I had a question which seems to be straightforward but I wasn't able to figure it out. In page 13 of this paper a conjecture of beilison is mentioned that if $\mathcal{X}_{\mathbb{Z}}$ is a proper flat model of $X_{\mathbb{Q}}$ then the image of $K_*'(\mathcal{X})\otimes \mathbb{Q} \rightarrow K_*(X)\otimes \mathbb{Q}$ does not depend on the choice of $\mathcal{X}$. Then it is proved for the case that $\mathcal{X}$ is regular in the following lines. I do not understand where the regularity condition is being used. It is probably related to the commutativity of the right part of the diagram i.e. $\pi_*'$ and the pull-backs to $K_i(X)$. So why does it fail to be commutative if we do not assume that the integral models are regular? Can't we just replace all $K$ by $K'$ with the same argument?
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$\begingroup$ My assumption was to replace the proper regular with proper flat and all $K$ with $K'$ and repeat the same argument. In order the pullback to make sense in $K'$-theory the morphism needs to be flat. So $\pi$ has to be flat. I thought it is clear but right now I can't figure it out, probably that's the problem. I'm not even sure whether proper flatness of $\mathcal{X}$ and $\mathcal{X}'$ implies the flatness of $\mathcal{X}''$ $\endgroup$– user127776Commented Jun 3, 2019 at 5:17
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3$\begingroup$ $\pi$ and $\pi'$ will almost never be flat (since the maps are birational). Also, the conjecture is now known to be false! Explicit counterexamples were given by Rob de Jeu. $\endgroup$– nafCommented Jun 3, 2019 at 6:47
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