[proofplan]
The proof has three parts. First, the $\partial\bar\partial$ lemma is used to replace an arbitrary $\bar\partial$-closed representative of a Dolbeault class by a $d$-closed representative and to prove that the resulting de Rham class is independent of choices. Second, injectivity is proved by taking a de Rham-exact sum of $d$-closed pure-type representatives and peeling off its highest holomorphic-degree component using the $\partial\bar\partial$ lemma. Finally, surjectivity is proved by the same filtration argument applied to an arbitrary $d$-closed degree-$k$ form, successively correcting it by $d$-exact terms until each pure-type component is separately $d$-closed.
[/proofplan]
[step:Use the $\partial\bar\partial$ lemma to choose $d$-closed representatives of Dolbeault classes]
Let $p,q\ge 0$ and let $[\alpha]_{\bar\partial}\in H^{p,q}_{\bar\partial}(X)$ be represented by a form $\alpha\in A^{p,q}(X)$ satisfying $\bar\partial\alpha=0$. By [citetheorem:8046], $\partial^2=0$ and $\partial\bar\partial+\bar\partial\partial=0$, so
\begin{align*}
\partial(\partial\alpha)=0.
\end{align*}
Also
\begin{align*}
\bar\partial(\partial\alpha)=-\partial(\bar\partial\alpha)=0.
\end{align*}
Thus $\partial\alpha\in A^{p+1,q}(X)$ is $d$-closed. It is also $\partial$-exact by definition. Since $X$ satisfies the $\partial\bar\partial$ lemma and $A^{p,q-1}(X)=0$ when $q=0$, there exists a form $\beta\in A^{p,q-1}(X)$ such that
\begin{align*}
\partial\alpha=\partial\bar\partial\beta.
\end{align*}
Define
\begin{align*}
\alpha_0:=\alpha-\bar\partial\beta\in A^{p,q}(X).
\end{align*}
Then $\alpha_0$ represents the same Dolbeault class as $\alpha$, because $\alpha_0-\alpha=-\bar\partial\beta$. Moreover,
\begin{align*}
\bar\partial\alpha_0=\bar\partial\alpha-\bar\partial^2\beta=0
\end{align*}
and
\begin{align*}
\partial\alpha_0=\partial\alpha-\partial\bar\partial\beta=0.
\end{align*}
Using [citetheorem:7004], $d=\partial+\bar\partial$, hence $d\alpha_0=0$.
[guided]
We start with a Dolbeault class $[\alpha]_{\bar\partial}\in H^{p,q}_{\bar\partial}(X)$, so by definition it has a representative $\alpha\in A^{p,q}(X)$ with $\bar\partial\alpha=0$. To send this class to de Rham cohomology, we need a representative that is closed for the full [exterior derivative](/theorems/1525) $d$, not just for $\bar\partial$.
By [citetheorem:7004], the exterior derivative on a pure-type form decomposes as $d=\partial+\bar\partial$. Since $\bar\partial\alpha=0$, the only obstruction to $d\alpha=0$ is the term $\partial\alpha$. We therefore study
\begin{align*}
\partial\alpha\in A^{p+1,q}(X).
\end{align*}
Using the Dolbeault relations from [citetheorem:8046], we have
\begin{align*}
\partial(\partial\alpha)=\partial^2\alpha=0.
\end{align*}
The same relation gives
\begin{align*}
\bar\partial(\partial\alpha)=-\partial(\bar\partial\alpha)=0.
\end{align*}
Thus $\partial\alpha$ is closed under both $\partial$ and $\bar\partial$, hence is $d$-closed. It is also $\partial$-exact, because it is the $\partial$ of $\alpha$.
Now apply the $\partial\bar\partial$ lemma to the pure-type form $\partial\alpha$. The hypotheses required are exactly the ones just verified: the form is pure-type, $d$-closed, and exact for one of the operators in the $\partial,\bar\partial,d$ system. Therefore there exists $\beta\in A^{p,q-1}(X)$ such that
\begin{align*}
\partial\alpha=\partial\bar\partial\beta.
\end{align*}
Define a corrected representative by
\begin{align*}
\alpha_0:=\alpha-\bar\partial\beta.
\end{align*}
This correction does not change the Dolbeault class, because it changes $\alpha$ by a $\bar\partial$-exact form. It kills the obstruction:
\begin{align*}
\partial\alpha_0=\partial\alpha-\partial\bar\partial\beta=0.
\end{align*}
It also preserves $\bar\partial$-closedness:
\begin{align*}
\bar\partial\alpha_0=\bar\partial\alpha-\bar\partial^2\beta=0.
