For regular schemes $X$ and $Y$, and a faithfully flat morphism $f:Y \to X$, there is a flat pullback map of Grothendieck groups: $$ f^*:K^0(X) \to K^0(Y). $$
Is this map injective?
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3$\begingroup$ Even for finite etale morphisms, the pullback is typically not injective, e.g., for the "multiplication by $n$" isogeny of an Abelian variety to itself, the kernel of the pullback map on Picard groups is the $n$-torsion subgroup of the Picard group. $\endgroup$– Jason StarrCommented Apr 1, 2022 at 23:42
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It is not. For example, take a surjective morphism $f:\mathbb{A}^1\to\mathbb{P}^1$. $K^0(\mathbb{P}^1)=\mathbb{Z}\oplus\mathbb{Z}$, while $K^0(\mathbb{A}^1)=\mathbb{Z}$.
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1$\begingroup$ $f$ is not surjective (and so not faithfully flat).. $\endgroup$ Commented Apr 2, 2022 at 9:23
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4$\begingroup$ @DenisNardin There are many surjective maps of the kind I describe. For example take a two to one map from $\mathbb{P}^1$ to itself and remove a general point from the domain. $\endgroup$– MohanCommented Apr 2, 2022 at 14:05