Let $V$ be a complex manifold and $D \subset V$ a smooth divisor. 


**Question 1** Is $H^i(V \setminus D, \mathbb{C}) \simeq 
\mathbb{H}^i ( V, \Omega^{\bullet}\_V(\log D)) $ ? 


**Question 2**(Edited) Ok, 1 is true. 
Is it possible to define naturally a homomorphism $H^2(V \setminus D, \mathbb{C}) \rightarrow H^1(V, \Omega\_V^1 (\log D))$? 


(In my case $V$ is of the form $U \setminus p $ where $U$ is a $3$-dimensional smooth Stein space and $D$ is of the form $\Delta \setminus p$ where $\Delta \subset U$ is a divosor with an isolated singularity at $p$. 
Then $H^1(V, \Omega\_V^1 (\log D))$ is the set of 1st order deformations of the pair $(U, \Delta)$. Since $\Delta$ has only isolated singularities, this is finite dimensional.) 


I think it is true when $V$ is compact. How about non-compact case? 



hypercohomology of logarithmic de Rham complex of complement of smooth divisor in smooth variety
**Question 1:** Sure, this is true. Another reference (beyond what Donu pointed out) is chapter 8 of Claire Voisin's *Hodge theory of ...*. But the point is $\Omega\_X^{\bullet}(\text{log} D)$ is quasi-isomorphic to the pushforward of a resolution of $\mathbb{C}$ on $X \setminus D$. By the way, this holds not just for smooth $D$, but also for normal crossing $D$. 


**Question 2:** I don't think you have maps of complexes as you describe. For example, why do we have the map $\Omega\_V^1(\text{log} D) \to \Omega^{\bullet}\_V(\text{log} D)$? If I had a map of complexes, then the image of $d : \Omega\_V^0(\text{log} D) \to \Omega\_V^1(\text{log} D)$ would be zero (ie, the diagram would commute). 


**EDIT:** Whoops, it looks like Donu beat me to this in the comments.


**Revised Question 2:** I don't see why this should hold in general. However, if you write down the relevant spectral sequence, and enough terms vanish (maybe the spectral sequence degenerates), you can be ok.


**EDIT (Response to the comment below):** No, there isn't a map in general. Even for a projective variety and $D = 0$, we only have an $E^1$ degeneration of the spectral sequence. This means that in some sense, $H^2(X, O\_X)$, $H^1(X, \Omega\_X^1)$ and $H^0(X, \Omega\_X^2)$ make up ${H}^2(X, \mathbb{C})$ (there is a filtration of the latter such that these terms make up the filtration). But we have maps:


$$H^2(X, \mathbb{C}) \to H^2(X, O\_X), \text{ and } H^0(X, \Omega\_X^2) \to H^2(X, \mathbb{C}).$$


There isn't going to be a map to the $H^1(X, \Omega\_X^1)$ term in general, unless for some reason $H^2(X, O\_X) = 0$ (in the non-compact/non-Kahler setting, things get more complicated as Donu mentioned above). Anyways, if you read a little about spectral sequences and do a couple examples from that perspective, I'm sure you'll see what's going on.


