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The basic question was whether there is a ]: The basic question was whether there is a ]:
:<math>A^*(X) \to \operatorname{H}^*(X, \mathbb{Z}).</math> :<math>A^*(X) \to \operatorname{H}^*(X, \mathbb{Z}).</math>

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The basic question was whether there is a cycle map:

A ( X ) H ( X , Z ) . {\displaystyle A^{*}(X)\to \operatorname {H} ^{*}(X,\mathbb {Z} ).}

If X is smooth, such a map exists since A ( X ) {\displaystyle A^{*}(X)} is the usual Chow ring of X. (Totaro 2014) harv error: no target: CITEREFTotaro2014 (help) has shown that rationally there is no such a map with good properties even if X is a linear variety, roughly a variety admitting a cell decomposition. He also notes that Voevodsky’s motivic cohomology ring is "probably more useful " than the operational Chow ring for a singular scheme (§ 8 of loc. cit.)


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