I have a complex smooth projective scheme $X$ with the sheaf of Kähler differentials $\Omega_{X/\mathbb{C}}$ (or only $\Omega$). Denote its analytification $X^{an}$ with analytification morphism $h:X^{an} \rightarrow X$. I already know that $h^*\Omega_{X/\mathbb{C}} \cong \Omega_{X^{an}}$ the sheaf of holomorphic $1$-forms on $X^{an}$. My goal is to define a wedge product:

\begin{align*} H^{q_1}(X, \Omega^{p_1}) \times H^{q_2}(X,\Omega^{p_2}) &\rightarrow H^{q_1+q_2}(X, \Omega^{p_1+p_2}) \\ (\alpha, \beta) \rightarrow \alpha \wedge \beta \end{align*}

My first attempt is to use Čech cohomology and mimic the cup product of singular cohomology. In our setting Čech cohomology is the same as sheaf cohomology, moreover there is a finite standard affine covering of $X$, on which we can compute the cohomology. Let $\check{\alpha} \in \check{C}^{p_1}(X, \Omega^{q_1})$ and $\check{\beta} \in \check{C}^{p_2}(X, \Omega^{q_2})$ be the representative of $\alpha, \beta$, respectively, then we can define a represetative of $\alpha \wedge \beta$ on the intersection $i_0,i_1, \ldots, i_{p_1+p_2}$ as $\check{\alpha}_{i_0,i_1, \cdots, i_{q_1}} \wedge \check{\beta}_{i_{q_1}, i_{q_1 + 1}, \cdots, i_{q_1 + q_2}}$ where the wedge here means:

\begin{align*} \Omega^{p_1}(U_{ i_0,i_1, \cdots, i_{q_1}}) \times \Omega^{p_2}(U_{i_{q_1}, i_{q_1+1}, \cdots, i_{q_1 + q_2}}) &\rightarrow \Omega^{p_1 + p_2}(U_{i_0,i_1, \cdots, i_{q_1+q_2}}) \\ \gamma_1 \times \gamma_2 &\mapsto \gamma_1|_{U_{i_0,i_1, \cdots, i_{q_1+q_2}}} \wedge \gamma_2|_{U_{i_0,i_1, \cdots, i_{q_1+q_2}}}. \end{align*}

I can prove that this is well-defined. The same thing can be done to get a wedge product:

$$ H^{q_1}(X^{an}, \Omega^{p_1}_{X^{an}}) \times H^{q_2}(X^{an},\Omega^{p_2}_{X^{an}}) \rightarrow H^{q_1+q_2}(X^{an}, \Omega^{p_1+p_2}_{X^{an}}) $$

But we also know that $H^{q}(X^{an}, \Omega^{p}_{X^{an}}) = H^{p,q}_{\bar\partial}(X^{an},\mathbb{C})$ the Dolbeault cohomology and here we have the usual wedge product between differential forms. So my questions are:

1. Is there a better way to define the wedge product between the cohomology groups categorically without using Čech cohomology

2. With either definition how can we prove the compatibility between the second and third row:

$\require{AMScd}$ \begin{CD} H^{q_1}(X, \Omega^{p_1}) \times H^{q_2}(X,\Omega^{p_2})@>>> H^{q_1+q_2}(X, \Omega^{p_1+p_2})\\ @VVV @VVV\\ H^{q_1}(X^{an}, \Omega^{p_1}_{X^{an}}) \times H^{q_2}(X^{an},\Omega^{p_2}_{X^{an}}) @>>> H^{q_1+q_2}(X^{an}, \Omega^{p_1+p_2}_{X^{an}})\\ @VVV @VVV\\ H^{p_1,q_1}_{\bar{\partial}}(X^{an}, \mathbb{C}) \times H^{p_2,q_2}_{\bar{\partial}}(X^{an},\mathbb{C}) @>>> H^{p_1+p_2,q_1+q_2}_{\bar{\partial}}(X^{an},\mathbb{C}) \end{CD}

Thank you in advance.