I had been reading an article by Spencer Bloch. There is a remark in this text which states the surjectivity of a particular map between cohomology groups without explaining further. I had been trying to work it out without much success. I elaborate on this map: Let $X$ be a smooth projective hypersurface of degree $d$ in $\mathbb{P}^{2n+1}$ for some integer $n \ge 1$. Let $Z$ be a smooth subscheme in $X$ of codimension $n$. Using the natural map, say $\epsilon: \mathcal{I}\_Z/\mathcal{I}\_Z^2 \to \Omega^1\_X \otimes \mathcal{O}\_Z$, where $\mathcal{I}\_Z$ is the ideal sheaf of $Z$ in $X$, one gets a map $\wedge^{n-1}\epsilon:\wedge^{n-1}\mathcal{I}\_Z/\mathcal{I}\_Z^2 \to \Omega^{n-1}\_X \otimes \mathcal{O}\_Z$. Taking dual and tensoring with the canonical sheaf, say $K\_X$, one gets a map:$\epsilon':\Omega^{n+1}\_X \otimes \mathcal{O}\_Z \to \wedge^{n-1} \mathcal{N}\_{Z|X} \otimes K\_X$, where $\mathcal{N}\_{Z|X}$ is the normal sheaf which is the dual of $\mathcal{I}\_Z/\mathcal{I}\_Z^2$. Bloch claims that for $d \gg 0$, the induced map $$H^{n-1}(\Omega^{n+1}\_X \otimes \mathcal{O}\_Z) \to H^{n-1}(\wedge^{n-1} \mathcal{N}\_{Z|X} \otimes K\_X)$$ is surjective. I do not understand why this is true. Any help on this question will be very much appreciated. This question is very closely related to the study of Hodge locus and deformation of algebraic cycles.



Surjectivity of certain cohomology groups on hypersurfaces of high degree
I assume that this is example 1.4 of Bloch's semi-regularity paper. The problem is that you've stated this in a way that makes it harder than it is. He says that you choose $Z$ *first* and then you choose $X\supset Z$ with $d\gg 0$. 
So now it's just a matter of killing the appropriate cohomology group using Serre vanishing. 


Since I really don't want to put too much effort into this, say $n=2$.
Your map $\epsilon'$ is part of the sequence
$$0 \to T\_Z\otimes K\_X\to \Omega\_X^3|\_Z\to N\_{Z/X}\otimes K\_X\to 0$$
Then $H^n(T\_Z\otimes K\_X) = 
 H^2(T\_Z(d-2(2)-2))=0$ for $d\gg 0$. Which implies what you want in this case.


