Proof. By Proposition 3.1, if $ S\subset \mathcal {X}$ is any deformation of $S_0$ not in $\mathcal {X}_0$ , then all irreducible components of the general fibre of S have, in a sufficiently small neighborhood of p, only isolated singularities. We want to prove that if the $T_1$ singularity of $S_0$ at p is limit of an isolated singularity, then this is a node. We argue as in the proof of Proposition 3.1.

Let $\mathbb {D}_\epsilon =\mathbb {D}(\underline 0,\epsilon )\subset \mathbb {A}^1$ be the open disc with center at the origin and radius $\epsilon $ and let ${\mathbb {D}_\epsilon ^o=\mathbb {D}(\underline 0,\epsilon )\setminus \{a+i0\,|\, 0<a<\epsilon \}}$ . We denote by $t=a+ib$ the coordinate on $\mathbb {D}_\epsilon $ and by $(x,y,z,u)$ the coordinates in $\mathbb {A}^4$ . In $\mathbb {A}^4\times \mathbb {D}_\epsilon ^o$ , we consider a one parameter family of $3$ -folds

$$ \begin{align*}\mathcal S_t: p(x,y,z,u;t)=0,\,\, t\in \mathbb{D}_\epsilon^o, \end{align*} $$

where $p(x,y,z,u;t)$ is a polynomial in $x,y,z,u$ whose coefficients are holomorphic functions on $\mathbb {D}_\epsilon ^o\setminus \underline 0$ , continuous in $\underline 0$ , and the one parameter family of $3$ -folds

$$ \begin{align*}\mathcal{X}_t: xy=t,\,\, t\in \mathbb{D}_\epsilon^o. \end{align*} $$

Assume that the surface

(3.1) $$ \begin{align} S_0=\mathcal S_0\cap \mathcal{X}_0: \left\{\begin{array}{l} p(x,y,z,u;0)= x+y+p_2(x,y,z,u)+o(3)=0\\ xy=0\,\, \end{array}\right. \end{align} $$

has a $T_1$ singularity at $\underline 0$ , where $p_2(x,y,z,u)$ is the homogeneous part of degree $2$ of $p(x,y,z,u;0)$ , where $o(3)$ is the sum of terms of degree at least $3$ in $p(x,y,z,u;0)$ , and where, by assumption, $p_2(0,0,z,u)$ has nonzero discriminant.

Assume that, for $t\neq 0$ , there exists $q(t)=(x(t), y(t),z(t),w(t);t)\in S_t=\mathcal S_t\cap \mathcal {X}_t$ specializing to $\underline 0$ , as t goes to $\underline 0$ and such that $S_t$ has a singular point at $q(t)$ . Thus, $\mathcal S_t$ is smooth at $q(t)$ since $\mathcal S_0$ is smooth at $q(0)=\underline 0$ , and we have that

$$ \begin{align*}T_{q(t)}(\mathcal S_t)= T_{q(t)}(\mathcal{X}_t). \end{align*} $$

In particular, there exists a function $c(t)$ , which is nonzero if $t\neq 0$ , such that

$$ \begin{align*} y(t)(x-x(t))+x(t)(y-y(t))& = c(t)\frac{\partial p}{\partial x}|_{q(t)}(x-x(t))+c(t)\frac{\partial p}{\partial y}|_{q(t)}(y-y(t))\\ &\quad + c(t)\frac{\partial p}{\partial z}|_{q(t)}(z-z(t))+c(t)\frac{\partial p}{\partial u}|_{q(t)}(u-u(t)), \end{align*} $$

from which we deduce that

(3.2) $$ \begin{align} y(t)=c(t)\frac{\partial p}{\partial x}|_{q(t)},\,\, x(t)=c(t)\frac{\partial p}{\partial y}|_{q(t)}, \end{align} $$

(3.3) $$ \begin{align} \frac{\partial p}{\partial z}|_{q(t)}=0\,\,\mathrm{and }\,\, \frac{\partial p}{\partial u}|_{q(t)}=0. \end{align} $$

As t goes to $\underline 0$ , $c(0)=0$ , since $x(t)\neq 0\neq y(t)$ if $t\neq 0$ but $x(0)=y(0)=0$ and $\frac {\partial p}{\partial x}|_{q(t)}\neq 0\neq \frac {\partial p}{\partial y}|_{q(t)}$ for any t in a neighborhood of $\underline 0$ . We now write down the local equations

(3.4) $$ \begin{align} \mathcal{X}_t: y(t)(x-x(t))+x(t)(y-y(t))+(x-x(t))(y-y(t))=0 \end{align} $$

of $\mathcal {X}_t$ at $q(t)$ , and the local equation

(3.5) $$ \begin{align} \,\,\,\,\,\, \mathcal S_t & : \frac{\partial p}{\partial x}|_{q(t)}(x-x(t))+\frac{\partial p}{\partial y}|_{q(t)}(y-y(t))\\ &+ \frac{\partial p}{\partial x\partial y}|_{q(t)}(x-x(t))(y-y(t))+\frac{\partial p}{\partial x\partial z}|_{q(t)}(x-x(t))(z-z(t))\nonumber \\ &+ \frac{\partial p}{\partial x\partial u}|_{q(t)}(x-x(t))(u-u(t))+\frac{\partial p}{\partial y\partial z}|_{q(t)}(y-y(t))(z-z(t))\nonumber \\ &+ \frac{\partial p}{\partial y\partial u}|_{q(t)}(y-y(t))(u-u(t))+\frac{\partial p}{\partial u\partial z}|_{q(t)}(u-u(t))(z-z(t))\nonumber \\ &+ \frac{1}{2}\frac{\partial p}{\partial x^2}|_{q(t)}(x-x(t))^2+\frac{1}{2}\frac{\partial p}{\partial y^2}|_{q(t)}(y-y(t))^2 + \frac{1}{2} \frac{\partial p}{\partial u^2}|_{q(t)}(u-u(t))^2\nonumber \\ &+ \frac{1}{2}\frac{\partial p}{\partial z^2}|_{q(t)}(z-z(t))^2 + o(3)=0\nonumber \end{align} $$

of $\mathcal S_t$ at $q(t)$ , where $o(3)\in (x-x(t), y-y(t), z-z(t), u-u(t))^3$ . By (3.4), one may write

