Codimension-one components of a maximal-order support

The local shape along the component. Fix $x\in C$, put $R={\mathcal{O}}_{X,x}$, and let $P=I(C{)}_x$ be the height-one component prime. By [F1], $P=(u)$ for a prime element $u$. Since $C$ lies in the marked support and $\mu \ge 1$, ${\mathcal{I}}_x\subseteq P$. Let $\eta$ be the generic point of $C$. By [F2], $R_P$ is a DVR with uniformizer $u$; since $\eta$ lies in the support and $\mathcal{I}$ has maximal order, [F3] gives ${\mathrm{ord}}_{R_P}({\mathcal{I}}_P)=\mu$, so ${\mathcal{I}}_P=(u^{\mu })$. For any $f\in {\mathcal{I}}_x$, this gives $sf=u^{\mu }g$ for some $s\mathrm{\notin }P$ and $g\in R$. The element $u$ is prime and does not divide $s$, so repeated cancellation gives $f\in (u^{\mu })$. Hence ${\mathcal{I}}_x=(u^{\mu })J$ for an ideal $J\subseteq R$. If ${\mathrm{ord}}_x(u)\ge 2$, then ${\mathcal{I}}_x\subseteq {\mathfrak{m}}_x^{2\mu }\subseteq {\mathfrak{m}}_x^{\mu +1}$, contradicting maximal order at $x$. Thus ${\mathrm{ord}}_x(u)=1$. If $J$ were proper, then $J\subseteq {\mathfrak{m}}_x$ and ${\mathcal{I}}_x\subseteq {\mathfrak{m}}_x^{\mu +1}$, again a contradiction. Therefore ${\mathcal{I}}_x=(u^{\mu })$. By [F5], $u$ is a regular parameter and $R\mathrm{/}(u)$ is regular, so $C$ is regular at $x$. Since $x$ was arbitrary, $C$ is regular, and locally along it the support is exactly $V(u)$; hence no other support component meets $C$.

The blowup is an isomorphism and resolves along the component. The regular component $C$ is an effective Cartier divisor by step 1.1, so its blowup is an isomorphism. Locally along $C$, write $y=u$ for its equation. Step 1.1 gives $\mathcal{I}=(u^{\mu })$, hence ${\sigma }^{\ast }\mathcal{I}=(y^{\mu })$ and $y^{-\mu }{\sigma }^{\ast }\mathcal{I}={\mathcal{O}}_{X^{\mathrm{\prime }}}$ along the exceptional divisor. If $C$ has SNC with $E$, this is an admissible marked-ideal transformation by [F4], so its controlled-transform support misses the divisor. The local quotient calculation itself needs no boundary hypothesis.