Let $f:X\to Y$ be a universal homeomorphism of schemes, $R$ a coefficient ring. Which assumptions on $f$ and $R$ are suffient to ensure that the pullback map $f^*$ of $R$linear Chow groups is bijective (clearly, it suffices to verify injectivity)? Actually, I am mostly interested in the case where $Y$ is a smooth variety over a field $k$ and $X=Y\times_{Spec\,k}Spec\,k^{perf}$; yet I would be deeply grateful for any references on this subject!

1$\begingroup$ I am afraid this will happen only in trivial cases. For instance if $Y$ is a smooth projective curve over $\mathbb{Q}$, with Jacobian variety $J$, $CH^1(Y)=\operatorname{Pic}(Y) $ is finitely generated, while $\operatorname{Pic}(X) =\mathbb{Z}\oplus J(\bar{\mathbb{Q}})$ is not. $\endgroup$– abxJun 19, 2020 at 8:28

$\begingroup$ Well, in characteristic $0$ only "trivial" universal homeomorphisms exist, yes.:) $\endgroup$– Mikhail BondarkoJun 19, 2020 at 8:50

$\begingroup$ @MikhailBondarko how about the resolution of a cusp? (Tag 0BRC) $\endgroup$– R. van Dobben de BruynJun 19, 2020 at 17:21
2 Answers
This is easy to answer in equicharacteristic under the additional assumption that $f$ is finite flat and $X$ and $Y$ are irreducible and generically reduced, using $\mathbf Z[\tfrac{1}{p}]$coefficients, where $p$ is the exponential characteristic (i.e. $1$ if we're in characteristic $0$).
Indeed, in this case push and pull shows that the composition $$\operatorname{CH}_*(Y) \stackrel{f^*}\to \operatorname{CH}_*(X) \stackrel{f_*}\to \operatorname{CH}_*(Y)$$ equals the degree of $f$ (see e.g. Fulton, Example 1.7.4). Since $f$ is radicial, the extension $K(Y) \to K(X)$ has degree $p^n$ for some $n \in \mathbf N$, hence the same goes for $f$. Thus $f^*$ is injective with $\mathbf Z[\tfrac{1}{p}]$coefficients.
Similarly, if $V \subseteq Y$ is an integral subscheme with image $W$, then $f_*[V] = p^r[W]$ for some $r \in \mathbf N$, and $f^*[W] = m[V]$ for some $m \in \mathbf Z_{>0}$. Pushing forward gives $$p^n[W] = f_*f^*[W] = mp^r[W],$$ so $m$ is a power of $p$ as well (these equalities hold in $Z_*(Y)$, so there is no torsion), showing surjectivity with $\mathbf Z[\tfrac{1}{p}]$coefficients. $\square$
This is enough to go from $k$ to $k^{\operatorname{perf}}$ with $\mathbf Z[\tfrac{1}{p}]$coefficients by a limit argument (see e.g. Tag 0FH6).
Example. Injectivity is not true integrally, even for divisors on smooth projective $\bar{\mathbf F}_p$varieties. For example, consider a $\pmb\mu_p$quotient $X \to Y$ of smooth projective varieties with $X$ a complete intersection of dimension $\geq 2$. Then $\mathbf{Pic}^\tau_X = 0$ but $\mathbf{Pic}^\tau_Y = (\pmb\mu_p)^\vee = \mathbf Z/p$. See for example Cor. 1.2 of this preprint of mine for a modern account (but the result is much older).
I'm not sure if the same phenomenon occurs for a base change along a purely inseparable field extension.
For surjectivity with $\mathbf Z$coefficients, by the argument above, it suffices to move a subscheme to a linearly equivalent one for which $[K(V):K(W)] = [K(X):K(Y)]$. This seems plausible at least in the quasiprojective case.
Remark. If you don't assume $f$ is flat or l.c.i. or $Y$ is regular, I'm not sure what the definition of $f^*$ is. (At least Fulton does not define this, nor does the Stacks project as far as I can tell.)
If you work in equal characteristic, you have a chance of survival. In characteristic zero, this amounts to check that Chow groups do see nilpotent extensions. In characteristic $p>0$, if $p$ is invertible in your coeffcient Ring $R$, what you want is true as well (in fact this is true for higher Chow groups). I do not know any explicit reference though (but this does not mean such thing does not exist). However, this is an easy consequence of combinations of results which are documented.
To see it first observe that higher Chow groups are representable in $DM^{eff}(k,R)$ at least for equidimensional affine schemes: this is Theorem 4.2 in Bivariant cycle cohomology, by Friedlander and Voevodsky (in the book "Triangulated categories of motives"). This extends to non affine schemes as follows: using Suslin's beautiful paper Motivic complexes over nonperfect fields, we may replace $k$ by its perfection without changing $DM^{eff}$ with $R$ coefficients, and then apply the main result from Kelly's book which allows to replace arguments of resolution singularities as in the paper of Friedlander and Voevodsky above by resolution by $\ell$alterations with $\ell\neq p$, which exist by a well known result of Gabber.
Finally, note that all motivic invariants tend to be invariant under universal homeomorphisms after you invert the exponential characteristic: this may be seen a joint paper of mine with Déglise for $DM$ (a way to summarize what I wrote above is that, using Suslin's paper, alle the representability results in section 8 of my joint paper with Déglise are true over arbitrary fields), and, more generally, in a paper of Khan and Elmanto for $SH$.
If you are happy to tensor with $\mathbf{Q}$, then all this extend to arbitrary schemes, since $DM$ satisfies $h$descent with rational coefficients.

1$\begingroup$ Thank you! Yet does there exist an easier proof? Is it really necessary to invert p? Since $X$ is universally homeomorphic to $Y$, any cycle in it is a pullback of a cycle in $Y$, and any cycle in $\mathbb{A}^1\times X$ is a pullback of a cycle in $\mathbb{A}^1\times Y$; doesn't this help? $\endgroup$ Jun 19, 2020 at 12:52

1$\begingroup$ I am pretty sure that the Frobenius is not invertible with integral coefficients: this would contradict the $p$adic version of BlochKato (obtained by replacing $\ell$adic cohomology with synthomic cohomology). As for the existence of an elementary proof away from $p$torsion, I do not see why not (not that you may use Suslin's paper as a basis to reach such a thing, since most of it is about cycles themselves). $\endgroup$ Jun 19, 2020 at 13:03