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My question concerns the proof of a part of Lemma IV.2.2 (pag. 168) of the book Geometry of Algebraic Curves. vol. I by Arbarello, Cornalba, Griffiths and Harris. In order to state my problem, let me first fix some notation: a degree-d Poincaré line bundle for a smooth curve $C$ over the complex numbers is a line bundle $\mathcal{P}$ over $C\times Pic^d(C)$ restricting to $L$ on $C\cong C\times\{L\}$ for each $L\in Pic^d(C)$. $C_d$ is the symmetric product $C^d/S^d$, i.e. the scheme parametrising effective divisors on $C$, and it comes with a morphism $u\colon C_d\to Pic^d(C)$ taking $D$ to $\mathcal{O}(D)$.

Lemma IV.2.2 states that such $\mathcal{P}$ satisfies the following universal property: for each analytic space $S$ and line bundle $\mathcal{L}$ over $C\times S$ such that $\mathcal{L}_{|_{C\times \{s\}}}$ has degree $d$ for each $s\in S$, $\exists! f\colon S\to Pic^d(C)$ such that $(1\times f)^*\mathcal{P}=\mathcal{L}\otimes\phi^*\mathcal{R}$, where $\phi\colon C\times S\to S$ is the projection and $\mathcal{R}$ is a line bundle on $S$.

At first the authors reduce to the case $d\geq 2g-1$ ($g$ being the genus of $C$), so that the fibres of $u\colon C_d\to Pic^d(C)$ are ($d-g$)-dimensional projective spaces; furthermore, one can assume $Pic(S)=0$.

Now if $S$ is reduced, uniqueness follows, I think, because any $f$ as in the statement satisfies $f(s)=\mathcal{L}_{|_{C\times \{s\}}}$ and morphisms from reduced schemes are determined by their values on points. Thus if $f'$ is another such map one can assume that

$f$ and $f'$ map into an open subset $U$ of $Pic^d(C)$ over which $C_d$ is a trivial $\mathbb{P}^{d-g}$-bundle [since equality of morphisms can be checked locally and $f$, $f'$ correspond locally to morphisms of rings which only differ by nilpotents, I guess].

My problem is in the following part:

Let $g,g'$ be liftings of $f,f'$ to maps into $C_d$. Also, let $D,D'$ be the corresponding liftings of $\mathcal{L}$ to relative divisors on $C\times S$ [How does the correspondence goes? I do not see what they mean] Now, to give a lifting of either $f$ or $\mathcal{L}$ corresponds [again, how?] to choosing a section over $S$ of $\mathbb{P}(\phi_*\mathcal{L})\cong\mathbb{P}^{d-g}\times S$. Thus there is a lifting $g''$ of $f$ whose corresponding relative divisor is $D'$. But then, by the universal property of the universal divisor, $g'=g''$ [this seems fine] and hence $f=f'$.

I would be grateful to anyone who could answer my questions in brackets and more generally shed any light on how the overall argument works.

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  • $\begingroup$ As you note, $C_d$ parameterizes effective divisors of degree $d$ on $C$. Given $g: S \to C_d$, we want a divisor on $S \times C$ whose fiber over a point $s$ in $S$ is $g(s)$. A good way to construct this is to first construct a divisor on $C_d \times C$ from the irreducible closed set where the last point of $C$ is equal to one of the first $d$ points of $C$, and then pullback from $C_d \times C$ to $S \times C$. This is the lift they are talking about. $\endgroup$
    – Will Sawin
    Commented Jan 10, 2021 at 15:16
  • $\begingroup$ A general trick is that given a map $Y \to Z$ and a map $f \colon X \to Z$, lifting $f$ to a map $X \to Y$ is equivalent to finding a section of the natural map $X \times_Z Y \to X$. This is a restatement of the universal property of the fiber product. So whenever someone claims that a lift is equivalent to a section of something, that thing should be the fiber product. Because $C_d$ is a trivial $\mathbb P^{d-g}$-bundle, the fiber product is again a trivial $\mathbb P^{d-g}$-bundle. $\endgroup$
    – Will Sawin
    Commented Jan 10, 2021 at 15:20
  • $\begingroup$ Many thanks, now I get it $\endgroup$
    – Vanni
    Commented Jan 11, 2021 at 11:10

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