(p,q)$ and the [decomposition of complex differential forms by type](/theorems/7975) [citetheorem:7975], a $(0,0)$-form is a smooth section of $\Lambda^{0,0}T^*M$. For every point $p\in M$, the fiber $\Lambda^{0,0}T_p^*M$ is canonically $\mathbb C$, so $\Lambda^{0,0}T^*M$ is the product complex line bundle $M\times\mathbb C$. Hence there is a natural complex-linear identification \begin{align*} \Omega^{0,0}(M)=C^\infty(M;\mathbb C). \end{align*} Thus an element of $\Omega^{0,0}(M)$ is a smooth map $f:M\to\mathbb C$. [/step] [step:Remove the incoming image term in degree $(0,0)$] Let \begin{align*} \bar\partial_{0,0}:\Omega^{0,0}(M)\to\Omega^{0,1}(M) \end{align*} be the Dolbeault operator in bidegree $(0,0)$, and let \begin{align*} \bar\partial_{0,-1}:\Omega^{0,-1}(M)\to\Omega^{0,0}(M) \end{align*} be the zero incoming map from the zero [vector space](/page/Vector%20Space) $\Omega^{0,-1}(M)=\{0\}$. By the definition of Dolbeault cohomology, \begin{align*} H^{0,0}_{\bar\partial}(M)=\ker(\bar\partial_{0,0})\big/\operatorname{im}(\bar\partial_{0,-1}). \end{align*} There are no forms of type $(0,-1)$, so $\Omega^{0,-1}(M)=\{0\}$ and the incoming image is $\{0\}$. Therefore \begin{align*} H^{0,0}_{\bar\partial}(M)=\ker\left(\bar\partial:C^\infty(M;\mathbb C)\to\Omega^{0,1}(M)\right). \end{align*} [guided] Dolbeault cohomology is defined exactly like cohomology of a cochain complex: take closed objects and quotient by exact objects. In bidegree $(0,0)$ the closed objects are smooth functions killed by \begin{align*} \bar\partial:\Omega^{0,0}(M)\to\Omega^{0,1}(M). \end{align*} The exact objects would have to come from the previous bidegree: \begin{align*} \bar\partial:\Omega^{0,-1}(M)\to\Omega^{0,0}(M). \end{align*} But negative anti-holomorphic degree is not part of the exterior algebra decomposition, so $\Omega^{0,-1}(M)=\{0\}$. Hence the image of this incoming map is $\{0\}$, and quotienting by it changes nothing. Consequently \begin{align*} H^{0,0}_{\bar\partial}(M)=\ker\left(\bar\partial:C^\infty(M;\mathbb C)\to\Omega^{0,1}(M)\right). \end{align*} Thus the theorem reduces to identifying which smooth complex-valued functions have vanishing $\bar\partial$ derivative. [/guided] [/step] [step:Identify the kernel of $\bar\partial$ with holomorphic functions] Let $f:M\to\mathbb C$ be a smooth function. Let $(U,z)$ be a holomorphic coordinate chart on $M$, where $z:U\to z(U)\subset\mathbb C^n$ has coordinate functions $z_j=x_j+i y_j$ for $1\le j\le n$. Define the coordinate representative \begin{align*} f_z:z(U)\to\mathbb C,\qquad f_z=f\circ z^{-1}. \end{align*} In this chart, \begin{align*} \bar\partial f\big|_U=\sum_{j=1}^n \frac{\partial f_z}{\partial \bar z_j}\,d\bar z_j, \end{align*} where \begin{align*} \frac{\partial}{\partial \bar z_j}:=\frac{1}{2}\left(\frac{\partial}{\partial x_j}+i\frac{\partial}{\partial y_j}\right). \end{align*} Since the forms $d\bar z_1,\dots,d\bar z_n$ form a local frame for $(0,1)$-forms, $\bar\partial f=0$ holds on $U$ if and only if \begin{align*} \frac{\partial f_z}{\partial \bar z_j}=0 \end{align*} for every $1\le j\le n$. These are precisely the [Cauchy-Riemann equations](/page/Cauchy-Riemann%20Equations) for the smooth function $f_z:z(U)\to\mathbb C$. By the smooth [Cauchy-Riemann characterization of holomorphic functions](/theorems/7013) in several complex variables, $f_z$ is holomorphic on $z(U)$ if and only if these equations hold throughout $z(U)$. Hence $f_z$ is holomorphic on $z(U)$ if and only if $\bar\partial f=0$ on $U$. Since holomorphicity on a complex manifold is defined chartwise and the chart $(U,z)$ was arbitrary, $f$ is holomorphic on $M$ if and only if $\bar\partial f=0$ on $M$. [/step] [step:Conclude the degree zero Dolbeault identification] From the previous steps, \begin{align*} H^{0,0}_{\bar\partial}(M)=\ker\left(\bar\partial:C^\infty(M;\mathbb C)\to\Omega^{0,1}(M)\right)=\mathcal O(M). \end{align*} The equality is complex-linear because $\bar\partial$ is a complex-linear operator and $\mathcal O(M)$ is closed under complex-linear combinations. This proves that the degree $(0,0)$ Dolbeault cohomology of $M$ is the vector space of holomorphic complex-valued functions on $M$. [/step]