# Continuous map with homeomorphic fibers whose associated $H^{k}_c$ sheaf is not a local system?

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:

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

Question: Must $\mathcal{H}^j$ be a local system on $Y$? If not what's a counterexample?

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.

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.

Here is a proper example. Define two functions $f^+,f^-\colon \mathbb R\to R$ by $x^2$ and $-x^2$. Then let $f=f^+\cup f^-\colon \mathbb R\cup\mathbb R\to \mathbb R$. Every fiber of $f$ is 2 points, so the pushforward $f_*\mathbb k$ has stalk $k^2$ at every point, but it is not a local system because it is the sum of the two components: $f_*\mathbb k=f^+_*k\oplus f^-_*k$, which are supported on $[0,\infty)$ and $(-\infty,0]$.

• I forgot to make it connected, but just glue the ends. In other words $S^1\to [0,1]\to S^1$ where the first map is cosine and the second map is gluing the endpoints of the interval. – Ben Wieland Jun 6 '18 at 19:34
• If you allow disconnected examples, and you allow the dimension to jump, then there is the holomorphic example of $\mathbb C\cup pt\to \mathbb C$, where the map on $\mathbb C$ is the squaring map and the point goes to $0$ to fix that fiber. – Ben Wieland Jun 6 '18 at 19:36