5
$\begingroup$

I have some confusion on the subject of sheaf cohomology on non-paracompact topological spaces, i hope you can help me.

My reference is Godement's book "Topologie algebrique et theorie dex faisceaux".

I know that, given a sheaf $\mathcal{F}$ on $X$ and $\phi$ a family of supports on $X$, I can define the cohomology $H^n_{\phi}(X,\mathcal{F})$ as $H^n(\Gamma_\phi(\mathcal{C}^*(X,\mathcal{F}))$ where $\mathcal{C}^*(X,\mathcal{F})$ is the canonical sequence and $\Gamma_\phi(\mathcal{F})=\{s\in\mathcal{F}(X)|supp(s)\in \phi\}$.

When Godement writes $H^n(X,\mathcal{F})$, without the $\phi$ being indicated, i think he means $H^n(\Gamma(\mathcal{C}^*(X,\mathcal{F}))$.

My questions are the following:

1) What is the difference between $H^n(\Gamma_\phi(\mathcal{C}^*(X,\mathcal{F}))$ and $H^n(\Gamma(\mathcal{C}^*(X,\mathcal{F}))$? What is the utility of the family of supports?

2) I've heard that when $X$ is not paracompact then $H^n(\Gamma(\mathcal{C}^*(X,\mathcal{F}))$ (but not $H^n(\Gamma_\phi(\mathcal{C}^*(X,\mathcal{F}))$) fails to be "functorial" (i think in the sense that short exact sequences don't go to long exact sequences). Can you explain this to me? What exactly doesn't work for $H^n(\Gamma(\mathcal{C}^*(X,\mathcal{F}))$?

Thank you very much

$\endgroup$
2
$\begingroup$

I hope the following is an answer to (a part) of your wondering : Stefan Schroër constructed a non-paracompact Hausdorf space for which Cech cohomology does not coincide with sheaf cohomology. Moreover, the sheaf of continuous real-valued functions is neither soft nor acyclic ... See : Top. and its Appl., vol.160, issue 13, 15/8/2013 (1809-1815). On the other hand, I recommend B. Iversen's book , "Cohomology of Sheaves".

$\endgroup$
2
$\begingroup$

2) Are you sure you are not confusing the sheaf cohomology $H^n(\Gamma(C^*(X,\mathcal F))$ with Čech cohomology$\check H^n(X,\mathcal F)$? The sheaf cohomology has long exact sequences for arbitrary topological space and short exact sequence of sheaves. For paracompact spaces the two are equal, there is a proof of it in Godement's book.

$\endgroup$

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.