(3.6) $$ \begin{align} x-x(t)=-\frac{x(t)(y-y(t))}{y-y(t)+y(t)}. \end{align} $$

Let d be the maximum degree of $x-x(t)$ in (3.5). By substituting in (3.5), by multiplying by ${y^d= (y-y(t)+y(t))^d}$ , and by using that $T_{q(t)}(\mathcal S_t)=T_{q(t)}(\mathcal {X}_t)$ (i.e., $y(t)\frac {\partial p}{\partial y}|_{q(t)}=x(t)\frac {\partial p}{\partial x}|_{q(t)}$ ) for any $t\neq 0$ , we find that the local equation of $S_t=\mathcal S_t\cap \mathcal {X}_t$ is given by

$$ \begin{align*} S_t & : \frac{\partial p}{\partial x}|_{q(t)}\Big(-x(t)(y-y(t))\Big)\Big(y-y(t)+y(t)\Big)^{d-1}\nonumber\\[3pt] & + \Big( \frac{\partial p}{\partial x\partial y}|_{q(t)}(y-y(t))\Big)\Big(-x(t)(y-y(t))\Big)\Big(y-y(t)+y(t)\Big)^{d-1}\nonumber\\[3pt] & + \Big(\frac{\partial p}{\partial x\partial z}|_{q(t)}(z-z(t))\Big)\Big(-x(t)(y-y(t))\Big)\Big(y-y(t)+y(t)\Big)^{d-1}\nonumber \\[3pt] & + \Big( \frac{\partial p}{\partial x\partial u}|_{q(t)}(u-u(t))\Big)\Big(-x(t)(y-y(t))\Big)\Big(y-y(t)+y(t)\Big)^{d-1} \\[3pt] &+ \frac{1}{2}\frac{\partial p}{\partial x^2}|_{q(t)}\Big(-x(t)(y-y(t))\Big)^2\Big(y-y(t)+y(t)\Big)^{d-2} \\[3pt] & + \frac{\partial p}{\partial y}|_{q(t)}\Big(y-y(t)\Big)\Big(y-y(t)+y(t)\Big)^{d}\\[3pt]& + \Big(\frac{\partial p}{\partial y\partial z}|_{q(t)}(z-z(t)) + \frac{\partial p}{\partial y\partial u}|_{q(t)}(u-u(t))\Big)\Big(y-y(t)\Big)\Big(y-y(t)+y(t)\Big)^d \\[3pt] & + \Big(\frac{1}{2}\frac{\partial p}{\partial y^2}|_{q(t)}(y-y(t))^2+\frac{\partial p}{\partial u\partial z}|_{q(t)}(u-u(t))(z-z(t))\Big)\Big(y-y(t)+y(t)\Big)^{d} \\[3pt] & + \Big(\frac{1}{2} \frac{\partial p}{\partial u^2}|_{q(t)}(u-u(t))^2 + \frac{1}{2}\frac{\partial p}{\partial z^2}|_{q(t)}(z-z(t))^2\Big)\Big(y-y(t)+y(t)\Big)^d + o(3)\\[3pt]& = y(t)^{d-2}\Big(\frac{\partial p}{\partial y}|_{q(t)}y(t)-\frac{\partial p}{\partial x\partial y}|_{q(t)}x(t)y(t) +\frac{1}{2}\frac{\partial p}{\partial x^2}|_{q(t)}x(t)^2+\frac{1}{2}\frac{\partial p}{\partial y^2}|_{q(t)}y(t)^2\Big)\Big(y-y(t)\Big)^2\\[3pt] & + y(t)^{d-1}\Big(y(t)\frac{\partial p}{\partial y\partial z}|_{q(t)}-x(t)\frac{\partial p}{\partial x\partial z}|_{q(t)}\Big)\Big(y-y(t)\Big)\Big(z-z(t)\Big)\\[3pt] & + y(t)^{d-1}\Big(y(t)\frac{\partial p}{\partial y\partial u}|_{q(t)}-x(t)\frac{\partial p}{\partial x\partial u}|_{q(t)}\Big)\Big(y-y(t)\Big)\Big(u-u(t)\Big)\\[3pt] & + y(t)^{d} \Big(\frac{1}{2}\frac{\partial p}{\partial z^2}|_{q(t)}(z-z(t))^2+\frac{1}{2}\frac{\partial p}{\partial u^2}|_{q(t)}(u-u(t))^2\Big)\\[3pt] & + y(t)^{d} \Big(\frac{\partial p}{\partial u\partial z}|_{q(t)}(u-u(t))(z-z(t))\Big)+o(3)=0. \end{align*} $$

Up to the irrelevant factor $y(t)^{d-2}$ , the Hessian matrix at $q(t)$ of the above polynomial is

(3.7) $$ \begin{align} H_{q(t)}=\left(\begin{array}{ccc}A_{11}(t) & A_{12}(t) & A_{13}(t) \\ A_{12}(t) & \frac{y(t)^2}{2}\frac{\partial p}{\partial z^2}|_{q(t)} & \frac{y(t)^2}{2}\frac{\partial p}{\partial z\partial u}|_{q(t)} \\ A_{13}(t) & \frac{y(t)^2}{2}\frac{\partial p}{\partial z\partial u}|_{q(t)} & \frac{y(t)^2}{2}\frac{\partial p}{\partial u^2}|_{q(t)}\end{array}\right), \end{align} $$

where

$$ \begin{align*} A_{11}(t) & = \frac{\partial p}{\partial y}|_{q(t)}y(t)-\frac{\partial p}{\partial x\partial y}|_{q(t)}x(t)y(t)+\frac{1}{2}\frac{\partial p}{\partial x^2}|_{q(t)}x(t)^2+\frac{1}{2}\frac{\partial p}{\partial y^2}|_{q(t)}y(t)^2,\\[3pt] A_{12}(t) & = \frac{y(t)}{2}\Big(y(t)\frac{\partial p}{\partial y\partial z}|_{q(t)}-x(t)\frac{\partial p}{\partial x\partial z}|_{q(t)}\Big),\\[3pt] A_{13}(t) & = \frac{y(t)}{2}\Big(y(t)\frac{\partial p}{\partial y\partial u}|_{q(t)}-x(t)\frac{\partial p}{\partial x\partial u}|_{q(t)}\Big). \end{align*} $$

Now $S_t$ has a node at $q(t)$ if and only if $\det (H_{q(t)})\neq 0$ . If we substitute the equalities (3.2) in $H_{q(t)}$ , we see that this matrix has the first column divisible by $c(t)$ and the second and third columns divisible by $c(t)^2$ . Let $B_{q(t)}$ be the matrix obtained by $H_{q(t)}$ by dividing the first column by $c(t)$ and the second and third columns by $c(t)^2$ . We have that

