Proof.
For $z'\in W'$, the polynomial $P(z',\cdot )$ has at most $\mu $ roots, all in $\Delta $; so the fibers of $Z$ over $W'$ are finite. We first describe $Z$ near a fiber. Let $b\in W'$, let $w_1,\ldots ,w_ l$ be the distinct roots of $P(b,\cdot )$, and put $z_ i=(b,w_ i)$. Let $\delta >0$ be so small that the closed discs $|w-w_ i|\leq 2\delta $ are pairwise disjoint and contained in $\Delta $. The function $|P|$ is continuous and positive on the compact set $\{ b\} \times K$, where $K=\{ |w|\leq \rho \} \setminus \bigcup _ i\{ |w-w_ i|<\delta \} $; hence it is positive on $V_ b\times K$ for a neighborhood $V_ b$ of $b$, and for $z'\in V_ b$ all roots of $P(z',\cdot )$ lie in $\bigcup _ i\{ |w-w_ i|<\delta \} $. Put $B_ i=V_ b\times \{ |w-w_ i|<2\delta \} $. These open sets are pairwise disjoint, and $Z\cap (V_ b\times \Delta )\subset \bigcup _ iB_ i$. Moreover, given neighborhoods $N_ i$ of the points $z_ i$, we can achieve $V_ b\times \{ |w-w_ i|\leq 2\delta \} \subset N_ i$ for all $i$: first decrease $\delta $ so that $\{ b\} \times \{ |w-w_ i|\leq 2\delta \} \subset N_ i$, and then decrease $V_ b$, using the compactness of these discs. We call such $\delta $ and $V_ b$ adapted to $b$; a smaller neighborhood of $b$ remains adapted.
(1) Let $s$ be a section of $\mathcal G$ over $V'\times \Delta $ whose germs at all the $z_ i$ vanish. Then $s$ vanishes on neighborhoods $N_ i$ of the $z_ i$; choose $\delta $ and $V_ b\subset V'$ adapted to $b$ with $B_ i\subset N_ i$. The open sets $B_1,\ldots ,B_ l$ and $(V_ b\times \Delta )\setminus Z$ cover $V_ b\times \Delta $, and $s$ vanishes on each of them, on the last one because all stalks of $\mathcal G$ vanish there. Hence $s=0$ on $V_ b\times \Delta $, and the map is injective. Conversely, let $g_ i\in \mathcal G_{z_ i}$ be given, represented by sections $t_ i$ on neighborhoods $N_ i$ of the $z_ i$, and choose $\delta $ and $V_ b$ adapted to $b$ with $B_ i\subset N_ i$. On the open cover of $V_ b\times \Delta $ formed by the $B_ i$ and $(V_ b\times \Delta )\setminus Z$, consider the sections $t_ i|_{B_ i}$ and $0$. They agree on the overlaps: the $B_ i$ are pairwise disjoint, and $t_ i$ vanishes on $B_ i\setminus Z$, since $\mathcal G$ has zero stalks there. Gluing them, we obtain a section over $V_ b\times \Delta $ with germs $g_ i$ at the $z_ i$. Hence the map is surjective.
(2) The map $\Phi $ is a morphism of sheaves of $\mathcal O_{W'}$-modules. It is injective on sections: suppose that the class of $R=\sum _{k<\mu }c_ kw^ k$ vanishes on $V'\times \Delta $, and let $z'\in V'$. If $w_0$ is a root of $P(z',\cdot )$ of multiplicity $m$, then $R=Pq$ near $(z',w_0)$ with $q$ holomorphic, so the polynomial $R(z',\cdot )=P(z',\cdot )q(z',\cdot )$ vanishes to order at least $m$ at $w_0$. Thus $R(z',\cdot )$, a polynomial of degree less than $\mu $, has at least $\mu $ zeros counted with multiplicity, and therefore vanishes; that is, $c_ k(z')=0$ for all $k$. To prove surjectivity on stalks, let $s$ be a section of $\mathcal O_ D/P\mathcal O_ D$ over $V'\times \Delta $ and $b\in V'$. Near each $z_ i$, $s$ is the class of a holomorphic function $f_ i$ on a neighborhood $N_ i$ of $z_ i$. Choose $\delta $ and $V_ b\subset V'$ adapted to $b$ with $V_ b\times \{ |w-w_ i|\leq 2\delta \} \subset N_ i$. On the circles $|\zeta -w_ i|=2\delta $ the function $P(z',\zeta )$ does not vanish for $z'\in V_ b$, since these circles have distance at least $\delta $ from all roots. For $z'\in V_ b$ and $w\in \mathbb C$ put
\[ R(z',w)=\frac1{2\pi \mathrm i}\sum _{i=1}^ l\int _{|\zeta -w_ i|=2\delta } \frac{f_ i(z',\zeta )}{P(z',\zeta )}\, \frac{P(z',\zeta )-P(z',w)}{\zeta -w}\, \mathrm d\zeta . \]
The difference quotient is a polynomial in $w$ of degree $\mu -1$ whose coefficients are polynomials in $\zeta $ and in the $a_ k(z')$; hence $R$ is a polynomial in $w$ of degree less than $\mu $ whose coefficients are holomorphic on $V_ b$, by differentiation under the integral sign. For $z'\in V_ b$ and $|w-w_ i|<2\delta $ put
\[ q_ i(z',w)=\frac1{2\pi \mathrm i}\sum _{j=1}^ l\int _{|\zeta -w_ j|=2\delta }\frac{f_ j(z',\zeta )}{P(z',\zeta )(\zeta -w)}\, \mathrm d\zeta , \]
a holomorphic function on $B_ i$. Adding the two integrals, we obtain
\[ P(z',w)q_ i(z',w)+R(z',w)=\frac1{2\pi \mathrm i}\sum _{j=1}^ l\int _{|\zeta -w_ j|=2\delta }\frac{f_ j(z',\zeta )}{\zeta -w}\, \mathrm d\zeta =f_ i(z',w), \]
since the term $j=i$ equals $f_ i(z',w)$ by Cauchy's formula, and the terms $j\neq i$ vanish by Cauchy's theorem, the point $w$ lying outside the closed disc $|\zeta -w_ j|\leq 2\delta $. Hence $f_ i\equiv R$ modulo $P$ on $B_ i$, so $s$ and the class of $R$ agree on each $B_ i$; on $(V_ b\times \Delta )\setminus Z$ both vanish, since $P$ is a unit there. Thus $s=\Phi (c)$ on $V_ b\times \Delta $, where $c$ is the tuple of coefficients of $R$. Being injective, and surjective on all stalks, $\Phi $ is an isomorphism.
