If $\Omega^p$ is the sheaf of holomorphic $p$-forms on a complex manifold $M$, then Dolbeault's theorem states
$$
H^{p,q}(M) \cong H^q(M,\Omega^p)
$$
Setting $p=0,q=1$, we get
$$
H^{0,1}(M) \cong H^1(M,\cal O) \cong \check{H}^1(\cal O)
$$
where $\cal O$ is the structure sheaf, and $\check{H}$ is Cech cohomology. Is there an explicit way to see the isomorphism between Cech cohomology and Dolbeault cohomology in this case?
Explicitly Understanding a Case of Dolbeault’s Theorem.
algebraic-geometrycomplex-geometrysheaf-cohomology
Related Solutions
New answer:
I'm going to explain in what sense the answer is yes, and why the "counterexample" I gave in my last answer was contrived.
First, recall that the cohomology $H^i(X, F) = R^i\Gamma(X, F)$ is only defined up to natural isomorphism, and depends on the choice of injective resolution of $F$. So, let's denote by $H^i(F,I)$ the cohomology with respect to an injective resolution $I$ of $F$.
If we pick a different resolution $J$ there is a natural map $H^i(F,I) \cong H^i(F,J)$ which doesn't depend on all the choices you make to construct it. So, if $I$ happens to be a resolution by $\mathcal O$-modules, this makes $H^i(F,I)$ an $\mathcal O$-module, and even if $J$ is not a resolution by $\mathcal O$-modules the natural isomorphism gives its cohomology an $\mathcal O$-module structure.
Since the map from the Cech sheaf to an injective resolution also gives natural maps $\check H^i(X,F) \to H^i(F,I)$ and $\check H^i(X,F) \to H^i(F,J)$, these will respect both the $\mathcal O$-action on $I$ and on $J$.
So in this sense the answer is yes, the map is always an isomorphism of $\mathcal O$-modules.
The example I gave originally is actually pretty contrived. It arose from this observation: If you take a complex of $\mathcal O$-modules $J$ which is an injective resolution of $F$ as abelian groups, but for which the augmentation $\varepsilon : F \to J$ is not $\mathcal O$-linear, then the map $\check H^i(X,F) \to H^i(F,J)$ will not be $\mathcal O$-linear, nor will the natural isomorphism $H^i(F,I) \cong H^i(F,J)$ with respect to be $\mathcal O$-actions.
Original answer:
How do we prove that sheaf cohomology and cech cohomology are the same anyway? For reference see a nice note on this here: pub.math.leidenuniv.nl/~edixhovensj/teaching/2011-2012/AAG/lecture_14.pdf or see Hartshorne.
Basically you can build a resolution of $\mathcal F$ by making a sheaf-version of the Cech complex (whose global sections is the usual Cech complex and which is a sheaf of $\mathcal O_X$-modules). After you check this is a resolution, it is a fact that any resolution will map to an injective resolution. This involves making choices, but since the choices are unique up to homotopy you get the well-defined map from Cech cohomology to sheaf cohomology. But what if you switch the category from $\mathcal O_X$-modules to sheaves of abelian groups - are the choices still unique up to homotopies taken from the other category?
The answer is no. Consider the case $X$ is a point. Let $\mathcal F = \mathcal O_X = \underline{k}$ be the constant sheaf. If we compute in the category of sheaves of abelian groups then we have the freedom to replace $\mathcal F$ with an injective resolution. One injective resolution is $\phi: \underline{k} \to \underline{k}$ taken to be any isomorphism chosen specifically not to be $k$-linear but only additive. Then in this case the map between sheaf cohomology and cech cohomology is not $\mathcal O$-linear by design because it coincides with $\phi$ itself.
So, if you want an $\mathcal O$ linear map you can get one for free, and I think it is pretty standard to assume this map is $\Gamma(\mathcal O,X)$-linear.
