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Let $X$ be a simplicial complexe and we assume it localy finite and finite dimensional. We suppose taht there exist a simplicial complexe $Y$ and a map assigning to each vertex $s\in Y$ a finite subcomplex $X_{s}$ of $X$ such that the collection $\mathcal{Y}=(X_{s})_{s\in Y}$ is a covering by closed subsets of $X$ and the nerve of this covering $\mathcal{Y}$ is isomorphic to the simplicial complexe $Y$.

My question is : It is true that $H_{c}^{p}(X,\mathbb{Z})\simeq\bigoplus_{s\in Y}H^{p}(X_{s},\mathbb{Z})\oplus H_{c}^{p}(Y,\mathbb{Z})$ ?

where $H_{c}^{\bullet}$ is the cohomoloy with compact support.

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    $\begingroup$ 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. $\endgroup$
    – user9072
    Commented Mar 11, 2013 at 17:19
  • $\begingroup$ 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 $Y$. $\endgroup$
    – Ricardo Andrade
    Commented Mar 11, 2013 at 23:05
  • $\begingroup$ 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. $\endgroup$
    – Ricardo Andrade
    Commented Mar 11, 2013 at 23:10
  • $\begingroup$ 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 $PGL(2,F)$. $\endgroup$
    – Rajkarov
    Commented Mar 12, 2013 at 14:08
  • $\begingroup$ 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. $\endgroup$
    – Rajkarov
    Commented Mar 12, 2013 at 14:26

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