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I am trying to $H_c^n(X\times \Bbb R;G)$ is isomorphic to $H_c^{n-1}(X;G)$ for all $n$. Here $H_c^n(X;G)$ is just $n$-th cohomology with compact support.

At first I tried to find a compact supported version for Kunneth formula, but quickly I give up this. Then I tried to prove this by choose some specific compact subset and see what happens about the limit but still get nowhere. Could someone give me a hint? Thanks!

Pedro
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  • may i suppose $X$ is a manifold, or it is arbitrary topological space? – Andrey Ryabichev Mar 27 '16 at 15:22
  • This is actually problem 22 of the duality section in hatcher's book. I am looking a solution by algebraic topology, so if you have one for a manifold please show it to me! –  Mar 27 '16 at 15:38
  • There is no additional hypothesis to the space X, but only a arbitrary topological space. So I believe the proof is based on definition of compact cohomology. And I am looking for an answer to solve it. – AG learner Apr 06 '16 at 13:22
  • @YilongZhang Maybe we can use some relative version of kunneth formula, then apply the fact that $H_c^0(R;G)$=0 but $H_c^1(R;G)$=G. –  Apr 06 '16 at 14:50
  • @Learning That's good intuition. But in compact cohomology case, I didn't find any Kunneth formula. Though, suppose $X$ is compact, I can prove it, just by definition and Kunneth formula for relative cohomology. But for $X$ non- compact case, I cannot go further. – AG learner Apr 11 '16 at 03:52

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Partial solution: When $X$ is a $G$-orientable manifold, the claim is a rather easy consequence of General Poincaré-Duality. Suppose $X$ has dimension $k$ so that $X\times \Bbb R$ has dimension $k+1$. Then two applications of PD yield

$H^n_c(X\times \Bbb R;G)\cong H_{k+1-n}(X\times \Bbb R; G)\cong H_{k+1-n}(X; G)\cong H^{n-1}_c(X; G)$.

iwriteonbananas
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    the equation $H^k_c(M)=H_{m-k}(M)$ for open $m$-dimensional manifold $M$ can be proved using dual triangulation (as in compact case) – Andrey Ryabichev Mar 28 '16 at 15:47