## Tag `03SL`

## 53.86. Frobenii

In this section we will prove a ''baffling'' theorem. A topological analogue of the baffling theorem is the following.

Exercise 53.86.1. Let $X$ be a topological space and $g : X \to X$ a continuous map such that $g^{-1}(U) = U$ for all opens $U$ of $X$. Then $g$ induces the identity on cohomology on $X$ (for any coefficients).

We now turn to the statement for the étale site.

Lemma 53.86.2. Let $X$ be a scheme and $g : X \to X$ a morphism. Assume that for all $\varphi : U \to X$ étale, there is an isomorphism $$ \xymatrix{ U \ar[rd]_\varphi \ar[rr]^-\sim & & {U \times_{\varphi, X, g} X} \ar[ld]^{\text{pr}_2} \\ & X } $$ functorial in $U$. Then $g$ induces the identity on cohomology (for any sheaf).

Proof.The proof is formal and without difficulty. $\square$Please see Varieties, Section 32.35 for a discussion of different variants of the Frobenius morphism.

Theorem 53.86.3 (The Baffling Theorem). Let $X$ be a scheme in characteristic $p>0$. Then the absolute frobenius induces (by pullback) the trivial map on cohomology, i.e., for all integers $j\geq 0$, $$ F_X^* : H^j (X, \underline{\mathbf{Z}/n\mathbf{Z}}) \longrightarrow H^j (X, \underline{\mathbf{Z}/n\mathbf{Z}}) $$ is the identity.

This theorem is purely formal. It is a good idea, however, to review how to compute the pullback of a cohomology class. Let us simply say that in the case where cohomology agrees with Čech cohomology, it suffices to pull back (using the fiber products on a site) the Čech cocycles. The general case is quite technical, see Hypercoverings, Theorem 24.10.1. To prove the theorem, we merely verify that the assumption of Lemma 53.86.2 holds for the frobenius.

Proof of Theorem 53.86.3.We need to verify the existence of a functorial isomorphism as above. For an étale morphism $\varphi : U \to X$, consider the diagram $$ \xymatrix{ U \ar@{-->}[rd] \ar@/^1pc/[rrd]^{F_U} \ar@/_1pc/[rdd]_\varphi \\ & {U \times_{\varphi, X, F_X} X} \ar[r]^{\text{pr}_1} \ar[d]^{\text{pr}_2} & U \ar[d]^\varphi \\ & X \ar[r]^{F_X} & X. } $$ The dotted arrow is an étale morphism which induces an isomorphism on the underlying topological spaces, so it is an isomorphism. $\square$Definition 53.86.4. Let $k$ be a finite field with $q = p^f$ elements. Let $X$ be a scheme over $k$. The

geometric frobeniusof $X$ is the morphism $\pi_X : X \to X$ over $\mathop{\rm Spec}(k)$ which equals $F_X^f$.Since $\pi_X$ is a morphism over $k$, we can base change it to any scheme over $k$. In particular we can base change it to the algebraic closure $\bar k$ and get a morphism $\pi_X : X_{\bar k} \to X_{\bar k}$. Using $\pi_X$ also for this base change should not be confusing as $X_{\bar k}$ does not have a geometric frobenius of its own.

Lemma 53.86.5. Let $\mathcal{F}$ be a sheaf on $X_{\acute{e}tale}$. Then there are canonical isomorphisms $\pi_X^{-1} \mathcal{F} \cong \mathcal{F}$ and $\mathcal{F} \cong {\pi_X}_*\mathcal{F}$.

This is false for the fppf site.

