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Let $X = \left(L \times [0, 1]\right) / \left(L \times \{0\}\right)$ be the closed cone over a closed smooth $d$-dimensional manifold $L^{d}$. Let $i \colon Y \hookrightarrow X$ denote the inclusion of the open cone $Y = \left(L \times [0, 1)\right) / \left(L \times \{0\}\right)$.

Let $\boldsymbol{IC}^{\bullet}_{\overline{p}}(Y)$ denote the perversity $\overline{p}$ intersection chain complex of the open cone $Y$ (using $\mathbb{Q}$-coefficients) as introduced in M. Goresky and R.D. MacPherson's paper Intersection homology II, Invent. Math. 71 (1983), 77-129, then it is well-known that the perversity $\overline{p}$ intersection homology of $Y$ can be computed to be the following cotruncation of $H_{\ast}(L)$:

$IH_{i}^{\overline{p}}(Y) \cong \mathcal{H}^{-i}(Y; \boldsymbol{IC}^{\bullet}_{\overline{p}}(Y)) \cong \begin{cases} 0, \quad i \leq d - \overline{p}(d+1), \\ H_{i-1}(L), \quad i > d - \overline{p}(d+1). \end{cases}$

My question: Is there a similar formula for $\mathcal{H}^{\ast}(X; i_{\ast}\boldsymbol{IC}^{\bullet}_{\overline{p}}(Y))$, i.e. the hypercohomology of the push-forward of $\boldsymbol{IC}^{\bullet}_{\overline{p}}(Y)$ to $X$? And if so, how can it be derived?

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I just realized I made an error in my previous answer here, so I've updated it:

Do you mean $Ri_*$ or literally $i_*$? $Ri_*$ would be the more usual thing to ask about in this context. Then, in general, if $f:X\to Y$ we have $\mathcal H^i(Y;Rf_*S^*)\cong \mathcal H^i(X;S^*)$. So in this case you'd get again the same hypercohomology groups.

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  • $\begingroup$ Dear Greg, thanks a lot for your answer! Yes, let's take $R i_{\ast}$, and then it is good to know that the hypercohomology will transform as you say! $\endgroup$ – Rahmpilz Nov 6 '18 at 8:32

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