\end{align*}
Therefore $d\alpha_0=(\partial+\bar\partial)\alpha_0=0$. This proves that every Dolbeault class admits a $d$-closed representative.
[/guided]
[/step]
[step:Show the de Rham class does not depend on the chosen $d$-closed representative]
Let $\alpha_0,\alpha_1\in A^{p,q}(X)$ be two $d$-closed representatives of the same Dolbeault class. Then there exists $\eta\in A^{p,q-1}(X)$ such that
\begin{align*}
\alpha_1-\alpha_0=\bar\partial\eta.
\end{align*}
The form $\alpha_1-\alpha_0$ is $d$-closed because both $\alpha_0$ and $\alpha_1$ are $d$-closed. It is also $\bar\partial$-exact. By the $\partial\bar\partial$ lemma, there exists $\theta\in A^{p-1,q-1}(X)$ such that
\begin{align*}
\alpha_1-\alpha_0=\partial\bar\partial\theta.
\end{align*}
Since $\bar\partial^2=0$,
\begin{align*}
d(\bar\partial\theta)=\partial\bar\partial\theta+\bar\partial^2\theta=\alpha_1-\alpha_0.
\end{align*}
Thus $\alpha_0$ and $\alpha_1$ define the same class in $H^k_{dR}(X;\mathbb C)$, where $k=p+q$. Hence the map $\Phi_k$ is well-defined.
[/step]
[step:Prove injectivity by checking each pure-type summand is $\bar\partial$-exact]
Let
\begin{align*}
\sum_{p+q=k}[\alpha^{p,q}]_{\bar\partial}\in \bigoplus_{p+q=k}H^{p,q}_{\bar\partial}(X)
\end{align*}
be an element whose image under $\Phi_k$ is zero, where each $\alpha^{p,q}\in A^{p,q}(X)$ is $d$-closed. Define
\begin{align*}
\alpha:=\sum_{p+q=k}\alpha^{p,q}\in A^k(X;\mathbb C).
\end{align*}
The assumption $\Phi_k(\sum[\alpha^{p,q}]_{\bar\partial})=0$ means that there exists $\gamma\in A^{k-1}(X;\mathbb C)$ such that
\begin{align*}
\alpha=d\gamma.
\end{align*}
Write the type decomposition of $\gamma$ as
\begin{align*}
\gamma=\sum_{a+b=k-1}\gamma^{a,b},
\end{align*}
where $\gamma^{a,b}=0$ if $a<0$ or $b<0$.
Fix $p$ and put $q=k-p$. The $(p,q)$-component of $d\gamma=\alpha$ is
\begin{align*}
\alpha^{p,q}=\partial\gamma^{p-1,q}+\bar\partial\gamma^{p,q-1}.
\end{align*}
Define
\begin{align*}
\delta^{p,q}:=\partial\gamma^{p-1,q}\in A^{p,q}(X).
\end{align*}
Then $\delta^{p,q}$ is $\partial$-exact and $\partial$-closed because $\partial^2=0$. To verify that it is $\bar\partial$-closed, use the neighbouring component
\begin{align*}
\alpha^{p-1,q+1}=\partial\gamma^{p-2,q+1}+\bar\partial\gamma^{p-1,q}.
\end{align*}
Since $\alpha^{p-1,q+1}$ is $d$-closed, its $\partial$-component vanishes, so
\begin{align*}
0=\partial\alpha^{p-1,q+1}=\partial\bar\partial\gamma^{p-1,q}.
\end{align*}
Using $\partial\bar\partial+\bar\partial\partial=0$, we get
\begin{align*}
\bar\partial\delta^{p,q}=\bar\partial\partial\gamma^{p-1,q}=-\partial\bar\partial\gamma^{p-1,q}=0.
\end{align*}
Thus $\delta^{p,q}$ is $d$-closed and $\partial$-exact. By the $\partial\bar\partial$ lemma, there exists $\rho^{p-1,q-1}\in A^{p-1,q-1}(X)$ such that
\begin{align*}
\delta^{p,q}=\partial\bar\partial\rho^{p-1,q-1}.
\end{align*}
Therefore
\begin{align*}
\alpha^{p,q}=\bar\partial\gamma^{p,q-1}+\partial\bar\partial\rho^{p-1,q-1}=\bar\partial(\gamma^{p,q-1}-\partial\rho^{p-1,q-1}).
\end{align*}
This proves $[\alpha^{p,q}]_{\bar\partial}=0$ for every $p+q=k$, so $\Phi_k$ is injective.
