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Consider the (complex) geometric realization of the motivic cohomology theory on simplicial presheaves over complex smooth schemes, which is a functorial homomorphism of $R$-modules, where $R$ is the coefficient ring: \begin{equation}\label{eq} \varphi: H^{s,t}_{mot}(X;R)\to H^s(X(\mathbb{C});R). \end{equation}

When restricted to the Chow ring $\varphi:H^{2*,*}_{mot}(X;\mathbb{Z})\to H^{2*}(X(\mathbb{C});\mathbb{Z})$ and $X$ a complex algebraic variety, it is a classical result that $\varphi$ is the cycle class map ring homomorphism.

Generally, is the map $\varphi: H^{*,*}_{mot}(X;R)\to H^*(X(\mathbb{C});R)$ an R-algebra homomorphism? I was not able to find a direct proof of this result in the literature. Or perhaps this is an obvious fact that I am missing.

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Complex realization sends the motivic Eilenberg-MacLane spectrum to the classical Eilenberg-MacLane spectrum [Theorem 5.5 in Marc Levine's "A comparison of motivic and classical stable homotopy theories]. It is also a symmetric monoidal functor, hence preserves ring spectra. The result follows.

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    $\begingroup$ Just to add a few more details, the map $\varphi$ in the question is the map induced on homotopy sheaves by the unit $1\to B_*B^*$ where $B^*:\operatorname{SH}(\mathbb{C})\to \operatorname{Sp}$ is the colimit-preserving, symmetric monoidal Betti realization functor and $B_*$ is its right adjoint. Since $B^*$ is symmetric monoidal (basically by definition), the unit is a lax symmetric monoidal natural transformation and so, in particular, it induces a map of algebras when given an algebraic variety. $\endgroup$
    – Denis Nardin
    Commented Sep 2, 2021 at 21:47

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