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Let $k$ be an algebraically closed field, and $X$ be a smooth variety. For any compactification $i: X \hookrightarrow Y$ (so $X$ is a dense open subset of $Y$), consider the induced map $i_!: H^i_{et,c}(X, \mathbb Z_l) \rightarrow H^i_{et,c}(Y,\mathbb Z_l)$ between compact supported etale cohomology groups.

Is $Ker(i_!)$ independent of the compactification (at least for $\mathbb Q_l$ coefficients )?

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The answer is yes, assuming $Y$ is smooth: this follows from the identification $$ \operatorname{Ker}(H^n_c(X,\mathbf Q_\ell) \to H^n_c(Y,\mathbf Q_\ell)) = W_{n-1}H^n_c(X,\mathbf Q_\ell),$$ where $W$ denotes the weight filtration. Clearly any element of $W_{n-1}$ is in the kernel, since the image of the map is pure of weight $n$. For the reverse inclusion use the long exact sequence of the pair in compact support cohomology to see that the kernel is the image of $H^{n-1}_c(Y \setminus X,\mathbf Q_\ell)$, which is of course of weights $\leq n-1$.

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  • $\begingroup$ Thank you ! How about the integral coefficient cohomology group, which may contain some torsion elements? $\endgroup$
    – Zhiyu
    Commented Jun 3, 2019 at 17:41
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    $\begingroup$ I believe the statement is true also integrally, but much more nontrivial: one needs to use the weight filtration on integral compactly supported étale cohomology constructed by Gillet and Soulé ("Descent, motives and K-theory"). $\endgroup$
    – Dan Petersen
    Commented Jun 3, 2019 at 17:51

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