Giraud's tangent-direction lemma

$u^{\mathrm{\prime }}$ is a tangent direction of the controlled transform. By [F3] the section $u^{\mathrm{\prime }}=y^{-1}{\sigma }^{\ast }(u)$ is the controlled transform of the section $u$ of the marked ideal $({\mathcal{D}}^{\mu -1}(\mathcal{I}),1)$; since the generator $u$ lies in ${\mathcal{D}}^{\mu -1}(\mathcal{I})(U)$ and the controlled transform of a generated ideal is generated by the controlled transforms of generators, $u^{\mathrm{\prime }}\in {\sigma }^{\mathrm{c}}({\mathcal{D}}^{\mu -1}(\mathcal{I}),1)(U^{\mathrm{\prime }})$. By [F2] with $r=\mu -1$ and the hypothesis $C\subseteq \mathrm{supp}(\mathcal{I},\mu )$, this ideal is contained in ${\mathcal{D}}^{\mu -1}({\sigma }^{\mathrm{c}}(\mathcal{I},\mu ))(U^{\mathrm{\prime }})$; hence $u^{\mathrm{\prime }}$ is a tangent direction of the controlled transform, which is (1).

Multiplicity one. Away from $D$, the map is an isomorphism and $y$ is a unit, so a zero of $u^{\mathrm{\prime }}$ has the same order as the corresponding zero of $u$, namely one. At a zero of $u^{\mathrm{\prime }}$ on $D$, its image $x$ lies in $C$. By [F1], $C\subseteq V(u)$, and the class of $u$ in ${\mathfrak{m}}_x\mathrm{/}{\mathfrak{m}}_x^2$ is nonzero by [F4]; hence $u$ can be chosen as one of the regular parameters generating ${\mathcal{I}}_C$. Since $u^{\mathrm{\prime }}=0$, the point is on a blowup chart whose exceptional parameter $y$ is a different normal parameter. There $u^{\mathrm{\prime }}=u\mathrm{/}y$ is a chart coordinate, so [F4] gives order one. Thus $u^{\mathrm{\prime }}$ has multiplicity one on $U^{\mathrm{\prime }}$, which is assertion (2).

The zero locus is the strict transform. By [F5] the strict transform of $V(u)$ on $U^{\mathrm{\prime }}$ is cut out by $({\sigma }^{\ast }(u):y^{\mathrm{\infty }})=(u^{\mathrm{\prime }})$; hence $V(u^{\mathrm{\prime }})$ is the strict transform of $V(u)$ restricted to $U^{\mathrm{\prime }}$, which is assertion (3).