[/step]
[step:Correct a closed form until all pure-type components are closed]
Let $[\omega]_{dR}\in H^k_{dR}(X;\mathbb C)$, and choose a representative $\omega\in A^k(X;\mathbb C)$ satisfying $d\omega=0$. Write its type decomposition as
\begin{align*}
\omega=\sum_{p+q=k}\omega^{p,q}.
\end{align*}
We modify $\omega$ by $d$-exact degree-$k$ forms, descending in the holomorphic degree $p$, until every component is separately $d$-closed.
Assume that for some $p$ all components of holomorphic degree greater than $p$ are already $d$-closed. Put $q=k-p$ and define the obstruction
\begin{align*}
\theta^{p,q+1}:=\bar\partial\omega^{p,q}\in A^{p,q+1}(X).
\end{align*}
The component of $d\omega=0$ in bidegree $(p,q+1)$ gives
\begin{align*}
\theta^{p,q+1}=-\partial\omega^{p-1,q+1},
\end{align*}
so $\theta^{p,q+1}$ is $\partial$-exact and $\bar\partial$-exact. Also $\bar\partial\theta^{p,q+1}=0$ by $\bar\partial^2=0$. The component of $d\omega=0$ in bidegree $(p+1,q)$ gives
\begin{align*}
\partial\omega^{p,q}+\bar\partial\omega^{p+1,q-1}=0.
\end{align*}
The term $\omega^{p+1,q-1}$ is $d$-closed by the descending induction hypothesis, so $\bar\partial\omega^{p+1,q-1}=0$ and therefore $\partial\omega^{p,q}=0$. Hence
\begin{align*}
\partial\theta^{p,q+1}=\partial\bar\partial\omega^{p,q}=-\bar\partial\partial\omega^{p,q}=0.
\end{align*}
Thus $\theta^{p,q+1}$ is $d$-closed. By the $\partial\bar\partial$ lemma, there exists a form $\beta^{p-1,q}\in A^{p-1,q}(X)$ such that
\begin{align*}
\theta^{p,q+1}=\partial\bar\partial\beta^{p-1,q}.
\end{align*}
Replace $\omega$ by
\begin{align*}
\omega+d\beta^{p-1,q}.
\end{align*}
This replacement changes $\omega$ by a $d$-exact degree-$k$ form and therefore preserves its de Rham class. Its only pure-type effects are to add $\partial\beta^{p-1,q}$ in bidegree $(p,q)$ and $\bar\partial\beta^{p-1,q}$ in bidegree $(p-1,q+1)$, so no component of holomorphic degree greater than $p$ is changed. The new $(p,q)$-component is $\omega^{p,q}+\partial\beta^{p-1,q}$, and
\begin{align*}
\bar\partial(\omega^{p,q}+\partial\beta^{p-1,q})=\theta^{p,q+1}+\bar\partial\partial\beta^{p-1,q}=\theta^{p,q+1}-\partial\bar\partial\beta^{p-1,q}=0.
\end{align*}
Since the total corrected form remains $d$-closed and all components of holomorphic degree greater than $p$ are $d$-closed, the component equation in bidegree $(p+1,q)$ also gives $\partial(\omega^{p,q}+\partial\beta^{p-1,q})=0$. Hence the corrected $(p,q)$-component is $d$-closed. Repeating this finite process for $p=k,k-1,\dots,0$ gives a $d$-closed form
\begin{align*}
\omega_0=\sum_{p+q=k}\omega_0^{p,q}
\end{align*}
cohomologous to $\omega$ such that every $\omega_0^{p,q}$ is separately $d$-closed.
[guided]
We want to represent the de Rham class $[\omega]_{dR}$ by a sum of pure-type forms that are individually $d$-closed. Start with a $d$-closed representative
\begin{align*}
\omega=\sum_{p+q=k}\omega^{p,q}.
\end{align*}
The equation $d\omega=0$ does not immediately say that each $\omega^{p,q}$ is $d$-closed. Since $d=\partial+\bar\partial$, it says that neighbouring type components cancel each other. The correction will remove these cancellations one holomorphic degree at a time while adding only $d$-exact degree-$k$ forms.
Suppose that all components with holomorphic degree greater than $p$ have already been made separately $d$-closed, and put $q=k-p$. The obstruction to $\bar\partial$-closedness of the $(p,q)$-component is the form
\begin{align*}
\theta^{p,q+1}:=\bar\partial\omega^{p,q}\in A^{p,q+1}(X).
\end{align*}
The bidegree $(p,q+1)$ component of $d\omega=0$ is
\begin{align*}
\bar\partial\omega^{p,q}+\partial\omega^{p-1,q+1}=0,
\end{align*}
so
\begin{align*}
\theta^{p,q+1}=-\partial\omega^{p-1,q+1}.
