Let $\Pi$ be a Poisson structure on a manifold $M$. Then we can define a differential $d$ on the complex $\Lambda^{\bullet}M$
$$
C^{\infty}M \to TM \to...\Lambda^kTM \to...
$$
in the following way:
$$
d = [\Pi,\cdot],
$$
where $[\cdot,\cdot]$ is the Schouten-Nijenhuis bracket. The cohomology of this complex is called Poisson cohomology.


I have several question about this construction:


1. Suppose $M=\mathbb R^{2d+s}$ with a Poisson structure $\sum\limits\_{i=1}^{d}\frac{\partial}{\partial q\_i}\wedge \frac{\partial}{\partial p\_i}$. Is it true that Poisson cohomology of $M$ are zero?
2. Is this construction functorial? I mean, let $f\colon M\to N$ be a poisson map, then do we have a map of complexes $f\_\*\colon \Lambda^{\bullet}M \to \Lambda^{\bullet}N$?
3. Let $F\colon M\times I \to N$ be a homotopy such that for each $t$ $F\_t$ is a Poisson map. Is it true that $F\_0$ and $F\_1$ induce the same map from Poisson cohomology of $M$ to Poisson cohomology of $N$?



Poisson cohomology
First of all, one can think of Poisson cohomology (or, rather, Poisson cochains) as functions on the $QP$-manifold $T^\*[1]M$ with the differential as you describe. In particular, it has a natural odd Poisson ($\mathbb{P}\_2$) structure given by the Schouten bracket.


1. You can split $T^\*[1](\mathbb{R}^{2d+s}) = T^\*[1](\mathbb{R}^{2d})\times T^\*[1](\mathbb{R}^s)$ and so the same decomposition applies to Poisson cohomology. Since $\mathbb{R}^{2d}$ is symplectic, its Poisson cohomology coincides with the de Rham cohomology which is quasi-isomorphic to $\mathbb{R}$ in degree 0. Therefore, Poisson cohomology of $\mathbb{R}^{2d+s}$ coincides with the Poisson cohomology of $\mathbb{R}^s$ with the zero Poisson structure, i.e. functions on $T^\*[1](\mathbb{R}^s)$.
2. Given a map of manifolds $M\rightarrow N$, there is no natural map on their cotangent bundles. Instead, there is a Lagrangian correspondence
$$T^\* M \leftarrow T^\*N\times\_N M \rightarrow T^\* N.$$
The same construction works with a shift and so instead of a map one has a dg bimodule between Poisson cohomologies of $M$ and $N$. The underlying graded bimodule is $C^\infty(T^\*[1] N\times\_N M)$. The differential is constructed as follows. Recall that for any coisotropic submanifold $C\hookrightarrow X$ the graded commutative algebra $\bigwedge^\bullet N\_{C/X}$ carries a natural differential (a version of the differential you wrote). The map $M\rightarrow N$ is Poisson iff its graph $M\rightarrow M \times N$ is coisotropic and taking the exterior algebra on the normal bundle of this coisotropic you recover $C^\infty(T^\*[1] N\times\_N M)$ and that's the required differential.
3. Given such a homotopy, you get a homotopy of (dg) bimodules from point 2, which seems to be the best you can hope for.


