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$\DeclareMathOperator\Alb{Alb}\DeclareMathOperator\CH{CH}$I'm wondering how the Albanese functor interacts with correspondences. Specifically if $X$, $Y$ are say smooth projective schemes over an algebraically closed field $k$ and $Z$ a correspondence between them. Does this induce a morphism of abelian varieties $$\Alb(X)\to \Alb(Y)?$$

If $X$, $Y$ are curves this is the case as the Albanese and Jacobian agree, then one can quote say Cor 6.3 of Milne's notes on Jacobian Varieties, which says that morphisms on Jacobians of curves are in bijection to those divisors (or line bundles) which are trivial after pulling back to either factor.

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  • $\begingroup$ Sorry, why does a correspondence give a morphism in the curve case? If $X$ and $Y$ are both elliptic curves, there are many correspondences that are not morphisms. Or do you mean whether a correspondence from $X$ to $Y$ induces a correspondence from $\operatorname{Alb}(X)$ to $\operatorname{Alb}(Y)$? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jun 21 at 16:36
  • $\begingroup$ @R.vanDobbendeBruyn I do mean to ask whether a correspondence from $X$ to $Y$ induces a morphism between the Albanese. Doesn't a correspondence induce a Galois-equivariant morphism of the $H^1$ which then (by Tate) induces a morphism in the isogeny category between the Albanese? (although this seems to work for elliptic curves and Albanese). Does this make sense? $\endgroup$
    – curious math guy
    Commented Jun 21 at 18:12
  • $\begingroup$ Yes, one does get an induced morphism on Albanese varieties. It might be easier to see this using the description of the Albanese variety as the dual of the Picard variety: by basic intersection theory one has a map (in the opposite direction) on Picard groups over any extension of $k$. This is enough to get the morphism. $\endgroup$
    – naf
    Commented Jun 28 at 1:37
  • $\begingroup$ If $X\to Y$ is a morphism of smooth projective schemes over $k$, then you get $Alb(X) \to Alb(Y)$ and $Alb(Y)\to Alb(X)$ as naf explains. So, if you have a correspondence $Z\to X, Z\to Y$, then you get a triangle of morphisms, hence a morphism from $Alb(X)$ to $Alb(Y)$. $\endgroup$
    – Ariyan Javanpeykar
    Commented Jun 28 at 12:14

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