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Mar 12, 2013 at 14:26 comment added Rajkarov I think if we pute the addionnal condition : a) For all $s,t\in Y$, the sub-complexes $X_{s}$ and $X{t}$ are isomorphic. b) $X$ is chamber complexe. Then the formula is true.
Mar 12, 2013 at 14:08 comment added Rajkarov Thank you Ricardo, but the formula is true for some particular examples. For example, if $X=\mathbb{Z}$ with vertex vertex $w\in\mathbb{Z}$ and edges $\{n,n+1\}$, and the recouvrement is by the edges $\{n,n+1\}$. The formula is also true for another complicated examples : one dimensional complexes above the Bruhat-Tits tree of the p-adique group $PGL(2,F)$.
Mar 11, 2013 at 23:10 comment added Ricardo Andrade By the way, here is an interesting common example. When each $X_s$ is contractible, then $X\simeq Y$ and so the cohomology of $X$ is isomorphic to the cohomology of $Y$. Nevertheless, their cohomologies with compact support may still be quite different.
Mar 11, 2013 at 23:05 comment added Ricardo Andrade Either I am misunderstanding something, or the answer is obviously no. Take $X=S^1$ triangulated with three 1-simplices. Consider $Y=\Delta^1$, let $X_0$ be some 1-simplex of $X$, and let $X_1$ be the union of the other two 1-simplices of $X$. Then $H^1_c(X)=\mathbb{Z}$, yet each of $X_0$, $X_1$, $X_0 \cap X_1$, and $Y$ has no cohomology in degree 1 (and $Y$ is compact). Nevertheless, for results related to what you are asking, perhaps you want to look at some kind of Mayer-Vietoris spectral sequence. However, such a spectral sequence usually does not directly involve the cohomology of $Y$.
Mar 11, 2013 at 22:30 history edited Ricardo Andrade
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Mar 11, 2013 at 21:31 history edited AFK CC BY-SA 3.0
replaced the french word "recouvrement" by "covering"; deleted 4 characters in body
Mar 11, 2013 at 17:19 comment added user9072 I retagged this question (not sure if overly well though), but then I still think it is an improvement over the old tag, which was '10'. I am not sure this is what happened, but perhaps: please do not use the numbers written next to the tags instaed of the name of the tag itself. The number (in the selection list) indicates only the number of times the tag was used so far it is not 'the number of the tag'. Using it as tag thus makes little sense.
Mar 11, 2013 at 17:15 history edited user9072
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Mar 11, 2013 at 14:57 history asked Rajkarov CC BY-SA 3.0