Proof.Let $\varphi : U \to X$ be étale. Recall that ${\pi_X}_* \mathcal{F} (U) = \mathcal{F} (U \times_{\varphi, X, \pi_X} X)$. Since $\pi_X = F_X^f$, it follows from the proof of Theorem 53.86.3 that there is a functorial isomorphism $$ \xymatrix{ U \ar[rd]_{\varphi} \ar[rr]_-{\gamma_U} & & U \times_{\varphi, X, \pi_X} X \ar[ld]^{\text{pr}_2} \\ & X } $$ where $\gamma_U = (\varphi, F_U^f)$. Now we define an isomorphism $$ \mathcal{F} (U) \longrightarrow {\pi_X}_* \mathcal{F} (U) = \mathcal{F} (U \times_{\varphi, X, \pi_X} X) $$ by taking the restriction map of $\mathcal{F}$ along $\gamma_U^{-1}$. The other isomorphism is analogous. $\square$Remark 53.86.6. It may or may not be the case that $F^f_U$ equals $\pi_U$.

We continue discussion cohomology of sheaves on our scheme $X$ over the finite field $k$ with $q = p^f$ elements. Fix an algebraic closure $\bar k$ of $k$ and write $G_k = \text{Gal}(\bar k/k)$ for the absolute Galois group of $k$. Let $\mathcal{F}$ be an abelian sheaf on $X_{\acute{e}tale}$. We will define a left $G_k$-module structure cohomology group $H^j (X_{\bar k}, \mathcal{F}|_{X_{\bar k}})$ as follows: if $\sigma \in G_k$, the diagram $$ \xymatrix{ X_{\bar k} \ar[rd] \ar[rr]^{\mathop{\rm Spec}(\sigma) \times \text{id}_X} & & X_{\bar k} \ar[ld] \\ & X } $$ commutes. Thus we can set, for $\xi \in H^j (X_{\bar k}, \mathcal{F}|_{X_{\bar k}})$ $$ \sigma \cdot \xi := (\mathop{\rm Spec}(\sigma) \times \text{id}_X)^*\xi \in H^j(X_{\bar k}, (\mathop{\rm Spec}(\sigma) \times \text{id}_X)^{-1} \mathcal{F}|{X_{\bar k}}) = H^j (X_{\bar k}, \mathcal{F}|_{X_{\bar k}}), $$ where the last equality follows from the commutativity of the previous diagram. This endows the latter group with the structure of a $G_k$-module.

Lemma 53.86.7. In the situation above denote $\alpha : X \to \mathop{\rm Spec}(k)$ the structure morphism. Consider the stalk $(R^j\alpha_*\mathcal{F})_{\mathop{\rm Spec}(\bar k)}$ endowed with its natural Galois action as in Section 53.55. Then the identification $$ (R^j\alpha_*\mathcal{F})_{\mathop{\rm Spec}(\bar k)} \cong H^j (X_{\bar k}, \mathcal{F}|_{X_{\bar k}}) $$ from Theorem 53.52.1 is an isomorphism of $G_k$-modules.

A similar result holds comparing $(R^j\alpha_!\mathcal{F})_{\mathop{\rm Spec}(\bar k)}$ with $H^j_c (X_{\bar k}, \mathcal{F}|_{X_{\bar k}})$.

Proof.Omitted. $\square$Definition 53.86.8. The

arithmetic frobeniusis the map $\text{frob}_k : \bar k \to \bar k$, $x \mapsto x^q$ of $G_k$.Theorem 53.86.9. Let $\mathcal{F}$ be an abelian sheaf on $X_{\acute{e}tale}$. Then for all $j\geq 0$, $\text{frob}_k$ acts on the cohomology group $H^j(X_{\bar k}, \mathcal{F}|_{X_{\bar k}})$ as the inverse of the map $\pi_X^*$.

The map $\pi_X^*$ is defined by the composition $$ H^j(X_{\bar k}, \mathcal{F}|_{X_{\bar k}}) \xrightarrow{{\pi_X}_{\bar k}^*} H^j(X_{\bar k}, (\pi_X^{-1} \mathcal{F})|_{X_{\bar k}}) \cong H^j(X_{\bar k}, \mathcal{F}|_{X_{\bar k}}). $$ where the last isomorphism comes from the canonical isomorphism $\pi_X^{-1} \mathcal{F} \cong \mathcal{F}$ of Lemma 53.86.5.

