Explicit examples presheaves associated to higher direct images which fail to be sheaves

So I would like to have a few simple examples where the presheaf associated to higher direct image of sheaf fails to be sheaf. So I'm looking for two (natural and simple) topological spaces $X$ and $Y$, a continuous map $f:X\rightarrow Y$ and a sheaf of abelian groups $\mathcal{F}$ on $X$ such that the presheaf $f_*^{i}(\mathcal{F})$ given by $$U\mapsto H^i(f^{-1}(U),\mathcal{F}|_{f^{-1}(U)})$$ (for some $i>0$) fails to be a sheaf on $Y$.

I'd like to see the two possible obstructions for $f_*^{i}(\mathcal{F})$ to be a sheaf namely

(i) the impossibility of gluing local sections of $f_*^{i}(\mathcal{F})$ to a global one,

(ii) the existence of a non-zero global section of $f_*^{i}(\mathcal{F})$ which is trivial at each stalk $y\in Y$.

For (i), take $X=S^3$, $Y= S^2$, $f$ the Hopf fibration. There is a local $H^1$ everywhere, coming from the fact that the fibers are circles. This local section is defined consistently everywhere (because there is no monodromy since $Y$ is simply-connected), but it does not glue to a global section because $H^1(X,\mathbb Z)=0$.
For (ii), simply take $X=Y$ a manifold space with nontrivial cohomology and $\mathcal F$ the constant sheaf. Then there is a global section corresponding to the higher cohomology classes. This section vanishes on every open ball, so it vanishes on every stalk.
• As to the first case, why don't the local sections glue to a global one? What's wrong with $H^1(X,\mathbb{Z})$? – user2013 Dec 17 '13 at 22:55
• Because the Leray spectral sequence does not degenerate at the second page. The $H^1(X,\mathbb Z)$ coming from $H^0 R^1$ cancels with the $H^2(X,\mathbb Z)$ coming from $H^2 R^0$, leading to vanishing cohomology groups. Less abstractly, the nontrivial $H_1$ class of a fiber is represented by the full fiber, a circle. This circle is a trivial homology class because it bounds a disc in $S^3$. This disc wraps all the way around $S^2$, becoming the nontrivial homology class in $S^2$, but in $S^3$ one bounds the other so they cancel each other out. – Will Sawin Dec 17 '13 at 22:59
• For (i), $f$ is a fibration with fibre $S^1$ and so $f^1_*(\underline{\mathbb{Z}})$ is a locally constant sheaf with fibre $\mathbb{Z}$. As $S^2$ is simply connected, we have $f_*^1$ is the constant sheaf, so I don't agree that (i) is an example. $f^3_*$ would however give an example for (ii). – Geordie Williamson Dec 17 '13 at 23:13
• The pushforward presheaf is locally constant but not a sheaf. The sheafification, as you point out, is the constant sheaf. So it has a global section. But the presheaf does not have any sections on all of $Y$, because these are just defined to be $H^1(X, \mathbb Z)$. – Will Sawin Dec 17 '13 at 23:18