1
$\begingroup$

I want to prove that that for $\mathcal{F}$ a sheaf on manifold $M$ we have $H^0(\{U\},\mathcal{F})=\mathcal{F}(M)$, where $\{U\}$ is any locally finite open cover of $M$. (Note that I think we really only need connectedness of $M$ but I'm not sure about this)

It is clear that we have an injection $i:\mathcal{F}\to Z^0(\{U_{\alpha}\}_{\alpha\in I},\mathcal{F}): x\mapsto (x|_{U_\alpha})_{\alpha\in I}$. However, I don't see how to prove this is a surjection.

i.e. how to prove that for $x=(x_\alpha)_{\alpha\in I}\in Z^0(\{U_{\alpha}\}_{\alpha\in I},\mathcal{F})$ where $x_\alpha\in \mathcal{F}(U_\alpha)$ and $\partial x=0$, then there exists a $g\in \mathcal{F}(M)$ s.t. $\forall \alpha: x_\alpha=g|_{U_\alpha}$. In some concrete sheaves this seems to more or less come down to a form of the pasting lemma. However I do not see how to do this in an arbitrary sheaf using just the axioms.

One could try something like picking any $x_\alpha$ and finding a $g\in\mathcal{F}(M)$ s.t. $g|_{U_\alpha}=x_\alpha$. I think that the condition $x\in Z^0(\{U_{\alpha}\}_{\alpha\in I},\mathcal{F})$, which means: \begin{equation} \forall \alpha,\beta:x_\alpha|_{U_\alpha \cap U_\beta}=x_\beta|_{U_\alpha \cap U_\beta} \tag{1} \end{equation} would force $g|_{U_\beta}=x_\beta$ for all $\beta$ using the local finiteness condition somehow.

One problem is that I don't see why such a $g$ and $x_\alpha$ should exist in the first place. There are possibly elements in $\mathcal{F}(U_\alpha)$ that are not the restriction of a an element in $\mathcal{F}(M)$. Now it seems intuitive that the condition that $\partial x=0$ and the compatibility condition $(1)$ would force that all $x_\alpha$ are in fact restrictions of an element of $\mathcal{F}(M)$, but here the reasoning becomes circular...

A hint for solving this would appreciated.


edit: note that for finite covers this is also clear from the sheaf axiom: $$U,V\subset M\text{ open }, x\in U,y\in V s.t. x|_{U\cap V}=y|_{U\cap V} \text{ then exists } v\in \mathcal{F}(U\cup V) s.t. v|_U=x, v|_V=y$$

$\endgroup$
0
$\begingroup$

This question is answered here. The formulation of the gluing axiom given by Griffiths and Harris is wrong, and seems to weak to prove this statement.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.