Let $ f: X \to Y$ be a continuous map between connected manifolds s.t. for all $y \in Y$ the fiber $f^{-1}(y)$ is homeomorphic to some fixed connected manifold $Z$.


Let $k$ be a ring and for every $j \ge 0$ let $\mathcal{H}^j:=R^{j}f\_!(k\_X)$, i.e. the shefification of the presheaf on $Y$ given by: 


$$U \mapsto H\_{c}^{j}(f^{-1}(U),k)$$



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> **Question:** Must $\mathcal{H}^j$ be a **local system** on $Y$? If not what's a **counterexample**?
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Continuous map with homeomorphic fibers whose associated $H^{k}_c$ sheaf is not a local system?
First of all, in case $f$ is not proper, the sheaf $\mathcal{H}^j = R^j f\_!(k\_X)$ is not defined as a sheaf associated to a presheaf $$U\mapsto H^j\_c(f^{-1}(U),k),$$ since that rule is not a presheaf. Compactly supported cohomology is **covariant** for open inclusions, it is not contravariant (presheaves are contravariant). 


Using the corrected definition, typically $\mathcal{H}^j$ is not a local system if $f$ is not proper. In case $f$ is a **proper** submersion of manifolds, then $f\_!$ equals $f\_\*$. The rule that you wrote down for usual cohomology (not compactly supported cohomology) is a presheaf whose associated sheaf equals $R^jf\_\*(k\_X).$ The fact that $R^jf\_\*(k\_X)$ is a local system in the proper case follows from Ehresmann's theorem, for instance. 


In the **non-proper** case, one family of counterexamples arises as in my answer to the following MathOverflow question. 


[Is an affine fibration over an affine space necessarily trivial?](https://mathoverflow.net/questions/136017/is-an-affine-fibration-over-an-affine-space-necessarily-trivial/136043#136043)   
 


Here is the construction. Let $Y$ equal the complex projective line as a complex manifold. Let $y$ and $y'$ be distinct points of $Y.$ Let $\overline{X}$ equal the compact complex manifold $Y\times Y.$ Let $\overline{f}:\overline{X}\to Y$ equal projection onto the first factor. Let $Z\subset Y\times Y$ equal the union of the following three irreducible closed, complex submanifolds: the diagonal $Z\_1=\Delta,$ the constant section $Z\_2=Y\times\{y\},$ and the singleton $Z\_3=\{(y,y')\}.$ Denote by $Z'$ the union $Z\_1\cup Z\_2.$ Let $X,$ resp. $X',$ denote the open complement in $\overline{X}$ of $Z,$ resp. of $Z'.$ Let $f:X\to Y$ denote the restriction to $X$ of $\overline{f}.$ This is a holomorphic submersion.


Every fiber of $f$ is a complex manifold that is biholomorphic to $\mathbb{C}^\times.$ The compactly supported cohomology equals the reduced cohomology of the one-point compactification (a "nodal plane cubic"), $$H^0\_c(\mathbb{C}^\times,\mathbb{Z}) = 0,\ \ H^1\_c(\mathbb{C}^\times,\mathbb{Z}) = \mathbb{Z}, \ \ H^2\_c(\mathbb{C}^\times,\mathbb{Z}) = \mathbb{Z}. $$ Restricted over $Y\setminus\{y\},$ the sheaves $\mathcal{H}^j$ on $Y$ are local systems. 


Now let $U\subset Y$ be an open disk centered at $y.$ The inclusion $Z\_3\subset X'$ with open complement $X$ induces a long exact sequence of cohomology with compact supports. In particular, $H^0\_c(Z\_3,\mathbb{Z})=\mathbb{Z}$ maps to a nonzero element in $\mathcal{H}^1(U)$ for every $U.$ In particular, the germ of this element in $\mathcal{H}^1\_y=H^1\_c(X\_y,\mathbb{Z})$ is nonzero. However, for every $z\in U\setminus\{y\},$ the image of this element in the stalk $\mathcal{H}^1\_z = H^1\_c(X\_z,\mathbb{Z})$ is zero. Thus, $\mathcal{H}^1$ is not a local system. 