Proof.The composition $X_{\bar k} \xrightarrow{\mathop{\rm Spec}(\text{frob}_k)} X_{\bar k} \xrightarrow{\pi_X} X_{\bar k}$ is equal to $F_{X_{\bar k}}^f$, hence the result follows from the baffling theorem suitably generalized to nontrivial coefficients. Note that the previous composition commutes in the sense that $F_{X_{\bar k}}^f = \pi_X \circ \mathop{\rm Spec}(\text{frob}_k) = \mathop{\rm Spec}(\text{frob}_k) \circ \pi_X$. $\square$Definition 53.86.10. If $x \in X(k)$ is a rational point and $\bar x : \mathop{\rm Spec}(\bar k) \to X$ the geometric point lying over $x$, we let $\pi_x : \mathcal{F}_{\bar x} \to \mathcal{F}_{\bar x}$ denote the action by $\text{frob}_k^{-1}$ and call it the

geometric frobenius^{1}We can now make a more precise statement (albeit a false one) of the trace formula (53.85.0.1). Let $X$ be a finite type scheme of dimension 1 over a finite field $k$, $\ell$ a prime number and $\mathcal{F}$ a constructible, flat $\mathbf{Z}/\ell^n\mathbf{Z}$ sheaf. Then \begin{equation} \tag{53.86.10.1} \sum\nolimits_{x \in X(k)} \text{Tr}(\pi_X | \mathcal{F}_{\bar x}) = \sum\nolimits_{i = 0}^2 (-1)^i \text{Tr}(\pi_X^* | H^i_c(X_{\bar k}, \mathcal{F})) \end{equation} as elements of $\mathbf{Z}/\ell^n\mathbf{Z}$. The reason this equation is wrong is that the trace in the right-hand side does not make sense for the kind of sheaves considered. Before addressing this issue, we try to motivate the appearance of the geometric frobenius (apart from the fact that it is a natural morphism!).

Let us consider the case where $X = \mathbf{P}^1_k$ and $\mathcal{F} = \underline{\mathbf{Z}/\ell\mathbf{Z}}$. For any point, the Galois module $\mathcal{F}_{\bar x}$ is trivial, hence for any morphism $\varphi$ acting on $\mathcal{F}_{\bar x}$, the left-hand side is $$ \sum\nolimits_{x \in X(k)} \text{Tr}(\varphi | \mathcal{F}_{\bar x}) = \#\mathbf{P}^1_k(k) = q+1. $$ Now $\mathbf{P}^1_k$ is proper, so compactly supported cohomology equals standard cohomology, and so for a morphism $\pi : \mathbf{P}^1_k \to \mathbf{P}^1_k$, the right-hand side equals $$ \text{Tr}(\pi^* | H^0 (\mathbf{P}^1_{\bar k}, \underline{\mathbf{Z}/\ell\mathbf{Z}})) + \text{Tr}(\pi^* | H^2 (\mathbf{P}^1_{\bar k}, \underline{\mathbf{Z}/\ell\mathbf{Z}})). $$ The Galois module $H^0 (\mathbf{P}^1_{\bar k}, \underline{\mathbf{Z}/\ell\mathbf{Z}}) = \mathbf{Z}/\ell\mathbf{Z}$ is trivial, since the pullback of the identity is the identity. Hence the first trace is 1, regardless of $\pi$. For the second trace, we need to compute the pullback $\pi^* : H^2(\mathbf{P}^1_{\bar k}, \underline{\mathbf{Z}/\ell\mathbf{Z}}))$ for a map $\pi : \mathbf{P}^1_{\bar k} \to \mathbf{P}^1_{\bar k}$. This is a good exercise and the answer is multiplication by the degree of $\pi$ (for a proof see Lemma 53.68.2). In other words, this works as in the familiar situation of complex cohomology. In particular, if $\pi$ is the geometric frobenius we get $$ \text{Tr}(\pi_X^* | H^2 (\mathbf{P}^1_{\bar k}, \underline{\mathbf{Z}/\ell\mathbf{Z}})) = q $$ and if $\pi$ is the arithmetic frobenius then we get $$ \text{Tr}(\text{frob}_k^* | H^2 (\mathbf{P}^1_{\bar k}, \underline{\mathbf{Z}/\ell\mathbf{Z}})) = q^{-1}. $$ The latter option is clearly wrong.

