Given a surjective homomorphism of abelian varieties $f:A\rightarrow B$ where $\text{dim}(A)>\text{dim}(B)$, does $f^*$ induce a rational injection of algebraic $K$-theory? According to the projection formula we have: $$f_*(x.f^*y)=f_*x.y$$ So if the image of the proper pushforward $f_*$ contains a rationally invertible element (integers) this holds. (This is true for example if $A$ contains a cycles that is finite and surjective over $B$). But this can also without these assumptions.
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2$\begingroup$ Yes. By the Poincare complete reducibility theorem, there exists an abelian subvariety $A'$ of $A$ such that the induced map $A' \to B$ is an isogeny, i.e., finite surjective. $\endgroup$– nafCommented Apr 12, 2021 at 11:36
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$\begingroup$ @naf Thanks for saving my life! $\endgroup$– user127776Commented Apr 12, 2021 at 11:40
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$\begingroup$ The injectivity is actually true for any surjective proper morphism of smooth varieties. The point is that one does not actually need a subvariety that is finite and surjective, generically finite and surjective suffices, and such a subvariety always exists. $\endgroup$– nafCommented Apr 13, 2021 at 2:03
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$\begingroup$ @naf Interesting. One thing to note: in the projection formula the lower star is actually proper pushforward not just the naïve pushforward. (So it might be better to write lower shriek rather than star). But in the case of abelian varieties since tangent bundles are trivial one can see that lower shriek of cycles coincides with lower star (by GRR). Without the assumption of abelian variety, this becomes complicated, not sure how that works. $\endgroup$– user127776Commented Apr 13, 2021 at 2:36
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1$\begingroup$ The fact that the tangent bundle won't be trivial in general doesn't really matter since the support of the higher derived functors (for a generically finite map) will be a proper subvariety (so GRR is actually irrelevant for the question). $\endgroup$– nafCommented Apr 13, 2021 at 4:05
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