I'm talking about etale sheaves.
For a morphism $f:X\to Y$ of schemes, there are two definition of $Rf_!:D(X)\to D(Y)$.
The usual derived functor:
$\forall\mathcal{F}\in Sh(X)$, let $f_!\mathcal{F}$ be the sheaf associated to the presheaf $U\mapsto\{s\in\Gamma(U\times_YX,\mathcal{F}):\operatorname{Supp}s$ is proper over $U\}$
$f_!:Sh(X)\to Sh(Y)$ is left exact, and we can consider its right derived functor $Rf_!$ (?)
Via compactification:
Under some finiteness assumptions, we can consider "compactifiable" morphisms
If $f:X\to Y$ is compactifiable (over $S$), let $X\overset{i}\hookrightarrow\bar{X}\overset{\bar{f}}{\to}Y$ be a compactification
i.e. $i:X\hookrightarrow\bar{X}$ open immersion, $\bar{f}:\bar{X}\to Y$ proper.
We can set $Rf_!=R\bar{f}_*i_!$, which make sense since extension by zero ($i_!$) is exact
(See for example Stack Project section 62.9)
In p357 of $\textit{Etale Cohomology Theory}$ of Lei Fu, he said that $Rf_!$ defined via compactification needn't be the derived functor of $f_!$. I didn't find further discussion about the relation between the two definition.
It seems that if the usual derived functor $Rf_!$ exists, $f=\bar{f}\circ i$, then $R\bar{f}_*\circ i_!\simeq R\bar{f}_!\circ Ri_!\simeq Rf_!$.
Is this a correct argument?