hypercohomology of logarithmic de Rham complex of complement of smooth divisor in smooth variety 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?   
 A: 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.
