very present in the literature, for which significant topological information is available.

Cohomology algebra and twisted cohomology {#sect:cohomology}
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Let $X$ be an arbitrary topological space and ${\mathbb{K}}$ a unitary commutative ring. The [*cohomology algebra*]{} of $X$ with (untwisted) coefficients in ${\mathbb{K}}$ is the graded ring $H^*(X, {\mathbb{K}})=\oplus_{i \geq 0}H^i(X, {\mathbb{K}})$ with a multiplication $H^i \otimes H^j \overset{\cup} \rightarrow H^{i+j}$, called cup product, which is commutative in the graded sense, that is, $a \cup b= (-1)^{ij} b\cup a$, for $a\in H^i$ and $b\in H^j$. See for instance [@H]. By convention, $H^*(X):=H^*(X,{\mathbb{Z}})$.

\[deRham\] The cohomology of a differentiable ($\mathcal{C}^{\infty}$) manifold $X$ with real coefficients, $H^*(X,{\mathbb{R}})$, is isomorphic to the cohomology of the de Rham algebra $\Omega^*_{dR}X$, where the cup product is induced by the exterior product on differential forms.

\[lineardecomp\] When $X=M_{{{\mathcal A}}}$ is the complement of a fiber-type arrangement, we have an additive combinatorial decomposition of $H^*(M_{{{\mathcal A}}})$. More precisely, in the notations of Definition \[fibertype\]: $$\label{factor}
H^*(M_{{{\mathcal A}}}) \cong H^*(F_1) \otimes \dots \otimes H^*(F_l)\, ,$$ as graded abelian groups.

Naturally, one may ask whether combinatorial determination holds for the cohomology algebra of an arbitrary arrangement complement. The answer is positive, as we shall see in Theorem \[thm:OS\]. To state 