Suppose everything below is defined over $k=\overline{\mathbb{F}}\_q$.


Let $H$ be a connected algebraic group acting on a separated variety $Y$. Denote the morphism $H\times Y\rightarrow H\times Y; (h,y)\mapsto (h,hy)$ by $f$, and denote by $\pi$ the left projection of $H\times Y$ to $H$. By applying proper base change to the Cartesian diagram
\begin{equation\*}
\require{AMScd}
\begin{CD}
 H\times Y @>>> Y\\
 @V \pi V V @VV V\\
 H @>>> \mathrm{Spec} (k)
\end{CD}
\end{equation\*}
we see $R^i\pi\_!\mathbb{Z}/n$ is the constant sheaf $H^i\_c(Y,\mathbb{Z}/n)$ on $H$. 


In the argument of Proposition 6.4 of Deligne--Lusztig's seminal [paper](http://www.jstor.org/stable/1971021), they assert that $f$ gives an endomorphism on this constant sheaf, in the way that at each stalk at $h$ the endomorphism at $H^i\_c(Y,\mathbb{Z}/n)\_h=H^i\_c(Y,\mathbb{Z}/n)$ is given by the induced map of $h$.


I understand that at each stalk at $h$, there is an induced action of $h$ on the cohomology group, but why they constitute a sheaf endomorphism?



A homotopy argument in etale topology
It follows from functoriality. For three varieties $X, Y, Z$ with maps $f:X \to Z$ and $g: Y\to Z$, for every isomorphism $h: X \to Y$ forming a commutative triangle with $f$ and $g$, we get an induced isomorphism between $R^i f\_! Y$ and $R^i g\_! Y$.


Apply this to your morphism $f$ and you get the desired action.


You can see it's the desired action because this isomorphism agrees with base change - because everything in base change is defined canonically so has to commute with isomorphisms.


