Let $S$ be a smooth complex projective surface. We let define an intersection form $()\cdot()$ on $\mathsf{Pic}(S)$ by setting $$D\cdot D':=\mathcal{O}_S(D)\cdot\mathcal{O}_S(D')$$ where the notations are the standard ones, used in the book Complex Algebraic Surfaces  Beauville. I've read in many textbooks that this form is induced by the cup product as follows. By taking the exponential sequence $0\longrightarrow \mathbb{Z}\longrightarrow \mathcal O_S\longrightarrow \mathcal O^\ast_S\longrightarrow 0$ we get a map $$c\colon H^1(S,\mathcal O^\ast_S)=\mathsf{Pic}(S)\longrightarrow H^2(S,\mathbb{Z}),$$ then $c(D)\smile c(D')=D\cdot D'$, where we use the canonical identification $H^4(S,\mathbb{Z})=\mathbb Z$. Anyway I've found no proof of this fact. Can you help me to check it or give me a reference for the proof?
I think that this can be extracted from the first chapter of Beauville's book on algebraic surfaces. First note that if D and D' are smooth irreducible curves that meet transversally, then the identity $C(D) \cup C(D') = D \cdot D'$ holds and is equal to the number of settheoretic intersection points of $D$ and $D'$. This is one of the standard ways to compute the intersection product in smooth topology, recall that transversal intersection points of complex submanifolds always contribute $+1$. The right hand side is equal to this by Definition 1.3 of the intersection form (which is proven to be equal to the sheaftheoretic definition in Theorem 1.4).
The issue of finding transversal representatives is treated by Fact 1.7 (which is originally a theorem of Serre). It states that any divisor $D \subset S$ may be written (up tp linear equivelance) as $D= AB$ where $A$ and $B$ are smooth curves, $A$ is a hyperplane section and $B$ is a positive multiple of a hyperplane section (for a possibly different projective embedding). The advantage of having hyperplane sections is that we can peturb them (by perturbing the corresponding hyperplane). This gives us enough freedom to find transversal representatives for any two divisors (up to a linearly equivalent, possibly noneffective representatives), and allows us to reduce to the above case.

$\begingroup$ The main issue is just proving that, for a prime divisor, the element $c(D)$ is the Poincarè dual of $D$. $\endgroup$ – Vincenzo Zaccaro Sep 29 '18 at 0:12

$\begingroup$ One may check that the map $c$ maps $D$ to $c_{1}(\mathcal{O}(D))$. If D is effective then there will be a section $s \in H^{0}(\mathcal{O}(D))$ whose zeroset is $D$. Then by definition of the first Chern class, this will be the Poincare dual of $D$. Now we can extend by linearity. $\endgroup$ – Nick L Sep 29 '18 at 7:47