(3.8) $$ \begin{align} B_{q(t)}=\left(\begin{array}{ccc}B_{11}(t) & B_{12}(t) & B_{13}(t) \\ B_{21}(t) & \frac{1}{2}(\frac{\partial p}{\partial x}|_{q(t)})^2\frac{\partial p}{\partial z^2}|_{q(t)} & \frac{1}{2}(\frac{\partial p}{\partial x}|_{q(t)})^2\frac{\partial p}{\partial z\partial u}|_{q(t)} \\ B_{31}(t) &\frac{1}{2}(\frac{\partial p}{\partial x}|_{q(t)})^2\frac{\partial p}{\partial z\partial u}|_{q(t)} & \frac{1}{2}(\frac{\partial p}{\partial x}|_{q(t)})^2\frac{\partial p}{\partial u^2}|_{q(t)}\end{array}\right), \end{align} $$

where

$$ \begin{align*} B_{11}(t) & = \frac{\partial p}{\partial y}|_{q(t)}\frac{\partial p}{\partial x}|_{q(t)}+ c(t)\Big(-\frac{\partial p}{\partial x\partial y}|_{q(t)}\frac{\partial p}{\partial x}|_{q(t)}\frac{\partial p}{\partial y}|_{q(t)}\\[3pt] &\quad+\frac{1}{2}\frac{\partial p}{\partial x^2}|_{q(t)}\Big(\frac{\partial p}{\partial y}|_{q(t)}\Big)^2+\frac{1}{2}\frac{\partial p}{\partial y^2}|_{q(t)}\Big(\frac{\partial p}{\partial x}|_{q(t)}\Big)^2\Big),\\[3pt] B_{21}(t) & = \frac{1}{2}\frac{\partial p}{\partial x}|_{q(t)}c(t)\Big(\frac{\partial p}{\partial x}|_{q(t)}\frac{\partial p}{\partial y\partial z}|_{q(t)}-\frac{\partial p}{\partial y}|_{q(t)}\frac{\partial p}{\partial x\partial z}|_{q(t)}\Big),\\[3pt] B_{31}(t) & = \frac{1}{2}\frac{\partial p}{\partial x}|_{q(t)}c(t)\Big(\frac{\partial p}{\partial x}|_{q(t)}\frac{\partial p}{\partial y\partial u}|_{q(t)}-\frac{\partial p}{\partial y}|_{q(t)}\frac{\partial p}{\partial x\partial u}|_{q(t)}\Big). \end{align*} $$

As t goes to $0$ , $c(t)$ goes to $0$ and the matrix $B_{q(t)}$ specializes to the matrix

(3.9) $$ \begin{align} B_{\underline 0}=\left(\begin{array}{ccc}\frac{\partial p}{\partial y}|_{\underline 0}\frac{\partial p}{\partial x}|_{\underline 0} & B_{12}(0) & B_{13}(0) \\ 0 & \frac{1}{2}(\frac{\partial p}{\partial x}|_{\underline 0})^2\frac{\partial p}{\partial z^2}|_{\underline 0} & \frac{1}{2}(\frac{\partial p}{\partial x}|_{\underline 0})^2\frac{\partial p}{\partial z\partial u}|_{\underline 0} \\ 0&\frac{1}{2}(\frac{\partial p}{\partial x}|_{\underline 0})^2\frac{\partial p}{\partial z\partial u}|_{\underline 0} & \frac{1}{2}(\frac{\partial p}{\partial x}|_{\underline 0})^2\frac{\partial p}{\partial u^2}|_{\underline 0}\end{array}\right). \end{align} $$

Using that $\frac {\partial p}{\partial x}|_{\underline 0}=1=\frac {\partial p}{\partial y}|_{\underline 0}$ , we see that $\det (B_{\underline 0})$ coincides with the discriminant of the degree $2$ homogeneous polynomial $p_2(0,0,z,u)$ , which is nonzero by the hypothesis that $S_0$ has a $T_1$ singularity at $\underline 0$ . We finally deduce that $\det (B_{q(t)})\neq 0\neq \det (H_{q(t)})$ and thus the surface $S_t$ has a node at $q(t)$ for $t\neq 0$ .

Proof. This is a straightforward consequence of Lemma 3.2.

Proof. Let $V\subset \mathcal H_C^R$ be the unique irreducible component of the locally closed set of curves on R with $\delta $ nodes that contains the point $[C]$ . Since $[C]$ sits in the image of $ \varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}$ , V intersects the image of $\varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}$ . Since $\varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}$ is dominant, the intersection of V with the image of $ \varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}$ is an open dense subset of V; hence, there is an irreducible component $\mathbb {T}$ of $\mathbb {T}_\delta $ containing $S_0$ such that the map

is dominant. Let a be the dimension of the general fibre of $\varphi _{|\mathbb {T}}$ and let b be the dimension of the general fibre of $\varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}$ . Of course, $a\geqslant b$ . We have

$$ \begin{align*}\dim(\mathbb{T})=\dim (V)+a, \,\, \text{and}\,\, \dim(\mathcal H^{\mathcal{X}_0}_{S_0})=\dim(\mathcal H_C^R)+b. \end{align*} $$

Hence,

$$ \begin{align*}\dim(\mathcal H^{\mathcal{X}_0}_{S_0})-\dim(\mathbb{T})=\dim(\mathcal H_C^R)-\dim (V)+b-a\leqslant \delta, \end{align*} $$

and the assertion follows.

Proof. Let $p=p_i$ , for $i=1,...,\delta $ , be a point where $S_0$ has a $T_1$ singularity. Consider the localized exact sequence

(3.13)

Let $(x,y,z,u,t)$ be an analytic coordinate system of $\mathcal {X}$ centered at p such that $\mathcal {X}$ is given by $xy=t$ and such that we have the following identifications:

• • the local ring $\mathcal {O}_{{S_0},\,p}=\mathcal {O}_{\mathcal {X},p}/\mathcal I_{{S_0}|\mathcal {X},p}$ of $S_0$ at p is identified with $\mathbb {C}[x,y,z,u]/(h_1,h_2)$ , localized at the origin, where $h_1(x,y,z,u)= x+y+h_{12}(x,y,z,u)$ , $h_{12}(x,y,z,u)\in (x,y,z,u)^2$ and $h_{12}(0,0,z,u)=0$ having a node at $\underline 0=p$ , and $h_2(x,y,z,u)=xy$ ;

• • the $\mathcal {O}_{{S_0},p}$ -module $\mathcal N_{{S_0}|\mathcal {X},\,p}$ is identified with the free $\mathcal {O}_{\mathcal {X},\,p}$ -module $\mathfrak {hom}_{\mathcal {O}_{\mathcal {X},\,p}}(\mathcal I_{{S_0}|\mathcal {X} ,\,p}, \mathcal {O}_{S_0,p}),$ generated by the morphisms $h_1^{*}$ and $h_2^{*}$ , defined by

  $$ \begin{align*}h_i^{*}(s_1(x,y,z,u)h_1(x,y,z,u)+s_2(x,y,z,u)h_2(x,y,z,u))=s_i(x,y,z,u),\,\mbox{for}\, i=1,\,2\end{align*} $$

and, finally,

• • the $\mathcal {O}_{S_0,p}$ -module

  $$ \begin{align*} (\Theta_{\mathcal{X}}|_{S_0})_{\, p}&\simeq \Theta_{\mathcal{X},p}\otimes \mathcal{O}_{S_0,p}\\ & \simeq \langle \partial /{\partial x},\partial /{\partial y},\partial / {\partial z},\partial / {\partial u},\partial /{\partial t} \rangle _{\mathcal{O}_{S_0,\,p}}/ \langle {\partial}/{\partial t}-x\partial /{\partial y}-y\partial /{\partial x}\rangle \end{align*} $$
  is identified with the free $\mathcal {O}_{\mathcal {X},\,p}$ -module generated by the derivatives $\partial /{\partial x},\partial / {\partial y},\partial /{\partial z}, \partial / {\partial u}.$

