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!



Are Chow groups invariant under universal homeomorphisms?
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](https://stacks.math.columbia.edu/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](https://arxiv.org/abs/2001.02787) 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 quasi-projective 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.)


