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### Is this a true weakening of the quasi-coherence property?

Any submodule of a quasicoherent $O_X$-module satisfies (#): this is clear via reduction to principal open sets, and the fact that localization is exact. More generally, as Neil observes, if $F$ ...

### Is the ideal product presheaf a sheaf? Do we have any reasons to believe it will be / it won't?

So here is a counterexample which is qcqs: Take $X$ the affine line with double origin $a_1$ and $a_2$, then take $I_1$ and $I_2$ the ideal of functions vanishing each at one of the origins ...

### Is the ideal product presheaf a sheaf? Do we have any reasons to believe it will be / it won't?

It need not be a sheaf. As an example, consider a space $X$ which is a disjoint union of open subspaces $X_n$, and pick $\mathcal O_X,\mathcal I,\mathcal J$ with the property that some element $c_n$ ...

### Can one glue De Rham cohomology classes on a differential manifolds?

Here is the great answer given by another of my brilliant friends: Let $X$ be $\mathbb C, U_0$ be the open complement in $X$ of the closed disk $\bar D=\{z\in \mathbb C\vert \vert z\vert \leq1 \}$ and ...

### Relation between sheaf and group cohomology

Short answer: I believe the answer is you can always do this, as long as you do it in the right way (which isn't always by taking global sections, so this complements the previous answer). I will ...

### Can one glue De Rham cohomology classes on a differential manifolds?

Here is the solution obtained by one of my brilliant geometer friends evoked in the question: Take $X=S^2$, the unit $2$-sphere with equation $x_1^2+x_2 ^2+x_3^2=1$, and cover it by the three open ...

### Can one glue De Rham cohomology classes on a differential manifolds?

This answer provides a positive answer to a refinement of the original question. Recall that two closed differential $k$-forms $ω_0$, $ω_1$ on a smooth manifold $M$ have the same de Rham cohomology ...
No. Make $M$ by gluing three strips to two discs to form a thrice-punctured sphere. Take three open sets $U_\lambda$, each made by both discs and two of the strips. Then each $U_\lambda$ is ...
We have $q_1 = p_1 \circ j$ and $q_2 = p_2 \circ j$ so  {q_2}_! q_1^*(\mathcal F) \cong {p_2}_! j_! j^* p_1^*(\mathcal F) \cong {p_2}_! j_! ( j^* p_1^*(\mathcal F) \otimes \mathbb Q_U) \ \cong {...