Let $H$ be a hypeplane in $\mathbb{P}^3$ containing a point $p$ and $I\_p$ be the ideal sheaf corresponding to $p$. Consider the natural exact sequence : 


$0 \to \mathcal{O} \to \mathcal{O}(H) \to \mathcal{O}(H) \mid\_H \to 0$.


Is it true that the tensoring the exact sequence by $I\_p$ remains exact ?
I guess not, because if it is exact then we get the following exact sequence:


$0 \to I\_p \to I\_p(1) \to I\_p(1) \mid\_H \to 0$.


Note that $h^0(I\_p) = h^1(I\_p) = 0$. Thus considering the long exact sequence of cohomology of the above sequence, gives that 
 $H^0(I\_p(1)) \cong H^0(I\_p(1)\mid\_H)$, which is a contradiction as their dimensions are $3$ and $2$ respectively. Please correct me if i am wrong. 



Restricting sheaves in projective space
Essentially, you are asking whether the map $I\_p \to I\_p(1)$ induced by the multiplication with the equation of a hyperplane is injective. Since $I\_p$ is a torsion-free sheaf, it is enough for this to check the map at the generic point of $\mathbb{P}^3$. But there $I\_p$ agrees with $\mathcal{O}$, hence injectivity follows from injectivity of your first sequence, so the answer to your question is yes.


As for the contradiction you mention, it arises from your wrong interpretation of $I\_p(1)\vert\_H = I\_p(1) \otimes \mathcal{O}\_H$. One has, in fact, an exact sequence
$$
0 \to \mathcal{O}\_p \to I\_p(1) \otimes \mathcal{O}\_H \to I\_{p,H}(1) \to 0,
$$
which shows that $\dim H^0(I\_p(1) \otimes \mathcal{O}\_H) = 3$.