With these identifications, the localization $\alpha _p:(\Theta _{\mathcal {X}}|_{S_0})_{\, p}\rightarrow \mathcal N_{S_0|\mathcal {X},\,p}$ of the sheaf map $\alpha $ from (3.10) is defined by

$$ \begin{align*} \alpha_p(\partial /{\partial x})&=\Big(s=s_1h_1+s_2h_2\rightarrowtail \partial s/\partial x=_{\mathcal{O}_{S_0,p}} s_1\partial h_1 /{\partial x} +s_2\partial h_2 /{\partial x}\Big)\\ &= (1+\partial h_{12}/\partial x ) h_1^{*}+yh_2^{*},\\ \alpha_p(\partial /{\partial y})&= (1+\partial h_{12}/\partial y ) h^{*}_1+xh^{*}_2,\\ \alpha_p(\partial /{\partial z})&= (\partial h_{12}/\partial z) h^{*}_1 \,\,\,\textrm{and}\\ \alpha_p(\partial /{\partial u})&=(\partial h_{12}/\partial u) h^{*}_1. \end{align*} $$

By definition of $\mathcal N^\prime _{S_0|\mathcal {X}}$ , a local section s of $\mathcal N^\prime _{S_0|\mathcal {X},\,p}$ is such that there exists a local section v of ${\Theta _{\mathcal {X}}|_{S_0}}_{\, p}$ with

$$ \begin{align*}v=v_x(x,y,z,u)\partial /{\partial x}+v_y(x,y,z,u)\partial /{\partial y}+v_z(x,y,z,u)\partial /{\partial z}+v_u(x,y,z,u)\partial /{\partial u},\end{align*} $$

such that $s=\alpha _p(v)$ . Hence, locally at p, first-order equisingular deformations of $S_0$ in $\mathcal {X}$ have equations

(3.14) $$ \begin{align} \left\{\begin{array}{lll} x+y+h_{12}(x,y,z,u)& + & \epsilon \Big(v_x(1+\partial h_{12}/\partial x)+v_y(1+\partial h_{12}/\partial y)\\ & + &\,\,\,\,\, v_z(\partial h_{12}/\partial z)+v_u(\partial h_{12}/\partial u)\Big)=0\\ xy+\epsilon (yv_x+xv_y) & = & 0. \end{array}\right. \end{align} $$

The first equation above gives a first-order infinitesimal deformation of the Cartier divisor cutting $S_0$ on $\mathcal {X}_0$ , while the second equation gives a first-order infinitesimal deformation of $\mathcal {X}_0$ in $\mathcal {X}.$ More precisely, by the exact sequence

one sees that $xy+\epsilon (yv_x+xv_y)=0$ is the local equation at p of a first-order equisingular deformation of $\mathcal {X}_0$ in $\mathcal {X}$ . But $H^0(\mathcal {X}_0, \mathcal N^{\prime }_{\mathcal {X}_0|\mathcal {X}})=H^0(\mathcal {X}_0, \mathcal I_{R|\mathcal {X}_0})=0.$ It follows that the polynomial $yv_x(x,y,z,u)+xv_y(x,y,z,u)$ in the second equation of (3.14) must be identically zero, proving the first assertion of the lemma. In particular, by expanding $v_x$ and $v_y$ in Taylor series, we see that

$$ \begin{align*}v_x(\underline 0)=v_y(\underline 0)=0.\end{align*} $$

Looking at the first equation of (3.14), we have that $\frac {\partial h_{12}}{\partial z}(\underline 0)=\frac {\partial h_{12}}{\partial u}(\underline 0)=0$ since $h_{12}(x,y,z,u)\in (x,y,z,u)^2$ . This shows the inclusion (3.12).

Proof. Consider the exact sequence

(3.17)

from which one deduces that $h^1(S_0, \mathcal N_{ S_0|\mathcal {X}_0}\otimes I_{\{p_1,\dots ,p_\delta \}|\mathcal {X}_0})=0$ if and only if $h^1(S_0, \mathcal N_{ S_0|\mathcal {X}_0})=0$ and

$$ \begin{align*}h^0(S_0,\mathcal N_{ S_0|\mathcal{X}_0}\otimes I_{\{p_1,\dots,p_\delta\}|\mathcal{X}_0})= h^0(S_0,\mathcal N_{S_0|\mathcal{X}_0})-\delta. \end{align*} $$

Assume that $h^1(S_0, \mathcal N_{ S_0|\mathcal {X}_0}\otimes I_{\{p_1,\dots ,p_\delta \}|\mathcal {X}_0})=0$ . Thus, $[S_0]$ is a smooth point of $\mathcal H^{\mathcal {X}_0}$ . In particular, there exists a unique component $\mathcal H_{S_0}^{\mathcal {X}_0}$ of $\mathcal H^{\mathcal {X}_0}$ containing $[S_0]$ and having dimension $h^0(S_0, \mathcal N_{ S_0|\mathcal {X}_0})$ at $[S_0]$ . Now, by Lemma 3.4, one has that

$$ \begin{align*}h^0(S_0, \mathcal N_{ S_0|\mathcal{X}_0})-\delta=\dim(\mathcal H_{S_0}^{\mathcal{X}_0})-\delta\leqslant\dim_{[S_0]}(\mathcal {ES}_{[S_0]}^{\mathcal{X}_0}))\leqslant\dim(T_{[S_0]}(\mathcal {ES}_{[S_0]}^{\mathcal{X}_0})).\end{align*} $$

However, by (3.12), one has that

$$ \begin{align*}\dim(T_{[S_0]}(\mathcal {ES}_{[S_0]}^{\mathcal{X}_0}))\leqslant h^0(S_0,\mathcal N^{\prime}_{S_0|\mathcal{X}})\leqslant h^0(S_0,\mathcal N_{ S_0|\mathcal{X}_0}\otimes I_{\{p_1,\dots,p_\delta\}|\mathcal{X}_0})=h^0(S_0, \mathcal N_{ S_0|\mathcal{X}_0})-\delta.\end{align*} $$

The corollary follows.

