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In Huybrechts' book 'Complex Geometry', page 80, he considers a holomorphic map $f: X \to Y$ and an Weil divisor $D = [Z]$, with $Z \subset Y$ an irreducible analytic hypersurface. Using a local defining function $g$ for $Z$, he takes the local factorization of the map $g\circ f = \prod g_j^{n_j}$ and defines $f^{*}([Z]) = \sum n_j[Y_j]$, where $Y_j \subset f^{-1}(Z)$ are the irreducible analytic hypersurfaces of $f(Z)^{-1}$ (which is also an analytic hypersurface) and $g_j$ is the irreducible local defining function of $Y_j$.

What I wasn't able to see is why the $n_j$'s is the same for every point in $f^{-1}(Z)$.

Thank you very much.

Kaitei
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  • You have typos here. Several places you've put $X$ where you mean $Z$. And your equation for $f^*([Z])$ is sloppy, as well. How are the $n_j$ actually computed? – Ted Shifrin Dec 04 '21 at 20:32
  • @TedShifrin Sorry, I fixed the typos. We are passing the functions to its germs right? By the Weierstrass preparation theorem, we can show that the germ of function $\mathcal{O}_{\mathbb{C}^n,0}$ is an UFD, and the factorization of holomorphic function comes from there. We can bring the defining functions to the same hyperplane in $\mathbb{C}^n$ at $0$ through a chart, but the defining functions of $Z$ can be different, even if they are zero in the hyperplane, so I don't know how to compare the $n_j$'s from different defining functions. – Kaitei Dec 05 '21 at 01:16
  • I am not thinking about all this from scratch again, but you might want to check out this discussion about a mistake in Huybrechts. – Ted Shifrin Dec 05 '21 at 01:36
  • If you have $g$ and $g'$ defining $Z$ on two opens $U$ and $U'$, then $g$ and $g'$ agree on $U \cap U'$ up to a unit. So $g \circ f$ and $g' \circ f$ agree on $f^{-1}(U \cap U')$ up to (the pull-back of) a unit. Hence the factorizations should be somewhat compatible? – red_trumpet Dec 09 '21 at 13:35

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