Proper morphism induces a map between compact support etale cohomology groups

Question

Let $f : Y \to X$ be a proper morphism between noetherian separated schemes of finte type (over a noetherian scheme $S$), and $\mathscr{F}$ a sheaf on the small etale site on $X$. Then does $f$ induce $f : H_c^i(X, \mathscr{F}) \to H_c^i (Y, f^* \mathscr{F})$?

The compact support cohomology is not a derived functor. So the canonical map between $H_c^0$ does not induce maps of $\delta$-functors.

Here's what I tried so far: Let $j : X \to \overline{X}$ be a compactification. Then since $f$ is separated and of finite type, $j \circ f$ is so. Therefore there is a compactification $i : Y \to \overline{Y}$ over $\overline{X}$. (Nagata compactification) So we have a commutative diagram: $\require{AMScd}$ \begin{CD} Y @>{i}>> \overline{Y}\\ @V{f}VV @V{\overline{f}}VV\\ X @>{j}>> \overline{X} \end{CD} Now $H_c^i(X, \mathscr{F}) = H^i(\overline{X}, j_!\mathscr{F}), H_c^i (Y, f^* \mathscr{F}) = H^i(\overline{Y}, i_! f^*\mathscr{F})$. But since the diagram is not Cartesian, it seems that $i_!f^* \neq \overline{f}^*j_!$. And even if these two coincide, since $j_!$ does not takes injectives to injectices, the canonical map between $H^0$ does not induce the required maps.

Thank you very much!

Answer 1

There seem to be two concerns :

• How to define the base-change map
• How to extend the base-change from the cohomological degree $0$ to the degree $n$

Let me answer first the second concern. If you are familiar with total derived functor, then this can be answer really easily. By adjunction, there is a natural morphism $\mathcal{F}\to R\overline{f}_*\overline{f}^*\mathcal{F}$ in $D(\overline{X}_{et})$. There is also a natural identification $H^*(\overline{X},R\overline{f}_*\mathcal{G})=H^*(\overline{Y},\mathcal{G})$ for any $\mathcal{G}\in D(\overline{Y})$. Composing these two, we get the usual pullback in cohomology : $$ H^*(\overline{X},\mathcal{F})\to H^*(\overline{X},R\overline{f}_*\overline{f}^*\mathcal{F})=H^*(\overline{Y},\overline{f}^*\mathcal{F})$$ From this, assuming we have our base-change map $(*)$, we can define : $$H^*_c(X,\mathcal{F})=H^*(\overline{X},j_!\mathcal{F})\to H^*(\overline{X},R\overline{f}_*\overline{f}^*j_!\mathcal{F})= H^*(\overline{Y},\overline{f}^*j_!\mathcal{F})\overset{(*)}\to H^*(\overline{Y}, i_!f^*\mathcal{F})=H^*_c(Y,f^*\mathcal{F})$$

So, the first concern remains : we have to construct the map $(*)$, or rather the map $\overline{f}^*j_!\to i_!f^*$. Consider $Y'=X\times_{\overline{X}}\overline{Y}$, so we have the commutative diagram : $$\require{AMScd} \begin{CD} Y@>i_1>>Y'@>i_2>>\overline{Y}\\ @VfVV@VVf'V@VV\overline{f}V\\ X@=X@>>j>\overline{X} \end{CD}$$ With $i=i_2i_1$ and $f=f'i_1$. So we have : $$\overline{f}^*j_!\overset{(1)}\simeq {i_2}_!f'^*\overset{(2)}\to {i_2}_!{i_1}_!i_1^*f'^*=i_!f^*$$ Where :

• the isomorphism labelled $(1)$ follows from base change isomorphism induced by the right square, which is cartesian.
• the morphism labelled $(2)$ follows from the unit of adjunction $1\to {i_1}_*i_1^*={i_1}_!i_1^*$ because $i_1$ is a closed immersion (this is where we use the properness of $f$. We have indeed $f=f'i_1$, therefore, since $f$ is proper, so is $i_1$. Because $i_1$ is clearly a quasi-finite, this is a closed immersion).