\end{align*}
Thus $\theta^{p,q+1}$ is both $\bar\partial$-exact by definition and $\partial$-exact by this component equation.
We must also verify the $d$-closedness hypothesis in the $\partial\bar\partial$ lemma. First,
\begin{align*}
\bar\partial\theta^{p,q+1}=\bar\partial^2\omega^{p,q}=0.
\end{align*}
For the $\partial$-part, inspect the bidegree $(p+1,q)$ component of $d\omega=0$:
\begin{align*}
\partial\omega^{p,q}+\bar\partial\omega^{p+1,q-1}=0.
\end{align*}
By the descending induction hypothesis, $\omega^{p+1,q-1}$ is already $d$-closed, hence $\bar\partial\omega^{p+1,q-1}=0$. Therefore $\partial\omega^{p,q}=0$, and the anticommutation relation from [citetheorem:8046] gives
\begin{align*}
\partial\theta^{p,q+1}=\partial\bar\partial\omega^{p,q}=-\bar\partial\partial\omega^{p,q}=0.
\end{align*}
So $\theta^{p,q+1}$ is $d$-closed, $\partial$-exact, and $\bar\partial$-exact. The hypotheses of the $\partial\bar\partial$ lemma are now verified, and we obtain $\beta^{p-1,q}\in A^{p-1,q}(X)$ with
\begin{align*}
\theta^{p,q+1}=\partial\bar\partial\beta^{p-1,q}.
\end{align*}
Now replace $\omega$ by
\begin{align*}
\omega+d\beta^{p-1,q}.
\end{align*}
This is the degree-correct replacement: $\beta^{p-1,q}$ has total degree $k-1$, so $d\beta^{p-1,q}$ has total degree $k$. It preserves the de Rham class because it adds a $d$-exact form. The replacement changes the $(p,q)$-component by $\partial\beta^{p-1,q}$ and the $(p-1,q+1)$-component by $\bar\partial\beta^{p-1,q}$, and therefore leaves all components of holomorphic degree greater than $p$ unchanged. For the new $(p,q)$-component we compute
\begin{align*}
\bar\partial(\omega^{p,q}+\partial\beta^{p-1,q})=\bar\partial\omega^{p,q}+\bar\partial\partial\beta^{p-1,q}.
\end{align*}
Using $\bar\partial\partial=-\partial\bar\partial$ and $\bar\partial\omega^{p,q}=\partial\bar\partial\beta^{p-1,q}$, this becomes
\begin{align*}
\bar\partial(\omega^{p,q}+\partial\beta^{p-1,q})=\partial\bar\partial\beta^{p-1,q}-\partial\bar\partial\beta^{p-1,q}=0.
\end{align*}
The total corrected form is still $d$-closed because $d(\omega+d\beta^{p-1,q})=d\omega+d^2\beta^{p-1,q}=0$. Since the higher holomorphic-degree components are unchanged and already $d$-closed, the bidegree $(p+1,q)$ component equation forces $\partial(\omega^{p,q}+\partial\beta^{p-1,q})=0$. Hence the corrected $(p,q)$-component is separately $d$-closed. Descending through the finitely many values $p=k,k-1,\dots,0$, we obtain a representative
\begin{align*}
\omega_0=\sum_{p+q=k}\omega_0^{p,q}
\end{align*}
of the same de Rham class such that each $\omega_0^{p,q}$ is separately $d$-closed.
[/guided]
[/step]
[step:Assemble the inverse image and conclude surjectivity]
For the form $\omega_0$ constructed above, each component $\omega_0^{p,q}$ is $d$-closed and has pure type. In particular $\bar\partial\omega_0^{p,q}=0$, so it defines a Dolbeault class
\begin{align*}
[\omega_0^{p,q}]_{\bar\partial}\in H^{p,q}_{\bar\partial}(X).
\end{align*}
By construction,
\begin{align*}
\Phi_k\left(\sum_{p+q=k}[\omega_0^{p,q}]_{\bar\partial}\right)
=
\left[\sum_{p+q=k}\omega_0^{p,q}\right]_{dR}
=
[\omega_0]_{dR}
=
[\omega]_{dR}.
\end{align*}
Thus $\Phi_k$ is surjective. Together with injectivity and well-definedness, this proves that
\begin{align*}
\Phi_k:\bigoplus_{p+q=k}H^{p,q}_{\bar\partial}(X)\longrightarrow H^k_{dR}(X;\mathbb C)
\end{align*}
is an isomorphism for every $k\ge 0$.
[/step]