Remark 53.86.11. The computation of the degrees can be done by lifting (in some obvious sense) to characteristic 0 and considering the situation with complex coefficients. This method almost never works, since lifting is in general impossible for schemes which are not projective space.

The question remains as to why we have to consider compactly supported cohomology. In fact, in view of Poincaré duality, it is not strictly necessary for smooth varieties, but it involves adding in certain powers of $q$. For example, let us consider the case where $X = \mathbf{A}^1_k$ and $\mathcal{F} = \underline{\mathbf{Z}/\ell\mathbf{Z}}$. The action on stalks is again trivial, so we only need look at the action on cohomology. But then $\pi_X^*$ acts as the identity on $H^0(\mathbf{A}^1_{\bar k}, \underline{\mathbf{Z}/\ell\mathbf{Z}})$ and as multiplication by $q$ on $H^2_c(\mathbf{A}^1_{\bar k}, \underline{\mathbf{Z}/\ell\mathbf{Z}})$.

The code snippet corresponding to this tag is a part of the file `etale-cohomology.tex` and is located in lines 15683–15989 (see updates for more information).

```
\section{Frobenii}
\label{section-frobenii}
\noindent
In this section we will prove a ``baffling'' theorem.
A topological analogue of the baffling theorem is the following.
\begin{exercise}
\label{exercise-baffling}
Let $X$ be a topological space and $g : X \to X$ a continuous map such that
$g^{-1}(U) = U$ for all opens $U$ of $X$. Then $g$ induces the identity on
cohomology on $X$ (for any coefficients).
\end{exercise}
\noindent
We now turn to the statement for the \'etale site.
\begin{lemma}
\label{lemma-baffling}
Let $X$ be a scheme and $g : X \to X$ a morphism. Assume that for all
$\varphi : U \to X$ \'etale, there is an isomorphism
$$
\xymatrix{
U \ar[rd]_\varphi \ar[rr]^-\sim & & {U
\times_{\varphi, X, g} X} \ar[ld]^{\text{pr}_2} \\
& X
}
$$
functorial in $U$. Then $g$ induces the identity on cohomology (for any sheaf).
\end{lemma}
\begin{proof}
The proof is formal and without difficulty.
\end{proof}
\noindent
Please see Varieties, Section \ref{varieties-section-frobenius}
for a discussion of different variants of the Frobenius morphism.
\begin{theorem}[The Baffling Theorem]
\label{theorem-baffling}
Let $X$ be a scheme in characteristic $p>0$. Then the absolute frobenius
induces (by pullback) the trivial map on cohomology, i.e., for all
integers $j\geq 0$,
$$
F_X^* : H^j (X, \underline{\mathbf{Z}/n\mathbf{Z}}) \longrightarrow H^j (X,
\underline{\mathbf{Z}/n\mathbf{Z}})
$$
is the identity.
\end{theorem}
\noindent
This theorem is purely formal. It is a good idea, however, to review how to
compute the pullback of a cohomology class. Let us simply say that in the case
where cohomology agrees with {\v C}ech cohomology, it suffices to pull back
(using the fiber products on a site) the {\v C}ech cocycles. The general case is
quite technical, see
Hypercoverings, Theorem \ref{hypercovering-theorem-cohomology-hypercoverings}.
To prove the theorem, we merely
verify that the assumption of Lemma \ref{lemma-baffling}
holds for the frobenius.
\begin{proof}[Proof of Theorem \ref{theorem-baffling}]