Proof. By the hypothesis, we have $\mathrm {Ext}^1(\mathcal {O}_{S_0},\mathcal N_{{S_0}|\mathcal {X}_0})\simeq H^1(S_0,\mathcal N_{S_0|\mathcal {X}_0})=0$ , and by the exact sequence

(3.18)

we have that

$$ \begin{align*}\mathcal N_{S_0|\mathcal{X}}\simeq \mathcal N_{S_0|\mathcal{X}_0}\oplus \mathcal{O}_{S_0}. \end{align*} $$

The statement then follows by standard deformation theory.

Proof. One has $\dim (\mathcal H)\geqslant \dim (\mathcal H^{\mathcal {X}_0}_{[S_0]})+1=h^0(S_0, \mathcal N_{S_0|\mathcal {X}_0})+1$ . However, $\dim (\mathcal H)\leqslant h^0(S_0,\mathcal N_{S_0|\mathcal {X}})=h^0(S_0, \mathcal N_{S_0|\mathcal {X}_0})+1$ by Lemma 3.8. The assertion follows.

Proof. We first prove the assertion for $r=1$ . Let $\mathbb {T}$ be the closure of the subset of the Hilbert scheme $\mathcal H_{S_0}^{\mathcal {X}_0}$ consisting of all surfaces $S_0^{\prime }$ such that the intersection curve of $S_0^{\prime }$ with $R=A\cap B$ is singular with at most nodes. Observe that $[S_0]\in \mathbb {T}$ .

We claim that any irreducible component of $\mathbb {T}$ that contains $[S_0]$ has exactly codimension 1 in $\mathcal H_{S_0}^{\mathcal {X}_0}$ . Indeed, let $\mathbb {T}'$ be such a component. Consider the dominant map $\varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}$ as in (3.11), which is defined at a general point of $\mathbb {T}'$ .

By our hypotheses, the general element in $\mathcal H_C^R$ is a smooth curve; hence, $\mathbb {T}'$ has codimension at least 1 in $\mathcal H_{S_0}^{\mathcal {X}_0}$ . Let $\mathcal T$ be the image of the restriction of $\varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}$ to $\mathbb {T}'$ . Then $\mathcal T$ has codimension 1 in $\mathcal H_C^R$ .

Let $\alpha $ be the dimension of the general fibre of $\varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}$ and let $\beta $ be the dimension of the general fibre of the restriction of $\varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}$ to $\mathbb {T}'$ . One has $\alpha \leqslant \beta $ . Then

$$ \begin{align*}\dim (\mathcal H_{S_0}^{\mathcal{X}_0})=\dim(\mathcal H_C^R)+\alpha \end{align*} $$

and

$$ \begin{align*}\dim (\mathbb{T}')=\dim(\mathcal T)+\beta=\dim(\mathcal H_C^R)-1+\beta\geqslant \dim(\mathcal H_C^R)-1+\alpha=\dim (\mathcal H_{S_0}^{\mathcal{X}_0})-1. \end{align*} $$

Since $\dim (\mathbb {T}')<\dim (\mathcal H_{S_0}^{\mathcal {X}_0})$ , we have $\dim (\mathbb {T}')=\dim (\mathcal H_{S_0}^{\mathcal {X}_0})-1$ and $\alpha =\beta $ , as claimed.

Now, we consider a suitably small analytic open neighborhood U of $[S_0]$ in $\mathbb {T}$ . Every surface $S_0^{\prime }$ such that $[S_0^{\prime }]\in U$ has at most $\delta $ singularities of type $T_1$ .

Consider the variety $I\subset U\times R$ consisting of all pairs $([S_0^{\prime }],q)$ with q a $T_1$ singularity of $S_0^{\prime }$ . Let $\pi _1: I\to U$ and $\pi _2: I\to R$ be the two projections. The former one has finite fibres, implying that every irreducible component of I has dimension $\dim (\mathcal H_{S_0}^{\mathcal {X}_0})-1$ . As for the latter, it is dominant because we assume the hypotheses of Lemma 3.4. Moreover, if $q\in R$ is a point, the fibre $\pi _2^{-1}(q)$ is the locally closed set of surfaces $S_0^{\prime }$ in $\mathcal H_{S_0}^{\mathcal {X}_0}$ having a $T_1$ singularity in q.

Let $V_i$ be a sufficiently small analytic neighborhood of $p_i$ in R for $i=1,\ldots , \delta $ . By the above considerations, $\pi _1(\pi _2^{-1}(V_i))\subseteq U$ is an analytic open set that parametrizes deformations of $S_0$ which are analytically equisingular at $p_i$ . Hence, by Remark 3.7, the tangent space to $\pi _1(\pi _2^{-1}(V_i))$ at $[S_0]$ is contained in $H^0(S,\mathcal N_{ S_0|\mathcal {X}_0}\otimes I_{p_i|\mathcal {X}_0})$ and

$$ \begin{align*}\dim_{[S_0]}(\mathcal H^{\mathcal{X}_0}_{S_0})-1=\dim_{[S_0]}(\pi_1(\pi_2^{-1}(V_i))) \leqslant h^0(S_0,\mathcal N_{ S_0|\mathcal{X}_0}\otimes I_{p_i|\mathcal{X}_0}). \end{align*} $$

If (3.15) holds, then $h^1(S_0,\mathcal N_{ S_0|\mathcal {X}_0}\otimes I_{p_i|\mathcal {X}_0})=0$ for all $i=1,\ldots , \delta $ , and one has

$$ \begin{align*}h^0(S_0,\mathcal N_{ S_0|\mathcal{X}_0}\otimes I_{p_i|\mathcal{X}_0})=h^0(S_0,\mathcal N_{ S_0|\mathcal{X}_0})-1=\dim_{[S_0]}(\mathcal H^{\mathcal{X}_0}_{S_0})-1.\end{align*} $$

Thus, $\pi _1(\pi _2^{-1}(V_i))$ , that is an open analytic subset of $\mathcal {ES}_{[S_0],p_i}^{\mathcal {X}_0}$ , is an analytic branch of U of dimension $h^0(S_0,\mathcal N_{ S_0|\mathcal {X}_0})-1$ , smooth at $[S_0]$ .

Next, we prove that the general element $[S_0^{\prime }]$ in $\pi _1(\pi _2^{-1}(V_i))$ has a unique $T_1$ singularity. We argue for the case $i=1$ , and the proof is analogous in the other cases. Suppose this is not the case, and that $S_0^{\prime }$ has s singularities $q_1,\ldots , q_s$ of type $T_1$ with $s>1$ . When $S_0^{\prime }$ specializes to $S_0$ , $q_1,\ldots , q_s$ specialize, say, to $p_1,\ldots , p_s$ . By the same argument as above, the tangent space to $\pi _1(\pi _2^{-1}(V_1))$ at $[S_0^{\prime }]$ is contained in $H^0(S_0^{\prime } ,\mathcal N_{ S_0^{\prime }|\mathcal {X}_0}\otimes I_{\{q_1,\ldots , q_s\}|\mathcal {X}_0})$ and, under the hypothesis (3.15), one has

$$ \begin{align*}h^0(S_0^{\prime},\mathcal N_{ S^{\prime}_0|\mathcal{X}_0}\otimes I_{\{q_1,\ldots, q_s\}|\mathcal{X}_0})\leqslant h^0(S_0,\mathcal N_{ S_0|\mathcal{X}_0}\otimes I_{\{p_1,\ldots, p_s\}|\mathcal{X}_0})= h^0(S_0,\mathcal N_{ S_0|\mathcal{X}_0})-s<h^0(S_0,\mathcal N_{ S_0|\mathcal{X}_0})-1 \end{align*} $$

and this is a contradiction. This proves the assertion for $r=1$ .

