An $\ell$-adic local system which is trivial on every fiber of a morphism Let $f: X \to Y$ be a morphism with connected fibers, where $X, Y$ are smooth algebraic varieties (I am specifically interested in the case when $f$ is a Zariski locally trivial fibration with connected fibers). Let $L$ be an $\ell$-adic local system on $X$ which is trivial when restricted to every fiber of $f$. Is it true that $L$ is isomorphic to the (non-derived) pullback of its (non-derived) pushforward $f^* f_* L$, that is the (non-derived) adjunction morphism is an isomorphism?
 A: Let $f : X \to Y$ be a morphism of smooth varieties with geometrically connected fibres. Consider a continuous map $\rho : \pi_1(X) \to G$ with $G$ finite such that $\pi_1(X_y) \to G$ is trivial for closed points $y \in Y$. Question: does $\rho$ factors through $\pi_1(X) \to \pi_1(Y)$?
This is a reformulation of the question as is clear from the equivalence between local systems and representations of the fundamental group.
Assume the ground field is the complex numbers. Then there is a nonempty Zariski open $V \subset Y$ such that $f^{-1}(V)^{an} \to V^{an}$ is a topological fibration (with connected fibres). Hence you see that the answer is yes after shrinking the base (use comparison topological fundamental group with the algebraic fundamental group, etc, etc).
I believe the thing you are asking can be wrong because $Y$ can be a smooth curve and $X \to Y$ can have multiple fibres. But if this kind of thing doesn't happen then you can show that the map $\pi_1(V) \to G$ we constructed above factors through $\pi_1(Y)$. I think it suffices that the smooth locus of $f$ surjects onto $Y$ but I haven't checked the details.
In positive characteristic, we need to replace the argument with the associated analytic space with a different one to show that the composition $\pi_1(X_{\overline{\eta}}) \to \pi_1(X) \to G$ is trivial, where $\eta \in Y$ is the generic point and $\overline{\eta}$ is the corresponding geometric point of $Y$. If the ground field is uncountable, you can do a kind of standard trick to get this. But I am currently blanking on what to do if the ground field is the algebraic closure of a finite field. But in the situation you need this for, you may actually already know what happens over the geometric generic fibre.
A: There is a sort of obvious proof when $f$ is proper and has geometrically connected fibres. This may or may not be sufficient for the application you have in mind. I have no idea whether the result remains true in the absence of either hypothesis (or the weakening to connected fibres).
Lemma. Let $f \colon X \to Y$ be proper with geometrically connected fibres. Let $F$ be an $\ell$-adic sheaf on $X$, which is trivial on the fibres of $f$. Then the natural map $\varepsilon \colon f^*f_* F \to F$ is an isomorphism.
Proof. It suffices to prove isomorphism on all stalks at geometric points $\bar x \colon \operatorname{Spec} \bar K \to X$. Let $\bar x$ be such a point, and let $\bar y$ be its image in $Y$ (i.e. the composition $f \circ \bar x$). We have a commutative diagram
$$\begin{array}{ccccc}\operatorname{Spec} \bar K & \to & X_{\bar y} & \xrightarrow{g} & \operatorname{Spec} \bar K \\ & \searrow & \ \ \downarrow \iota & &\ \ \downarrow \bar y \\ & & X & \xrightarrow{f} &\ Y,\end{array}$$
where the square is a pullback. We pull back the map $\varepsilon \colon f^* f_* F \to F$ along $\iota \colon X_\bar y \to X$; if we can prove that it is an isomorphism on $X_\bar y$, then it will also be an isomorphism after further pulling back to $\operatorname{Spec} \bar K$, i.e. on the stalk at $\bar x$.
On $X_\bar y$, the map $\varepsilon$ factors as
$$\iota^* f^* f_* F = g^* \bar y^* f_* F \longrightarrow g^* g_* \iota^* F \longrightarrow  \iota^* F,$$
where the first equality comes from commutativity of the diagram, the first map comes from the base change map of the diagram, and the second map from the counit of the adjunction $g^* \dashv g_*$.
Now the first map above is an isomorphism by proper base change. Thus, we only have to prove that the second map is an isomorphism as well, i.e. we have to prove the result for the sheaf $\iota^* F$ on $X_\bar y$. But this is trivial: by assumption, $\iota^* F$ is constant, $X_\bar y$ is connected, and a sheaf on $\operatorname{Spec} \bar K$ is determined by its global sections. $\square$