We need to verify the existence of a functorial isomorphism as above. For an
\'etale morphism $\varphi : U \to X$, consider the diagram
$$
\xymatrix{
U \ar@{-->}[rd] \ar@/^1pc/[rrd]^{F_U}
\ar@/_1pc/[rdd]_\varphi \\
& {U \times_{\varphi, X, F_X} X} \ar[r]^{\text{pr}_1}
\ar[d]^{\text{pr}_2} & U \ar[d]^\varphi \\
& X \ar[r]^{F_X} & X.
}
$$
The dotted arrow is an \'etale morphism which induces an isomorphism on the
underlying topological spaces, so it is an isomorphism.
\end{proof}
%10.22.09
\begin{definition}
\label{definition-geometric-frobenius}
Let $k$ be a finite field with $q = p^f$ elements. Let $X$ be a scheme
over $k$. The {\it geometric frobenius} of $X$ is the morphism
$\pi_X : X \to X$ over $\Spec(k)$ which equals $F_X^f$.
\end{definition}
\noindent
Since $\pi_X$ is a morphism over $k$, we can base change it to any scheme over
$k$. In particular we can base change it to the algebraic closure $\bar k$
and get a morphism $\pi_X : X_{\bar k} \to X_{\bar k}$. Using $\pi_X$ also
for this base change should not be
confusing as $X_{\bar k}$ does not have a geometric frobenius of its own.
\begin{lemma}
\label{lemma-sheaf-over-finite-field-has-frobenius-descent}
Let $\mathcal{F}$ be a sheaf on $X_\etale$.
Then there are canonical isomorphisms
$\pi_X^{-1} \mathcal{F} \cong \mathcal{F}$ and
$\mathcal{F} \cong {\pi_X}_*\mathcal{F}$.
\end{lemma}
\noindent
This is false for the fppf site.
\begin{proof}
Let $\varphi : U \to X$ be \'etale. Recall that
${\pi_X}_* \mathcal{F} (U) = \mathcal{F} (U \times_{\varphi, X, \pi_X} X)$.
Since $\pi_X = F_X^f$, it follows from the proof of
Theorem \ref{theorem-baffling} that there is a functorial isomorphism
$$
\xymatrix{
U \ar[rd]_{\varphi} \ar[rr]_-{\gamma_U}
& & U \times_{\varphi, X, \pi_X} X \ar[ld]^{\text{pr}_2} \\
& X
}
$$
where $\gamma_U = (\varphi, F_U^f)$. Now we define an
isomorphism
$$
\mathcal{F} (U) \longrightarrow {\pi_X}_* \mathcal{F} (U) =
\mathcal{F} (U \times_{\varphi, X, \pi_X} X)
$$
by taking the restriction map of $\mathcal{F}$ along $\gamma_U^{-1}$.
The other isomorphism is analogous.
\end{proof}
\begin{remark}
\label{remark-may-be-confusing}
It may or may not be the case that $F^f_U$ equals $\pi_U$.
\end{remark}
\noindent
We continue discussion cohomology of sheaves on our scheme $X$ over
the finite field $k$ with $q = p^f$ elements.
Fix an algebraic closure $\bar k$ of $k$ and write $G_k =
\text{Gal}(\bar k/k)$ for the absolute Galois group of $k$.
Let $\mathcal{F}$ be an abelian sheaf on $X_\etale$.
We will define a left $G_k$-module structure
cohomology group $H^j (X_{\bar k}, \mathcal{F}|_{X_{\bar k}})$
as follows: if $\sigma \in G_k$, the diagram
$$
\xymatrix{
X_{\bar k} \ar[rd] \ar[rr]^{\Spec(\sigma) \times \text{id}_X} & &
X_{\bar k} \ar[ld] \\
& X
}
$$
commutes. Thus we can set, for $\xi \in H^j (X_{\bar k}, \mathcal{F}|_{X_{\bar
k}})$
$$
\sigma \cdot \xi := (\Spec(\sigma) \times \text{id}_X)^*\xi \in
H^j(X_{\bar k}, (\Spec(\sigma) \times \text{id}_X)^{-1}
\mathcal{F}|{X_{\bar k}})
= H^j (X_{\bar k}, \mathcal{F}|_{X_{\bar k}}),
$$
where the last equality follows from the commutativity of the previous diagram.