Consider now the case $\delta>r>1$ . Fix $p_{i_1},\ldots , p_{i_r}$ distinct points among $p_1,\ldots , p_\delta $ . The intersection

$$ \begin{align*}\mathfrak T_{i_1,\ldots, i_r}:=\bigcap_{j=1}^r \pi_1(\pi_2^{-1}(V_{i_j})),\end{align*} $$

that is an analytic open subset of $\mathcal {ES}_{[S_0],p_{i_1},\ldots , p_{i_r}}^{\mathcal {X}_0}$ , is an analytic variety in $\mathcal H^{\mathcal {X}_0}_{S_0}$ parametrizing deformations of $S_0$ that are analytic equisingular at the points $p_{i_1},\ldots , p_{i_r}$ . With the same argument as above, one sees that, under the hypothesis (3.15), $\mathfrak T_{i_1,\ldots , i_r}$ is smooth of codimension r in $\mathcal H_{S_0}^{\mathcal {X}_0}$ , with tangent space at $[S_0]$ given by $H^0(S_0,\mathcal N_{ S_0|\mathcal {X}_0}\otimes I_{\{p_{i_1},\dots ,p_{i_r}\}|\mathcal {X}_0})$ .

Moreover, again by the same argument as above, the general element $S^{\prime }_0$ in $\mathfrak T_{i_1,\ldots , i_r}$ has exactly r singularities of type $T_1$ at points specializing to $p_{i_1}, \ldots , p_{i_r}$ when $S^{\prime }_0$ specializes to $S_0$ . So $\mathfrak T_{i_1,\ldots , i_r}$ is a smooth analytic branch of $\mathbb {T}_r$ containing $[S_0]$ , and this ends the proof of the corollary.

Proof. Let $S_0$ be a surface as in the statement. The inclusion (3.19) follows from Lemma 3.6, Remark 3.7 and well-known deformation theory of nodal surfaces (see [Reference Greuel and Karras11, §2.3]). It can be proved by using (3.10). In particular, if one localizes (3.10) at a node p of $S_0$ , then $H^0(S_0,T^1_{{S_0},p})\cong \mathbb {C}$ can be identified with the tangent space to the versal deformation space of a node.

Now we want to prove that, under the hypotheses of the corollary, the locally closed set $\mathcal {ES}^{\mathcal {X}_0}_{[S_0]}$ is smooth at $[S_0]$ of codimension $\delta $ in the Hilbert scheme $\mathcal H^{\mathcal {X}_0}_{S_0}$ .

Let $C\subset R$ be the $\delta _R$ -nodal curve cut out by $S_0$ on R. By the hypothesis (3.20), one has that $h^1(C, \mathcal N_{C|R}\otimes \mathcal I_{\{p_{R,1}, \ldots , p_{R,\delta _R}\}|R})=0.$ This implies that $[C]$ is a smooth point of the locally closed Severi variety $\mathcal V_\delta $ of $\delta $ -nodal curves in $\mathcal H^R_C$ . Let V be the unique irreducible component of $\mathcal V_\delta $ containing $[C]$ . Now, as we saw in Lemma 3.4, $ \varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}^{-1}(V)$ has at least one irreducible component of codimension at most $\delta _R$ in $\mathcal H^{\mathcal {X}_0}_{S_0}$ . However, by Lemma 3.6, we have that

$$ \begin{align*}T_{[{S_0}]}( \varphi_{|\mathcal H^{\mathcal{X}_0}_{S_0}}^{-1}(V))\subseteq H^0(S_0, \mathcal N_{S_0|\mathcal{X}_0}\otimes \mathcal I_{\{p_{R,1}, \ldots, p_{R,\delta_R}\}|\mathcal{X}_0}),\end{align*} $$

and by (3.20), we have $h^0(S_0, \mathcal N_{S_0|\mathcal {X}_0}\otimes \mathcal I_{\{p_{R,1}, \ldots , p_{R,\delta _R}\}|\mathcal {X}_0})=\dim (\mathcal H^{\mathcal {X}_0}_{S_0})-\delta _R$ . Thus, $\varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}^{-1}(V)$ is smooth at $[S_0]$ of codimension $\delta _R$ . We observe that $\varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}^{-1}(V)$ is an analytic open set of the variety $\mathcal {ES}^{\mathcal {X}_0}_{[S_0],p_{R,1}, \ldots , p_{R,\delta _R}}$ of deformations of $S_0$ in $\mathcal {X}_0$ that are locally trivial at every $T_1$ singularity $p_{R,i}$ , and we just proved that

$$ \begin{align*}T_{[S_0]}(\mathcal {ES}^{\mathcal{X}_0}_{[S_0],p_{R,1}, \ldots, p_{R,\delta_R}})=T_{[S_0]}( \varphi_{|\mathcal H^{\mathcal{X}_0}_{S_0}}^{-1}(V))= H^0(S_0, \mathcal N_{S_0|\mathcal{X}_0}\otimes \mathcal I_{\{p_{R,1}, \ldots, p_{R,\delta_R}\}|\mathcal{X}_0}).\end{align*} $$

We morever observe that the general element $[S^{\prime }_0]$ of $\varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}^{-1}(V)$ corresponds to a surface $S^{\prime }_0=S^{\prime }_A\cup S^{\prime }_B$ , where $S^{\prime }_A$ and $S^{\prime }_B$ intersect transversally along a curve $C'$ on R, except for $\delta _R$ points $p_{R,1}', \ldots , p_{R,\delta _R}'\in C'$ , which are singularities of type $T_1$ of $S^{\prime }_0$ , and specialize to $p_{R,1}, \ldots , p_{R,\delta _R}$ as $S_0^{\prime }$ specializies to $S_0$ .

