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I read this paragraph from Complex Geometry by Huybrechts:

...The pull-back of a divisor $D$ under a morphism $f:X\rightarrow Y$ is not always well-defined, one has to assume that the image of $f$ is not contained in the support of $D$. Thus, one usually considers only dominant morphisms...

So why would the pull-back be not well-defined if the image of $f$ is contained in the support of $D$? And why does dominant morphisms solve this issue?

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  • $\begingroup$ This is almost a duplicate of this other question, though in the algebraic world instead of the analytic one. Do you think the linked question solves your issues? $\endgroup$
    – KReiser
    May 26, 2020 at 20:17
  • $\begingroup$ @KReiser Yes that one explains well. Thanks! $\endgroup$
    – MathChopper
    May 28, 2020 at 3:04

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Suppose you are working with weil divisors on a regular variety. Then, effective divisor correspond to codimension 1 subvarieties. In this case, if the image of $f$ is contained in the support of $D$; then the pullack would be all of $X$, which is not a divisor.

Taking a dominant morphism solves the issue, because then the image of $f$ will be an open subset, and a codimension 1 subvariety of $Y$ cannot contain any open subset.

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    $\begingroup$ +1 for a good explanation. One very minor thing, though: it may not actually be the case that the image of $f$ is an open subset - take $\Bbb A^2\to \Bbb A^2$ by $(x,y)\mapsto (x,xy)$, for instance. Instead, one should say that the image of a dominant map is dense, and then your explanation is completely fine - no dense set is contained in a codimension one subvariety. $\endgroup$
    – KReiser
    May 27, 2020 at 4:20

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