Let $X$ be a smooth hypersurface of degree $d$in $\mathbb{P}^n$ for some $n \ge 3$ and $d>0$. Let $Z\_1, Z\_2$ be codimension $1$ closed subschemes in $X$ which are locally of complete intersection in $X$. Suppose that the natural morphism $\pi:H^0(\mathcal{O}\_X(d)) \to H^0(\mathcal{O}\_{Z\_1}(d))$ is surjective. Assume that, $Z\_1 \subset Z\_2$. Then the above morphism factors through $H^0(\mathcal{O}\_{Z\_2}(d))$. Denote by $\pi\_1$ (resp. $\pi\_2$) the morphism from $H^0(\mathcal{O}\_X(d))$ (resp. $H^0(\mathcal{O}\_{Z\_2}(d))$) to $H^0(\mathcal{O}\_{Z\_2}(d))$ (resp. $H^0(\mathcal{O}\_{Z\_1}(d))$). Is it true that the kernel of $\pi\_2$ is contained in the image of $\pi\_1$?



A Castelnuovo-Mumford regularity type statement
No, that is not true. Is there some reason you want that? Maybe if you explain more, then the MO community could help you find a theorem to help. Anyway, take $X$ to be a smooth quadric surface in $\mathbb{P}^3$. Let $e\geq 0$ be an integer, e.g., $e=d=2$. Take $Z\_1$ to be the union of $e+1$ disjoint lines contained in $X$. Take $Z\_2$ to be the union of $m\geq e+2$ disjoint lines contained in $X$ and containing $Z\_1$. Then the following restriction map is an isomorphism, $$ \pi\_e:H^0(X,\mathcal{O}\_X(e)) \to H^0(Z\_1,\mathcal{O}\_{Z\_1}(e)). $$  

On the other hand, the map $\pi\_1$ is not surjective, and the kernel $\pi\_2$ is not injective. Yet, since $\pi\_e$ is injective, and since $\pi\_2\circ \pi\_1$ equals $\pi\_e$, no element in the image of $\pi\_1$ can be in the kernel of $\pi\_2$.