We claim that $S_A'$ and $S_B'$ are smooth outside R. Indeed, since $[S_0]$ belongs to $ \varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}^{-1}(V)$ , the surface $S^{\prime }_0$ may have at most $\delta ^{\prime }_A\leqslant \delta _A$ nodes $p_{A,1},\ldots , p_{A,\delta ^{\prime }_A}$ on A and $\delta ^{\prime }_B\leqslant \delta _B$ nodes $p_{B,1},\ldots , p_{B,\delta ^{\prime }_B}$ on B, deformations of $\delta _A'$ nodes of $S_0$ on A and $\delta _B'$ nodes of $S_0$ on B. If this happens, denoting by $\mathfrak Z'$ the scheme of singular points of $S^{\prime }_0$ , then $ T_{[S^{\prime }_0]}(\mathcal {ES}^{\mathcal {X}_0}_{[S_0],p_{R,1}, \ldots , p_{R,\delta _R}})=T_{[S^{\prime }_0]}\varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}^{-1}(V)\subseteq H^0(S^{\prime }_0,\mathcal N_{S^{\prime }_0|\mathcal {X}_0}\otimes I_{\mathfrak Z'|\mathcal {X}_0}). $ But, once again by (3.20) and by semicontinuity, one has that $h^0(S_0^{\prime },\mathcal N_{S_0^{\prime }|\mathcal {X}_0}\otimes I_{\mathfrak Z'|\mathcal {X}_0})= H^0(S^{\prime }_0,\mathcal N_{S^{\prime }_0|\mathcal {X}_0})-\delta _R-\delta ^{\prime }_A-\delta _B'.$ It follows that $\delta ^{\prime }_A=\delta _B'=0$ (i.e., $S^{\prime }_A$ and $S^{\prime }_B$ are smooth off R and $\varphi _{|\mathcal H^{\mathcal {X}_0}_{S_0}}^{-1}(V)$ is a locally closed set in one irreducible component $\mathbb {T}\subset \mathbb {T}_{\delta _R}$ , smooth at $[S_0]$ ). We just proved that, under our hypotheses, one may deform $S_0$ in $\mathcal {X}_0$ by smoothing all nodes of $S_0$ and by preserving all $T_1$ singularities.

Let now $\mathfrak Z_A$ and $\mathfrak Z_B$ be, respectively, the scheme of nodes of $S_0$ on A and B. Let $\mathcal {ES}^{\mathcal {X}_0}_{[S_0],\mathfrak Z_A,\mathfrak Z_B}$ be the scheme of deformations of $S_0$ which are locally trivial at every node of $S_0$ . By standard deformation theory of nodal surfaces, one has that, under the hypothesis $h^1(S_0, \mathcal N_{S_0|\mathcal {X}_0}\otimes \mathcal I_{\mathfrak Z_A\cup \mathfrak Z_B|\mathcal {X}_0})=0$ (that holds by (3.20)), $\mathcal {ES}^{\mathcal {X}_0}_{[S_0],\mathfrak Z_A,\mathfrak Z_B}$ is smooth of codimension $\delta _A+\delta _B$ in $\mathcal H^{\mathcal {X}_0}_{S_0}$ at $[S_0]$ , and moreover,

$$ \begin{align*}T_{[S_0]}(\mathcal {ES}^{\mathcal{X}_0}_{[S_0],\mathfrak Z_A,\mathfrak Z_B})=H^0(S_0, \mathcal N_{S_0|\mathcal{X}_0}\otimes \mathcal I_{\mathfrak Z_A\cup \mathfrak Z_B|\mathcal{X}_0}).\end{align*} $$

With a similar argument as above, one sees that the general element $[\tilde S_0]$ of $\mathcal {ES}^{\mathcal {X}_0}_{[S_0],\mathfrak Z_A,\mathfrak Z_B}$ corresponds to a surface $\tilde S_0=\tilde S_A\cup \tilde S_B$ , with $\tilde S_A$ and $\tilde S_B$ intersecting transversally along a smooth curve $\tilde C\subset R$ and having, respectively, $\delta _A$ and $\delta _B$ nodes as singularities. In particular, $\mathcal {ES}^{\mathcal {X}_0}_{[S_0],\mathfrak Z_A,\mathfrak Z_B}$ is a locally closed set in an irreducible component $\tilde {\mathbb {T}}$ of $\mathbb {T}_{\delta _A,\delta _B,0}$ , of which $[S_0]$ is a smooth point.

Now the equisingular deformation locus $\mathcal {ES}^{\mathcal {X}_0}_{[S_0]}$ of $S_0$ in $\mathcal {X}_0$ is the intersection of the loci $ \mathcal {ES}^{\mathcal {X}_0}_{[S_0],p_{R,1}, \ldots , p_{R,\delta _R}}$ and $ \mathcal {ES}^{\mathcal {X}_0}_{[S_0],\mathfrak Z_A,\mathfrak Z_B}$ . Hence, $\mathcal {ES}^{\mathcal {X}_0}_{[S_0]}$ has codimension at most $\delta $ in $\mathcal H^{\mathcal {X}_0}_{S_0}$ because $[S_0]$ is a smooth point of $\mathcal H^{\mathcal {X}_0}_{S_0}$ .

However, one has

$$ \begin{align*} T_{[S_0]}(\mathcal {ES}^{\mathcal{X}_0}_{[S_0]}) & = T_{[S_0]}(\mathcal {ES}^{\mathcal{X}_0}_{[S_0],p_{R,1}, \ldots, p_{R,\delta_R}})\cap T_{[S_0]}(\mathcal {ES}^{\mathcal{X}_0}_{[S_0],\mathfrak Z_A,\mathfrak Z_B})\\ & = H^0(S_0, \mathcal N_{S_0|\mathcal{X}_0}\otimes \mathcal I_{\{p_{R,1}, \ldots, p_{R,\delta_R}\}|\mathcal{X}_0}) \cap H^0(S_0, \mathcal N_{S_0|\mathcal{X}_0}\otimes \mathcal I_{\mathfrak Z_A\cup \mathfrak Z_B|\mathcal{X}_0})\\ & = H^0(S_0, \mathcal N_{S_0|\mathcal{X}_0}\otimes \mathcal I_{\mathfrak Z|\mathcal{X}_0}). \end{align*} $$

By (3.20), we have

$$ \begin{align*}h^0(S_0, \mathcal N_{S_0|\mathcal{X}_0}\otimes \mathcal I_{\mathfrak Z|\mathcal{X}_0})=h^0(S_0, \mathcal N_{S_0|\mathcal{X}_0})-\delta. \end{align*} $$

This proves that $\mathcal {ES}^{\mathcal {X}_0}_{[S_0]}$ is smooth at $[S_0]$ of codimension exactly $\delta $ in $\mathcal H^{\mathcal {X}_0}_{S_0}$ , as wanted. This proves the first part of the corollary.

The second part is proved with analogous arguments as the ones used in the proof of Corollary 3.8.2.

