Induced resolution of singularities I am not a specialist in singularity theory but currently I have to touch resolution of singularities and I'd like to know whether I have understood Hironaka's theorem correctly.
Let $k$ be a field of characteristic zero and $X$ an $k$-variety and $Z$ a closed subscheme of $X$. Then Hironaka's theorem claims that there exists a log-resolution $h: Y \longrightarrow X$ of the pair $(X,Z)$, by this I mean:

*

*$Y$ is a smooth $k$-variety.

*$h$ is projective, proper birational morphism.

*$h$ induces an isomorphism outside $X_{\mathrm{sing}} \cup Z$.

*If $E$ is the exceptional locus then $h^{-1}(Z) + E$ has strict normal crossings.

and $h$ itself can be taken to be a composition of successive blowing-ups along smooth centers.
I am interested in the following situation: let's assume we still have $(X,k)$ as above and we are given a morphism $f: X \longrightarrow \mathbb{A}^1_k$. Let $U$ be any smooth open subscheme of $X$ and $F = X \setminus U$, hence $X_{\mathrm{sing}} \subset F$ so we could take a log-resolution of the pair $(X,F \cup f^{-1}(0))$ (one may assume $f^{-1}(0)$ is nowhere dense if necessary).
Let $E_i$ ($i \in I$) be the smooth irreducible components of $h^{-1}(F \cup f^{-1}(0))$ with corresponding multiplicities $N_i \neq $ ($i \in I$) , i.e.
$$h^{-1}(F \cup f^{-1}(0)) = \sum_{i \in I} N_i E_i.$$
What I want to know is can we control the multiplicities in the way that there exists $C \subset I$ such that
$$h^{-1}(f^{-1}(0)) = \sum_{i \in C} N_i E_i \ \ \ \text{and} \ \ \ h^{-1}(F) = \sum_{i \in I \setminus C} N_i E_i.$$ If it is the case, then does restricting to $U$ (which gives a new resolution since blowing-ups commute with flat base-changes) lose some multiplicities?
$\require{AMScd}$
\begin{CD}
Y_{\mid U} @>{}>> X\\
@VVV @VVhV\\
(U,f^{-1}(0) \cap U) @>{}>> (X,F \cup f^{-1}(0))
\end{CD}
 A: This has nothing to do with your morphism to $\mathbb A^1$. You are asking if there is another closed subset $F'$ (which is secretly equal to $h^{-1}(0)$), then can you separate the exceptional divisors according to the pre-images of the two sets $F$ and $F'$. This depends on how $F$ and $F'$ intersect with respect to the resolution. If none of the centers of the blow ups in the given resolution is contained in $F\cap F'$, then this holds. Then $C=\{i\in I\vert h(E_i)\subseteq F'\}$ (and then, of course, $I\setminus C=\{i\in I\vert h(E_i)\subseteq F\}$). In this case, when you restrict to $U$, you do not "lose" any multiplicities. That could only happen if you lose one of the $E_i$, but that would only happen if $h(E_i)\cap U=\emptyset$, but that would mean that $h(E_i)\subseteq F\cap F'$, which we assumed does not happen.
The problem is that if $h(E_i)\subseteq F\cap F'$, then that particular $E_i$ appears in both $h^{-1}(F)$ and $h^{-1}(F')$. This happens for sure if for instance $F\subseteq F'$.
