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To deform a map $f:X \to Y$ is equivalent to deforming its graph $\Gamma \subset X\times Y$. But a deformation of a subscheme $Z \subset V$ are governed by the normal bundle $N_{Z/V}$ (the tangent spa …

There is no such map. In fact, the spectral sequence gives you a filtration on $\mathrm{Ext}^2(i_*\mathcal{O}_Y, i_*\mathcal{O}_Y)$ with three factors: $$ \mathrm{Ext}^2(\mathcal{O}_Y, \mathcal{O}_Y), …

See Lemma B.5.6 in A. Kuznetsov, Yu. Prokhorov, C. Shramov, "Hilbert schemes of lines and conics and automorphism groups of Fano threefolds", Japanese Journal of Mathematics, V. 13 (2018), N. 1, pp. 1 …

The fiber of the map over a curve $C$ is just $P(H^0(P^3,I_C(d)))$. So, the sufficient and necessary condition is that $\dim H^0(P^3,I_C(d))$ is constant on $Hilb_P$.

Each conic in $P^3$ is contained in a plane $P^2 \subset P^3$, and the conic contained in a given plane are parameterized by $P^5$. Becaue of that the Hilbert scheme of conincs in $P^3$ is a $P^5$-fib …

For a general smooth base $S$ it is not true that $T$ is regular. Assume even that $\mathcal{C} = C \times S$ for a fixed smooth curve $C$. The line bundle $\mathcal{L}$ then defines a morphism $f \co …

Du Val singularities are hypersurface singularities, hence they can be smoothed --- just replace the defining equation $F(x,y,z) = 0$ by the equation $F(x,y,z) = \epsilon$.

Let $C$ be the union of 5 lines in general position in $\mathbb{P}^2$ (hence with 10 pairwise intersection points $P_{ij}$, $1 \le i < j \le 5$) and let $F$ be the equation of $C$. We have the standar …

This is a standard projective duality argument. Let $W = H^0(\mathcal{O}_{\mathbb{P}^8}(1))$. Consider the variety $X$ of tuples $$ (P,H_1,H_2,H_3) \in V \times W^{\oplus 3} $$ such that $V \cap H_1 \ …

A morphism $F \to E$ can be thought of as a representation of the quiver $$ A_2 = \{ \bullet \to \bullet \} $$ in the category $Coh(X)$. The category $Rep(A_2,Coh(X))$ is abelian, so what you need is …

Apart of the deformations of $X$ you can consider deformations of any subscheme supported on $X$ (that is of any subscheme $Y$ such that $Y_{red} = X$). Another choice is to consider deformations of a …

You can degenerate the covering $z^n \colon \mathbb{P}^1 \to \mathbb{P}^1$ to a morphism $$ \mathbb{P}^1 \cup \mathbb{P}^1 \to \mathbb{P}^1 $$ equal to $z^{n-2}$ and $z^2$ on the first and second copi …