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Hi, I posted something similar in learning: questions already.
I am trying to follow Bredon's proof, that Cech and Sheaf cohomology coincide, but I am struggling to see, where the space being Hausdorff is used. I think it is needed to prove that the stalks of the Čech cohomology of a presheaf are zero if the stalks of are zero everywhere.
To be precise, let be a presheaf of abelian groups on a Hausdorff paracompact space with zero stalks for all . If is an open subset and is a point with , then is the colimit zero? I think the stalk of the Cech cohomology (seen as the presheaf ) is a colimit of finite products of such colimits?
Sorry, if this is actually obvious...
so this hypothesis of paracompactness and Hausdorff is pretty nice because it tells you that Čech and sheaf cohomologies (aka "Grothendieck cohomology) agree for all sheaves, whereas lots of other theorems about cohomology theories agreeing place weaker conditions on the space itself and instead consider conditions on the sheaves in question
(I know this doesn't answer your question but I'm just thinking about it, and I though I would mention this in the mean time!)
one reason it's hard to know where exactly the Hausdorff assumption is being used is because Godement (who wrote the proof that I once read, a few years ago now), amongst others, defines a paracompact topological space to be, in particular, Hausdorff!
I don't know about your precise technical question with that boundary point... But a proof that a presheaf with zero stalks on a paracompact Hausdorff space has zero Cech cohomology is given in Brylinski," Loop spaces, characteristic classes and geometric quantization", Thm. 1.3.13 (3):
Brylinski.png
(for Brylinski "paracompact" means paracompact & Hausdorff)
For that proof he uses that for an open covering of a paracompact Hausdorff space, there exists an open covering such that for all (his Prop. 1.3.12):
Brylinski2.png
This statement (his Prop. 1.3.12) is proven as Lemma 41.6 (page 258) of Munkres, Topology and uses regularity of paracompact Hausdorff spaces.
The proof that paracompact Hausdorff spaces are regular is embedded in the proof of Munkres, Thm. 41.1 (page 253) and explicitly uses the Hausdorff property.
(for Munkres "paracompact" means paracompact and not necessarily Hausdorff)
Peter Arndt said:
I don't know about your precise technical question with that boundary point... But a proof that a presheaf with zero stalks on a paracompact Hausdorff space has zero Cech cohomology is given in Brylinski," Loop spaces, characteristic classes and geometric quantization", Thm. 1.3.13 (3):
Brylinski.png
(for Brylinski "paracompact" means paracompact & Hausdorff)
Thanks a lot - This is exactly what I was searching for! I somehow did not think about checking Brylinski's Book. I'm more interested in loop spaces now
I read that part and now actually think that the colimit is not zero...
Tim Hosgood said:
one reason it's hard to know where exactly the Hausdorff assumption is being used is because Godement (who wrote the proof that I once read, a few years ago now), amongst others, defines a paracompact topological space to be, in particular, Hausdorff!
Yeah right, this is exactly what puzzled me :upside_down:
I’m very out of practice with this sort of stuff, but I also don’t think the colimit is zero, and I think you could cook up some counterexample showing this on something like the Sierpinski space
Tim Hosgood said:
I’m very out of practice with this sort of stuff, but I also don’t think the colimit is zero, and I think you could cook up some counterexample showing this on something like the Sierpinski space
I will look that that... though I have no idea about Siepinski spaces tbh :)
The Sierpinski space is the most interesting topological space with 2 points.
A continuous map from a topological space X to the Sierpinski space is the same as an open subset of X, so we say the Sierpinski space is the 'classifying space' for open sets.
John Baez said:
A continuous map from a topological space X to the Sierpinski space is the same as an open subset of X, so we say the Sierpinski space is the 'classifying space' for open sets.
Well, that's a good primer!
Is there a well known theory of Cech cohomology with coefficients in a profinite group?
I have a cohomology theory for CH spaces and it naturally takes coefficients in a profinite group. I am curious about what the standard cohomology theories are that take coefficients in a profinite group as I have personally only worked with ordinary Abelian groups as coefficient groups.
I think I should compare my cohomology theory to other cohomology theories to see what it agrees with but I don't know what to compare it to!
https://twitter.com/CihanPostsThms/status/1567223117054971904
Click on the tweet for some more details.