This endows the latter group with the structure of a $G_k$-module.
\begin{lemma}
\label{lemma-two-actions-agree}
In the situation above denote $\alpha : X \to \Spec(k)$ the structure morphism.
Consider the stalk $(R^j\alpha_*\mathcal{F})_{\Spec(\bar k)}$ endowed with its
natural Galois action as in Section \ref{section-galois-action-stalks}. Then
the identification
$$
(R^j\alpha_*\mathcal{F})_{\Spec(\bar k)} \cong H^j (X_{\bar k},
\mathcal{F}|_{X_{\bar k}})
$$
from Theorem \ref{theorem-higher-direct-images} is an isomorphism of
$G_k$-modules.
\end{lemma}
\noindent
A similar result holds comparing
$(R^j\alpha_!\mathcal{F})_{\Spec(\bar k)}$ with
$H^j_c (X_{\bar k}, \mathcal{F}|_{X_{\bar k}})$.
\begin{proof}
Omitted.
\end{proof}
\begin{definition}
\label{definition-arithmetic-frobenius}
The {\it arithmetic frobenius} is the map
$\text{frob}_k : \bar k \to \bar k$, $x \mapsto x^q$ of $G_k$.
\end{definition}
\begin{theorem}
\label{theorem-geometric-arithmetic-inverse}
Let $\mathcal{F}$ be an abelian sheaf on $X_\etale$. Then for all
$j\geq 0$, $\text{frob}_k$ acts on the cohomology group $H^j(X_{\bar k},
\mathcal{F}|_{X_{\bar k}})$ as the inverse of the map $\pi_X^*$.
\end{theorem}
\noindent
The map $\pi_X^*$ is defined by the composition
$$
H^j(X_{\bar k}, \mathcal{F}|_{X_{\bar k}}) \xrightarrow{{\pi_X}_{\bar k}^*}
H^j(X_{\bar k}, (\pi_X^{-1} \mathcal{F})|_{X_{\bar k}}) \cong
H^j(X_{\bar k}, \mathcal{F}|_{X_{\bar k}}).
$$
where the last isomorphism comes from the canonical isomorphism
$\pi_X^{-1} \mathcal{F} \cong \mathcal{F}$ of
Lemma \ref{lemma-sheaf-over-finite-field-has-frobenius-descent}.
\begin{proof}
The composition $X_{\bar k} \xrightarrow{\Spec(\text{frob}_k)} X_{\bar k}
\xrightarrow{\pi_X} X_{\bar k}$ is equal to $F_{X_{\bar k}}^f$, hence the
result follows from the baffling theorem suitably generalized to nontrivial
coefficients. Note that the previous composition commutes in the sense that
$F_{X_{\bar k}}^f = \pi_X \circ \Spec(\text{frob}_k) =
\Spec(\text{frob}_k) \circ \pi_X$.
\end{proof}
\begin{definition}
\label{definition-geometric-frobenius-on-stalk}
If $x \in X(k)$ is a rational point and $\bar x : \Spec(\bar k) \to X$
the geometric point lying over $x$, we let $\pi_x : \mathcal{F}_{\bar x} \to
\mathcal{F}_{\bar x}$ denote the action by $\text{frob}_k^{-1}$ and call it the
{\it geometric frobenius}\footnote{This notation is not standard.
This operator is denoted $F_x$ in \cite{SGA4.5}. We will likely change
this notation in the future.}
\end{definition}
\noindent
We can now make a more precise statement (albeit a false one) of the trace
formula (\ref{equation-trace-formula-initial}). Let $X$ be a finite
type scheme of dimension 1
over a finite field $k$, $\ell$ a prime number and $\mathcal{F}$ a
constructible, flat $\mathbf{Z}/\ell^n\mathbf{Z}$ sheaf. Then
\begin{equation}
\label{equation-trace-formula-second}
\sum\nolimits_{x \in X(k)}
\text{Tr}(\pi_X | \mathcal{F}_{\bar x})
=
\sum\nolimits_{i = 0}^2
(-1)^i \text{Tr}(\pi_X^* | H^i_c(X_{\bar k}, \mathcal{F}))
\end{equation}
