Let $X$ be an algebraic variety over $\mathbb{C}$. If $X$ is smooth, the étale cohomology $H^p_{\textrm{ét}}(X,\mathbb{Z}/n)$ is isomorphic to the singular cohomology $H^p(X(\mathbb{C}),\mathbb{Z}/n)$. What is the situation if $X$ is not smooth? Are there counter-examples?
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In Katz's review of $\ell$-adic cohomology in the first "Motives" volume, this isomorphism is stated without any smoothness assumptions, with a reference to SGA4, XVI 4.1 (which I don't have easily available).
answered Jan 21, 2015 at 10:41
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$\begingroup$ Right, thanks! I had wrongly looked at SGA4, XI which treats only the smooth case. $\endgroup$– abxCommented Jan 21, 2015 at 11:03
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8$\begingroup$ The reliance on resolution of singularities (due to the lack of an excision sequence as robust as for $H^{\ast}_c$) can be avoided: Berkovich's proof of the comparison isomorphism over all non-archimedean fields $k$ in his IHES paper (assuming $\ell\ne {\rm{char}}(k)$), based on the ideas from Deligne's "generic base change" expose in SGA 4.5, and adapts without difficulty to the complex-analytic case. Passing to the $\ell$-adic limit (allowing constructible coefficients, as always) lies deeper if one wishes to avoid using triangulability of analytifications of non-smooth $\mathbf{C}$-schemes. $\endgroup$ Commented Jan 21, 2015 at 13:59
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2$\begingroup$ @user74230 — That deserves to be an answer in its own right. Or maybe Dan can incorporate it in his answer… $\endgroup$– jmcCommented Feb 3, 2015 at 14:16