Proof. From the hypotheses, it follows that $H^1(S_0,\mathcal N_{S_0|\mathcal {X}_0})=0$ . Then, by Corollary 3.8.1, $\mathcal H^{\mathcal {X}|\mathbb {D}}$ is smooth at $[S_0]$ with dimension $h^0(S_0,\mathcal N_{S_0|\mathcal {X}_0})+1=\dim (\mathcal H_{S_0}^{\mathcal {X}_0})+1$ . By standard deformation theory, there is an analytic neighborhood U of $[S_0]$ in $\mathcal H$ and a versal morphism

$$ \begin{align*}f: U\longrightarrow \prod_{p\in \mathfrak Z_A+\mathfrak Z_B} \Delta_p, \end{align*} $$

where $\Delta _p$ is the versal deformation space of a node, and therefore, it has dimension 1. Let $U'=U\cap \mathcal H_{S_0}^{\mathcal {X}_0}$ . Then f restricts to

$$ \begin{align*}g: U'\longrightarrow \prod_{p\in \mathfrak Z_A+\mathfrak Z_B} \Delta_p. \end{align*} $$

The differential of g at $[S_0]$ is

$$ \begin{align*}H^0(S_0, \mathcal N_{S_0|\mathcal{X}_0})\to \prod_{p\in \mathfrak Z_A+\mathfrak Z_B} T_{S_0,p}^1\cong \mathbb{C}^{\delta_A+\delta_B}, \end{align*} $$

and this map is surjective since its kernel is $H^0(S_0, \mathcal N_{S_0|\mathcal {X}_0}\otimes \mathcal I_{\mathfrak Z_A\cup \mathfrak Z_B|\mathcal {X}_0})$ , which has codimension $\delta _A+\delta _B$ in $H^0(S_0, \mathcal N_{S_0|\mathcal {X}_0})$ by (3.20). Hence, g has maximal rank at $[S_0]$ and therefore also f is of maximal rank at $[S_0]$ . Hence, $f^{-1}(0)$ and $g^{-1}(0)$ are analytic subvarieties of U and $U'$ , respectively, smooth at $[S_0]$ and of codimension $\delta _A+\delta _B$ in U and $U'$ , respectively. By versality, $f^{-1}(0)$ (resp. $g^{-1}(0)$ ) coincides with $\mathcal {ES}^{\mathcal {X}}_{[S_0],\mathfrak Z_A,\mathfrak Z_B}$ (resp. $\mathcal {ES}^{\mathcal {X}_0}_{[S_0],\mathfrak Z_A,\mathfrak Z_B}$ ). The statement follows.

Proof. (i) Let $I'$ be an irreducible component of I such that $I'$ is dominant to R via $\pi _2$ . If $p\in R$ is a general point, we know that $\pi _2^{-1}(p)$ can be identified with $\mathcal L_0(2,p)$ , and $\mathcal L_0(2,p)$ is a projective space with dimension s independent on the general point p. This clearly implies that $I'$ is unique.

(ii) Suppose the dominating component $I'$ exists. With the same notation as above, we have $\dim (I')=\dim (R)+s=s+2$ . However, by Remark 2.4, $s\geqslant \dim (|\mathcal L_0|)-3$ ; hence, $\dim (I')\geqslant \dim (|\mathcal L_0|)-1$ . By the hypotheses, the map $\pi _1$ , restricted to $I'$ , is generically finite onto the image, and this image cannot be dense in $|\mathcal L_0|$ by Bertini’s theorem. Hence, $\dim (I')\leqslant \dim (|\mathcal L_0|)-1$ , so the equality holds, and this implies that $s=\dim (|\mathcal L_0|)-3$ , as wanted.

(iii) Keep the same notation as above. The dimension of the fibre of $\pi _2$ over a general point of R is $r\geqslant \dim (|\mathcal L_0|)-3= \mathcal L_0(2,p)\geqslant 0$ . Moreover, there is an open dense subset U of R, containing p, such that for all $q\in U$ , one has that $\mathcal L_0(2,p)$ has dimension $ s = \dim (|\mathcal L_0|)-3$ . Hence, there is a component $I'$ of I dominating R via $\pi _2$ .

Proof of Proposition 3.10

We notice that by Corollary 3.8.1, $[S_0]$ is a smooth point for $\mathcal H^{\mathcal {X}|\mathbb {D}}$ . Part (i) follows by Corollary 3.7.1.

Let us prove part (ii). For this, we go back to the notation introduced before the statement of Proposition 3.10. Consider then the intersection $\bar Z_0$ of $\bar Z$ with $\bar \pi ^{-1}(0)\cong |\mathcal L_0|$ , such that any of its irreducible components has codimension 1 in $|\mathcal L_0|$ . By the hypotheses we made, if $\bar Z^{\prime }_0$ is any irreducible component of $\bar Z_0$ , its general element does not contain R; hence, it is a surface $S_0^{\prime }\in |\mathcal L_0|$ that intersects R along a curve $C'$ . By Proposition 3.1, the curve $C'$ is singular.

By the hypothesis (b) and by Lemma 3.11(iii), the pair $(S_0,p)$ belongs to the unique irreducible component $I'$ of I dominating R via $\pi _2$ , and $I'$ has dimension equal to $\dim (|\mathcal L_0|)-1$ . Consider now the subset $I"\subseteq I$ of the pairs $(S_0^{\prime },p')$ with $S_0^{\prime }\in \bar Z^{\prime }_0$ , where $\bar Z^{\prime }_0$ is as in Claim 3.12. We notice that $I"$ also dominates R via $\pi _2$ . So by Lemma 3.11(i), $I"$ coincides with $I'$ . This implies that $\bar Z^{\prime }_0=\pi _1(I')$ ; hence, $S_0\in \bar Z^{\prime }_0$ and therefore, the general surface $S^{\prime }_0\in \bar Z^{\prime }_0$ has a unique $T_1$ singularity. By Lemma 3.2, the assertion (ii) follows.

To prove (iii), we remark first of all that (ii) implies that $\mathcal V^{\mathcal {X}|\mathbb {D}}_1$ is nonempty and there is an irreducible component $\mathcal V$ of $\mathcal V^{\mathcal {X}|\mathbb {D}}_1$ that dominates $\mathbb {D}$ and contains $[S_0]$ . The general point in $\mathcal V$ corresponds to a surface $S_t$ with $t\neq 0$ , with a unique node at a general point $p_t\in \mathcal {X}_t$ . Moreover, since $[S_0]$ is a smooth point of $\mathcal H$ , we have that $[S_t]$ is a smooth point of $\mathcal H$ , and by the hypothesis (c) and by semicontinuity, we have that $h^1(S_t, \mathcal N_{S_t|\mathcal {X}_t}\otimes I_{p_t})=0$ . This yields that $\mathcal V\cap \mathcal H_t$ is smooth of dimension $\dim (\mathcal H_t)-1$ . Hence, $\mathcal V$ has dimension $\dim (\mathcal H)-1$ . To prove that $\mathcal V$ is reduced, it suffices to prove that $\mathcal V$ is smooth at $[S_t]$ . To see this, consider the exact sequence