I am reading the paper ["$p$-adic cohomology: from theory to practice"](http://swc.math.arizona.edu/aws/2007/KedlayaNotes11Mar.pdf) by K. S. Kedlaya. I have several naive questions about section 2: Frobenius action on de Rham cohomology. As a physicist, I lack proper backgrounds needed for this paper, and I am very grateful if someone could explain my questions carefully.


**Question 1**: In the remark 2.1.3 (page 16), (in the notation of this paper), a morphism $\overline{f}:(\overline{X},\overline{Z}) \rightarrow (\overline{X}',\overline{Z}')$ functorially induces a homomorphism 
\begin{equation}
H^i\_{dR}(X',Z') \rightarrow H^i\_{dR}(X,Z)
\end{equation}
why do we want $\overline{f}$ induces this morphism which is at a higher level (since there is a morphism we already easily get $H^i\_{dR}(\overline{X}',\overline{Z}') \rightarrow H^i\_{dR}(\overline{X},\overline{Z})$)? Is this because Theorem 2.1.2, i.e. the canonical comparison isomorphism is between $H^i\_{dR}(X,Z)$ and $H^i\_{crys}(\overline{X},\overline{Z})$?


**Question 2**: I guess the $q$-th power Frobenius map $F\_{q}$ is induced by the $\mathbb{F}\_q$-algebra homomorphism $A \rightarrow A$ which sends $x$ to $x^q$? The map $F\_q$ induces an endomorphism $F\_q^\*$ of $H^i\_{dR}(X,Z)$, under the comparison isomorphism, it induces an endomorphism of $H^i\_{crys}(\overline{X},\overline{Z})$. Is $F\_q^\*$ sent to ation of geometric Frobenius (the inverse of Frobenius element of $\text{Gal}(\overline{\mathbb{F}}\_q/\mathbb{F}\_q)$)? 



Frobenius actions on de Rham cohomology, clarify questions on a paper of Kedlaya
For question 1, you are correct that the map
$$\bar f^\* \colon H^i\_{\text{dR}}(X',Z') \to H^i\_{\text{dR}}(X,Z)$$
is the one obtained from
$$\bar f^\* \colon H^i\_{\text{crys}}(\bar X',\bar Z') \to H^i\_{\text{crys}}(\bar X,\bar Z)$$
using the comparison isomorphism. This is exactly the point of Remark 2.1.3: it is remarkable that the map $\bar f^\*$ on $H^i\_{\text{dR}}$ exists, because a priori de Rham cohomology is only functorial for morphisms $f \colon (X,Z) \to (X',Z')$ (think: a lift of $\bar f$).




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For question 2, note that the geometric Frobenius is not an element of $\operatorname{Gal}(\bar{\mathbb F}\_q/\mathbb F\_q)$. The composition of geometric Frobenius with arithmetic Frobenius is the absolute Frobenius. What happens is that on étale cohomology, the absolute Frobenius acts trivially, hence geometric Frobenius *acts* as the inverse of arithmetic Frobenius. See for example [this MO post](https://mathoverflow.net/a/30316/82179).


In the case of crystalline cohomology, one usually only studies the action of the absolute Frobenius. In the case of a variety over $\mathbb F\_q$, the $q$-power Frobenius acts trivially on $\mathbb F\_q$, hence geometric $q$-power Frobenius equals absolute $q$-power Frobenius, and the arithmetic $q$-power Frobenius is trivial. I guess one could study the arithmetic $p$-power Frobenius viewed as an element of $\operatorname{Gal}(\mathbb F\_q/\mathbb F\_p)$, but this is not usually considered.


