I am reading the paper "$p$-adic cohomology: from theory to practice" 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 $\text{Gal}(\overline{\mathbb{F}}_q/\mathbb{F}_q)$)?