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Timeline for Fine and acyclic sheaves on locales

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Mar 7, 2014 at 20:21 history edited user9072
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Nov 20, 2011 at 11:27 comment added Dmitri Pavlov @Mario: Yes, the sheaf of locally constant Y-valued (measurable) functions on the underlying measurable space of X is flasque. I have no idea what do you mean when you say that f is linear (f was an arbitrary mesuarable function on A, why should it become linear when we extend it to X?).
Oct 21, 2011 at 17:10 comment added Mario Carrasco Yeah, I was wrong, my bad, now check this out: Let X and Y be finite-dimensional topological vector spaces. Let F be the sheaf of Y-valued locally constant functions. Let S1 and S2 be their Borel sigma-algebras respectivelly. Now since every measurable function $f$ defined on a measurable subset A of a measurable space can be extended to a measurable function f′ on X\A by defining it to be 0 (hence linear) on X\A, this means f′ is continuous as linear maps between finite-dimensional TVSs are continuous, so the sheaf of locally constant functions on X is flasque? in this case no?
Oct 16, 2011 at 11:51 comment added Dmitri Pavlov @Mario: You asked a completely different question at MSE. I have never mentioned arbitrary topological spaces in my comments. Eric has also pointed this out at MSE in his reply to your comment.
Oct 16, 2011 at 3:38 comment added Mario Carrasco @Dimitri Pavlov: Are you meaning to say that the sheaf of locally constant functions on a topological space is flasque? I mean if I take a topological space and the Borel sigma-algebra generated by the topology and do what you're saying then every locally constant function would extend to a locally constant function outside of its definition domain and the sheaf would be flasque? I posted the question on Math Stack Exchange: bit.ly/pgyJbN. They're are saying that the sheaf of locally constant functions on a topological space is not necessarily flasque, I'm confused, thanks a bunch tho
Oct 2, 2011 at 22:55 comment added Dmitri Pavlov @Mario: I am not sure what do you mean by the extension property, but any measurable function defined on a measurable subset of a measurable space can be extended to a measurable function defined on the whole space by declaring it to be 0 outside of the original domain. Locally constant functions stay locally constant under this transformation.
Sep 25, 2011 at 15:33 comment added Mario Carrasco Oh yeah, you mean the extension property for measurable spaces? The sheaf I'm working with is the sheaf of locally constant functions? Do locally constant functions extend to locally constant functions according to the property?
Sep 23, 2011 at 23:07 comment added Dmitri Pavlov If you have a measurable space X and some subspace Y⊂X, then functions on X are precisely arbitrary pairs of functions on Y and X\Y. The same is true for morphisms of sheaves on X. Therefore any sheaf on the site that you described is fine, flabby, and soft for trivial reasons.
Sep 23, 2011 at 19:29 comment added Mario Carrasco Thank you both, I actually downloaded Matthew Jackson's thesis a while back, I know the discrete topological structure is in itself not interesting, but that's just the space I'm working on, really have to look deeper into that but it seems right from the point of view of what I need this whole construction for
Sep 23, 2011 at 11:24 comment added Peter Arndt I don't think it addresses your question, but there is a thesis on measure theory and topoi by Matthew Jackson which you might find interesting: andrew.cmu.edu/user/awodey/students/jackson.pdf
Sep 23, 2011 at 3:02 comment added David Roberts Starting from a discrete space is not very interesting. As far as fine sheaf on a locale goes, have a look at the Voisin definition at ncatlab.org/nlab/show/fine+sheaf. The definition of a partition of unity on a locale is probably what you are after, however, so this advice might be circular.
Sep 23, 2011 at 0:53 history edited Mario Carrasco CC BY-SA 3.0
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Sep 22, 2011 at 22:38 history asked Mario Carrasco CC BY-SA 3.0