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 )? 



Is the induced map between etale cohomology groups independent of the compactification?
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 $\leq n-1$.