as elements of $\mathbf{Z}/\ell^n\mathbf{Z}$. The reason this equation is wrong
is that the trace in the right-hand side does not make sense for the kind of
sheaves considered. Before addressing this issue, we try to motivate the
appearance of the geometric frobenius (apart from the fact that it is a natural
morphism!).
\medskip\noindent
Let us consider the case where $X = \mathbf{P}^1_k$ and $\mathcal{F} =
\underline{\mathbf{Z}/\ell\mathbf{Z}}$. For any point, the Galois module
$\mathcal{F}_{\bar x}$ is trivial, hence for any morphism $\varphi$ acting on
$\mathcal{F}_{\bar x}$, the left-hand side is
$$
\sum\nolimits_{x \in X(k)} \text{Tr}(\varphi | \mathcal{F}_{\bar x}) =
\#\mathbf{P}^1_k(k) = q+1.
$$
Now $\mathbf{P}^1_k$ is proper, so compactly supported cohomology equals
standard cohomology, and so for a morphism $\pi : \mathbf{P}^1_k \to
\mathbf{P}^1_k$, the right-hand side equals
$$
\text{Tr}(\pi^* | H^0 (\mathbf{P}^1_{\bar k},
\underline{\mathbf{Z}/\ell\mathbf{Z}})) + \text{Tr}(\pi^* | H^2
(\mathbf{P}^1_{\bar k}, \underline{\mathbf{Z}/\ell\mathbf{Z}})).
$$
The Galois module $H^0 (\mathbf{P}^1_{\bar k},
\underline{\mathbf{Z}/\ell\mathbf{Z}}) = \mathbf{Z}/\ell\mathbf{Z}$ is trivial,
since the pullback of the identity is the identity. Hence the first trace is 1,
regardless of $\pi$. For the second trace, we need to compute the pullback
$\pi^* : H^2(\mathbf{P}^1_{\bar k}, \underline{\mathbf{Z}/\ell\mathbf{Z}}))$
for a map $\pi : \mathbf{P}^1_{\bar k} \to \mathbf{P}^1_{\bar k}$. This is a
good exercise and the answer is multiplication by the degree of $\pi$
(for a proof see Lemma \ref{lemma-pullback-on-h2-curve}). In other
words, this works as in the familiar situation of complex cohomology. In
particular, if $\pi$ is the geometric frobenius we get
$$
\text{Tr}(\pi_X^* | H^2 (\mathbf{P}^1_{\bar k},
\underline{\mathbf{Z}/\ell\mathbf{Z}})) = q
$$
and if $\pi$ is the arithmetic frobenius then we get
$$
\text{Tr}(\text{frob}_k^* | H^2 (\mathbf{P}^1_{\bar k},
\underline{\mathbf{Z}/\ell\mathbf{Z}})) = q^{-1}.
$$
The latter option is clearly wrong.
\begin{remark}
\label{remark-compute-degree-lifting}
The computation of the degrees can be done by lifting (in some obvious sense)
to characteristic 0 and considering the situation with complex coefficients.
This method almost never works, since lifting is in general impossible for
schemes which are not projective space.
\end{remark}
\noindent
The question remains as to why we have to consider compactly supported
cohomology. In fact, in view of Poincar\'e duality, it is not strictly
necessary for smooth varieties, but it involves adding in certain powers
of $q$. For example, let us consider the case where
$X = \mathbf{A}^1_k$ and
$\mathcal{F} = \underline{\mathbf{Z}/\ell\mathbf{Z}}$.
The action on stalks is again trivial, so we only need look at the action
on cohomology. But then $\pi_X^*$ acts as the identity on
$H^0(\mathbf{A}^1_{\bar k}, \underline{\mathbf{Z}/\ell\mathbf{Z}})$
and as multiplication by $q$ on
$H^2_c(\mathbf{A}^1_{\bar k}, \underline{\mathbf{Z}/\ell\mathbf{Z}})$.
```

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