It is known that a flasque sheaf on a topological space has trivial cohomology. Suppose that we are in a relative situation of a smooth fibration $\pi: X \to S$ and $F$ is a sheaf on $X$. Is there are weaker condition on $F$ which implies that the higher direct images to $S$ are trivial (but not necessarily the cohomology groups of $F$ on $X$). 



Relative flasqueness?
Modulo quasi-compactness, the following seems to work:


**Definition.** Let $f:X\to S$ be a map of topological spaces and let $F$ be a sheaf on $X$. We call $F$ *flasque over $S$* if for any opens $U\subseteq V\subseteq X$, any $x\in U$, and any $\sigma\in\Gamma(U, F)$ there exists an open $W$ containing $f(x)$ such that $\sigma|\_{U\cap f^{-1}(W)}$ lies in the image of 
$$ \Gamma(V\cap f^{-1}(W), F) \to \Gamma(U\cap f^{-1}(W), F). $$


**Proposition.** Suppose that $X$ is quasi-compact and that $F$ is flasque over $S$. Then $R^i f\_\* F = 0$ for $i>0$.


This will follow from the fact that (1) injective sheaves are flasque, hence flasque over $S$, (2) quotient of a flasque sheaf by a subsheaf flasque over $S$ is flasque over $S$, (3) the following lemma: 


**Lemma.** With the hypotheses of the Proposition, for any short exact sequence 
$$ 0\to F\to F'\to F''\to 0$$
the map $f\_\* F'\to f\_\* F''$ is surjective.


*Proof.* Let $s\in S$ and let $\sigma$ be a section of $f\_\* F''$ on an open $W\_0$ containing $s$. We need to find an open $W\subseteq W\_0$ containing $s$ such that $\sigma|\_W$ comes from a section of $f\_\* F'$ on $W$.


By definition, $\sigma\in \Gamma(f^{-1}(W\_0), F'')$. We can cover $f^{-1}(W\_0)$ by *finitely many* opens $U\_i$ and find sections $\tilde\sigma\_i$ of $F'$ over $U\_i$ whose image in $F(U\_i)$ is $\sigma|\_{U\_i}$. Let $\tau\_{ij} = \tilde\sigma\_i-\tilde\sigma\_j \in \Gamma(U\_i\cap U\_j, F)$. 


By flasqueness of $F$ over $S$, can find $s\in W\_{ij}\subseteq W\_0$ and $\tau'\_{ij} \in \Gamma(f^{-1}(W\_{ij}, F)$ extending $\tau\_{ij}|\_{U\_{ij}\cap f^{-1}(W\_{ij}, F)}$. Since there were finitely many $i$, $j$, we can intersect them, obtaining the desired $W$.