Here is a version of the $\bar \partial$-Poincaré lemma as formulated in [1]
Proposition 2.31 Let $\alpha$ be a $C^1$-form of type $(p,q)$ with $q > 0$. If $\bar \partial \alpha = 0$, then there locally exists on $X$ a $C^1$-form $\beta$ of type $(p,q-1)$ such that $\alpha = \bar \partial \beta$.
Now to your question:
It's not entirely clear to me, however, if we also have a ∂-Poincarè Lemma, and assuming we have it, can we use it to create a "complex de rham cohomology" which is based entirely on holomorphic forms and doesn't use smooth ones?
This is not immediately possible, because if you formulate a similar $\partial$-Poincaré lemma, and apply it to a holomorphic form $\alpha$, you do not know if $\beta$ is holomorphic as well. However, it is still true that one can build a holomorphic de Rham complex, using the sheaves $\Omega^k$ of holomorphic differentials. It looks like one would expect: $$0 \to \mathbb C_X \to \mathcal O_X \xrightarrow{\partial} \Omega^1_X \xrightarrow{\partial} \Omega^2_X \to \dotsc \xrightarrow{\partial} \Omega^n_X \to 0$$ To show that this is an exact complex of sheaves, one can for each $k$ look at the resolution $$0 \to \Omega^k \xrightarrow{\bar \partial} \mathcal A^{k,0} \xrightarrow{\bar \partial} \mathcal A^{k,1} \dotsc$$ and those are exact by the $\bar \partial$-Poincaré lemma. Here $\mathcal A^{p,q}$ is the sheaf of $C^\infty$-forms of type $(p,q)$. Together they form a double complex $$(A^{\bullet, \bullet}, \partial, \bar \partial),$$ whose total complex is the de Rham complex $(\mathcal A^\bullet, d)$ with complex coefficients, as $\mathcal A^k = \bigoplus_{p+q=k} \mathcal A^{p,q}$. All of this shows that the holomorphic de Rham complex $(\Omega^\bullet, \partial)$ is quasi-isomorphic to the complexified de Rham complex. Hence it computes the de Rham cohomology $$H^k_{dRh}(X) \otimes \mathbb C = H^k(X, \mathbb C_X),$$ where on the right hand side I mean sheaf cohomology.
[1] Voisin, Hodge Theory and Complex Algebraic Geometry, I
Best Answer
Consider a class $[\alpha]\in H^{0,1}(M)$, we can find a fine enough open cover $\{U_i\}_{i\in I}$ so that $\alpha$ is locally $\bar\partial$-exact, so $\alpha|_{U_i}=\bar\partial \beta_i$ for some smooth functions $\beta_i$. Notice that on the intersection $U_i\cap U_j$, $\bar\partial(\beta_i-\beta_j)=0$, so $\beta_{ij}=\beta_i-\beta_j$ is holomorphic on $U_i\cap U_j$. You can verify that $U_{ij}$ defines a Čech 1-cocycle, this gives a Čech cohomology class. This does not depend on the choice of representative $\alpha$, because if we choose $\alpha +\bar\partial \gamma$ instead each $\beta_i$ will be changed to $\beta_i+\gamma$ and $\beta_{ij}$ will be unchanged.
For the reverse direction, by Leray's theorem the Čech cohomology can be represented by Čech cocycle in a Leray cover. Fix one such cover and a Čech 1-cocyle $\{f_{ij}\in \mathcal O(U_i\cap U_j\}_{i,j\in I}$, it is always possible to find smooth functions $f_i\in C^{\infty}(U_i)$ so that $f_{ij}=f_i-f_j$ due to partition of unity. (See for example, Griffiths-Harris.) Then $\{\bar\partial f_i\}$ glues together to give a $(0,1)$-form. This again is independent on the choice of $f_i$'s, if $f_{ij}=f_i-f_j=g_i-g_j$, then $f_i-g_i=f_j-g_j$ so $\{f_i-g_i\}_{i\in I}$ glues to a global function.