TITLE: Is the Chow ring of a wonderful model for a hyperplane arrangement isomorphic to the singular cohomology ring?
QUESTION [16 upvotes]: In the article "Hodge theory for combinatorial geometries" by Adiprasito, Huh and Katz, it it claimed in the proof of theorem 5.12 that there is a Chow equivalence between the de Concini-Processi wonderful model $Y$ of an arrangement, and a certain toric variety $X$.
However, the source they cite only proves that
$$H^*(Y)\cong Ch(X)$$
It seems that to draw the conclusion that $Ch(Y)\cong Ch(X)$ it is used that $Ch(Y)\cong H^*(Y)$, but I do not know why that is the case?

Here is what I know: $Y\subset X$ as a closed subset, and the relevant maps are (should be) induced by this inclusion. We have a diagram
$$
\require{AMScd}
\begin{CD}
Ch(X) @>{ch}>> H^*(X)\\
@VVV @VVV \\
Ch(Y) @>{ch}>> H^*(Y)
\end{CD}
$$
and we know that the diagonal arrow is an isomorphism. To establish that $Ch(X)\cong Ch(Y)$ it is enough to show that $Ch(Y)\cong H^*(Y)$. Since the diagonal is an isomorphism, the map $Ch(Y)\to H^*(Y)$ is surjective. I do not see how to prove injectiveness though.

REPLY [6 votes]: The isomorphism $H^\cdot(Y)\cong Ch^\cdot(X)$ is shown by 
Eva Feichtner and Sergey Yuzvinsky in Feichtner, E. & Yuzvinsky, S. Invent. Math. (2004) 155: 515. https://doi.org/10.1007/s00222-003-0327-2
It's worth emphasizing that the inclusion $Y\subset X$ is not a homotopy equivalence, and generally $H^\cdot(X)\not\cong H^\cdot(Y)$.
The property that $H^\cdot(Y)\cong Ch^\cdot(Y)$ for iterated blowups of projective spaces I think is due to Sean Keel, Trans. Amer. Math. Soc. 330 (1992), 545-574, 
https://doi.org/10.1090/S0002-9947-1992-1034665-0, in line with Libli's comment above.