$f^*f_*(\mathcal{F})$ is surfective if $\mathcal{F}$ is generated be global sections Suppose $\mathcal{F}$ is a sheaf of module on $X$,$f:X\to Y$,suppose $\mathcal{F}$ is generated by global sections. Is $f^*f_*(\mathcal{F})\to \mathcal{F}$ is surjective ? 
To check on stalks, $f^*f_*(\mathcal{F})_x \cong f_*(\mathcal{F})_{f(x)} \to \mathcal{F}_x$, and it became messy..
And is there counterexample or is this condition necessary?
 A: Let's call $F \in \mathsf{Mod}(X)$ good if $f^* f_* F \to F$ is an epimorphism.
$1.$ Step: $f^* G$ is good for every $G \in \mathsf{Mod}(Y)$. In fact, then the map is a split epimorphism by the triangle identities of the adjunction between $f_*$ and $f^*$.
$2.$ Step: If $F'$ is a quotient of $F$ and $F'$ is good, then $F$ is good. This follows from the obvious commutative diagram.
$3.$ Step: If $F$ is generated by global sections, then $F$ is good. This is because $F$ is a quotient of a direct sum of copies of $\mathcal{O}_X = f^* \mathcal{O}_Y$, so that we may apply Steps 1 and 2.
A: For a counterexample if $\mathcal{F}$ is not generated by global sections, take $f: X \rightarrow Y=\{pt\}$ and $\mathcal{F}$ a non-zero sheaf of $\mathcal{O}_X$-modules, which has no (non-trivial) global sections. Then the natural map
$$
f^*f_*\mathcal{F} \rightarrow \mathcal{F}
$$
isn't surjective.
Indeed, let $x \in X$ be such that $\mathcal{F}_x \neq 0$. We have
\begin{split}
(f^*f_*\mathcal{F})_x \cong (f_*\mathcal{F})_{pt} \otimes_{\mathcal{O}_{Y,pt}} \mathcal{O}_{X,x} = 0
\end{split}
since $(f_*\mathcal{F})_{pt} \cong \mathcal{F}(X)$, and the latter is zero by assumption. Hence $f^*f_*\mathcal{F} \rightarrow \mathcal{F}$ cannot be surjective at $x$.