(3) The sheaf $\mathcal M(\mathcal I)$ is a subsheaf of $\mathcal O_{W'}$-modules of $\mathcal O_{W'}^\mu $. Since $\mathcal O_{W'}$ is coherent (theorem 9.2.7), the last assertion of lemma 9.2.6 shows that it suffices to prove that $\mathcal M(\mathcal I)$ is locally finitely generated. Note that $\mathcal I/P\mathcal O_ D$ has zero stalks off $Z$, because $P$ is a unit there; so (1) applies to it as well as to $\mathcal O_ D/P\mathcal O_ D$.
Fix $b\in W'$, with $z_ i$ as above. Since $\mathcal I$ is coherent, there are neighborhoods $N_ i$ of $z_ i$ and sections $f_{i1},\ldots ,f_{i\kappa }$ of $\mathcal I$ over $N_ i$ generating all stalks of $\mathcal I$ on $N_ i$. Choose $\delta $ and $V_ b$ adapted to $b$ with $B_ i\subset N_ i$. For $1\leq i\leq l$, $1\leq \lambda \leq \kappa $ and $0\leq m<\mu $, let $\sigma _{i\lambda m}$ be the section of $\mathcal I/P\mathcal O_ D$ over $V_ b\times \Delta $ that equals the class of $w^ mf_{i\lambda }$ on $B_ i$ and $0$ on the other $B_ j$ and on $(V_ b\times \Delta )\setminus Z$. As in (1), these local sections glue, since the $B_ j$ are pairwise disjoint and the class of $w^ mf_{i\lambda }$ vanishes on $B_ i\setminus Z$, where $P$ is a unit.
We claim that the sections $\Phi ^{-1}(\sigma _{i\lambda m})$ of $\mathcal M(\mathcal I)$ over $V_ b$ generate all its stalks. Let $b'\in V_ b$, and let $y_1,\ldots ,y_ p$ be the points of $Z$ over $b'$; each of them lies in exactly one $B_ i$. By (1), every element of $\pi _*(\mathcal I/P\mathcal O_ D)_{b'}$ is a sum of elements $e$ whose germs vanish at all points of the fiber except one point $y_ q$, where the germ is some $g\in (\mathcal I/P\mathcal O_ D)_{y_ q}$. Let $y_ q\in B_ i$, and write $g$ as the class of $\sum _\lambda h_\lambda f_{i\lambda }$ with germs $h_\lambda $ at $y_ q$. By (1) for $\mathcal O_ D/P\mathcal O_ D$ and by (2), for each $\lambda $ there is $c^{(\lambda )}\in \mathcal O_{W',b'}^\mu $ such that $R_\lambda =\sum _ mc^{(\lambda )}_ mw^ m$ is congruent to $h_\lambda $ modulo $P$ at $y_ q$ and to $0$ modulo $P$ at the other points of the fiber. Then the element $\sum _{\lambda ,m}c^{(\lambda )}_ m\sigma _{i\lambda m}$ of $\pi _*(\mathcal I/P\mathcal O_ D)_{b'}$ has the germ $g$ at $y_ q$ and the germ $0$ at every other point of the fiber: at the other points in $B_ i$ its germ is the class of $\sum _\lambda R_\lambda f_{i\lambda }\in P\mathcal O_ D$, and at the points outside $B_ i$ it vanishes by construction. By the injectivity in (1), it equals $e$. This proves the claim, and $\mathcal M(\mathcal I)$ is coherent.
Finally, let $\mathcal M(\mathcal I)=\mathcal M(\mathcal I')$ on $V'$, and let $z\in V'\times \Delta $. If $z\notin Z$, then $\mathcal I_ z=\mathcal O_{D,z}=\mathcal I'_ z$, since $P$ is a unit at $z$. Let $z\in Z$, $b=\pi (z)$, and $f\in \mathcal I_ z$. By (1), there is an element of $\pi _*(\mathcal O_ D/P\mathcal O_ D)_ b$ whose germ at $z$ is the class of $f$ and whose germs at the other points of the fiber vanish. It belongs to $\pi _*(\mathcal I/P\mathcal O_ D)_ b=\Phi (\mathcal M(\mathcal I)_ b)=\Phi (\mathcal M(\mathcal I')_ b)=\pi _*(\mathcal I'/P\mathcal O_ D)_ b$, so its germ at $z$ belongs to $(\mathcal I'/P\mathcal O_ D)_ z$. Hence $f\in \mathcal I'_ z+P\mathcal O_{D,z}=\mathcal I'_ z$. By symmetry, $\mathcal I_ z=\mathcal I'_ z$.
$\square$
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