I am trying to compute the Chern classes of the restriction of a rank two vector bundle on $\mathbb{P}^3$, denoted by $E$, with fixed Chern classes, $c\_1(E) = c\_1$ and $c\_2(E) = c\_2$, to a hyperplane $H \subset \mathbb{P}^3$. For this let $s$ be the equation of the hiperplane $H$ and consider the exact sequence:


$$0 \to \mathcal{O}\_{\mathbb{P^3}}(-1) \to^{s} \mathcal{O}\_{\mathbb{P^3}} \to \mathcal{O}\_{\mathbb{P^3}}|\_{H} \to 0 $$


tensoring this sequence by $E(k)$ where $k$ is some integer, one has : 


$$0 \to E(k-1) \to^{s} E(k) \to E(k)|\_{H} \to 0 $$


and computing the Chern classes one sees that $c\_1(E(k)|\_{H}) = 2$ and $c\_2(E(k)|\_{H}) = 3-2k-c\_1$


But this seems to be very strange to me, once the first Chern class of the restriction does not change with the Chern classes of the bundle $E$, and the second Chern class of $E$ does not play any role in it. Moreover, if this was true, one will have that $c\_1(E(k-1)|\_{H}) = c\_1(E(k)|\_{H})=2$ for a rank two vector bundle on $\mathbb{P}^2$ which is clearly an absurd. 


Since I triple checked my computations, I would like to know if the multiplicative property of the Chern classes on the exact sequences does not hold here. 


Thanks in advance. 



Restriction of vector bundles
OK, I'll write my comment as an answer. The computation of the OP using the multiplicative property of exact sequences gives the Chern classes of the coherent sheaf $i\_\*(E\_{|H})$ on $\mathbb{P}^3$, where $i:H\hookrightarrow \mathbb{P}^3$ is the inclusion map. They are different from the Chern classes of the vector bundle $E\_{|H}$; they are related, but in a nontrivial way, through the Grothendieck-Riemann-Roch theorem. 


