Hilbert scheme is isomorphic to moduli space of sheaves over a CY3 I  a question regarding example 1.2 of the paper "Cohomological Donaldson-Thomas theory" by Balazs Szendroi, which I reproduce below: Let $Y$ be a CY3.

For the case of sheaves of rank 1, one can be more explicit about the stability condition involved. Given a torsion-free sheaf $E$ of rank 1 on $Y$, there is an injective map of sheaves $E\to\det E$ to the determinant of $E$, which must be an invertible sheaf. Up to tensoring by this invertible sheaf, we can assume that the determinant is trivial, and then we get an embedding $E\to\mathcal{O}_Y$ with cokernel the structure sheaf $Z\subset Y$ of a subscheme of $Y$, necessarily supported in codimension two. The parameter space of such embedded subschemes is the Hilbert scheme of codimension-two subschemes of $Y$; under the assumption $H^1(\mathcal{O}_Y)=0$, the moduli scheme of sheaves is isomorphic to the Hilbert scheme [...]

My main question is how to see the isomorphism mentioned at the end.
I think there is a bijection between both spaces. Indeed, given a codimension two subscheme $Z\in Hilb$ we map it to its ideal sheaf in $\mathcal{M}$, and conversely given a torsion-free rank 1 sheaf we apply the construction above to get a codimension-two subscheme. However, I don't know enough about the construction of these moduli spaces to know how to show these functions induce scheme maps.

(1) Can someone comment on this?

Once we show there is a scheme map defining a bijection I believe the standard method to show it's an isomorphism is showing the tangent spaces are isomorphic. I've read that the tangent space of the Hilbert scheme at some $Z\subset Y$ is $H^0(Z,N_{Z|Y})$ and the tangent space at $E$ of the moduli space of sheaves is $\operatorname{Ext}^1(E,E)$. Here's where I believe the equality $H^1(\mathcal{O}_Y)=0$ comes into play.

(2) In the same spirit as above, is there a straightforward way to present how the induced map on the tangent spaces looks?
(3) How does $H^1(\mathcal{O}_Y)=0$ come into play here?

I'm not looking for a fully detailed proof, some comments highlighting the main points of it would be great and much appreciated.
 A: I want to make the comment that this is a very interesting and nontrivial question. There is a long discussion on this in this MO post, which should clarify a lot of things. 
Let me show how to produce an explicit map between tangent spaces in the special case where $X\hookrightarrow Y$ is a local complete intersection (here $Y$ is nonsingular of arbitrary dimension). The spectral sequence $H^p(Y,\mathscr Ext^q(\mathscr I,\mathscr I))\Rightarrow\textrm{Ext}^{p+q}(\mathscr I,\mathscr I)$ gives us an exact sequence of low degree terms
which looks like
$$
0\to H^1(Y,\mathscr Hom(\mathscr I,\mathscr I))\to \textrm{Ext}^1(\mathscr I,\mathscr I)\to H^0(X,\mathscr Ext^1(\mathscr I,\mathscr I))\to
H^2(Y,\mathscr Hom(\mathscr I,\mathscr I))\to 0
$$
but by assumption we have the identification
$\mathscr Ext^i(\mathscr I,\mathscr I)=\bigwedge^iN_{X/Y}$, so the sequence becomes
$$
0\to H^1(Y,\mathscr O_Y)\to \textrm{Ext}^1(\mathscr I,\mathscr I)\to H^0(X,N_{X/Y})\to
H^2(Y,\mathscr O_Y)\to 0.
$$
So in the CY3 case, if you assume $H^1(Y,\mathscr O_Y)=0$ you get an isomorphism
$$
\textrm{Ext}^1(\mathscr I,\mathscr I)\,\widetilde{\to}\, H^0(X,N_{X/Y}),
$$
so
you can identify the moduli spaces to first order if $X\subset Y$ is lci. Following the references given in the paper you are reading, a full proof of the scheme isomorphism can probably be extracted from Theorem $2.7$ of this paper. 
