Does action on $X$ define an action on cohomology $H^i(X,F)$?

Question

Let $X$ be a topological space, $F$ a sheaf on $X$, given an automorphism $f\colon X\to X$, do we have an induced morphism $H^i(X,F)\to H^i(X,F)$ for any sheaf on $X$?(I can understand for $(X,f^*F)\to (Y,F)$, there's a map $H^i(Y,F)\to H^i(X,f^*F)$, given by $H^i(Y,F)\to H^i(Y,f_*f^*F)\to H^i(X,f^*F)$, where the last step is given by choosing an resolution of $f^*F\to I^\cdot$ and define by induced map on hypercohomology for $f_*f^*F\to f^*I^\cdot$, LHS viewed as complex concentrated at degree $0$)

Answer 1

Merely by transport of structure, there is a natural map (covariant in $f$) $H^i(X,F)\to H^i(X,f_*F)$; equivalently, replacing $f$ by its inverse, there is a natural map $H^i(X,f^*F)\to H^i(X,F)$ contravariant in $f$ (note that $f_*$ and $f^*$ are inverse functors when $f$ is a homeomorphism). However, there is no reason there should be a natural map $H^i(X,F)\to H^i(X,F)$ associated to $f$, because $f$ is only an automorphism of $X$, not of the pair $(X,F)$. I don't know how to prove rigorously that no such (sufficiently nice) natural map exists, but I see no reason at all to expect there to be one. If you are unconvinced, then I ask you: what do you think the map should be in the case $i=0$?