Let $X$ be compact and Hausdorff, $A\subseteq B\subseteq X$ both closed such that $X\setminus A$ is an open orientable $d$-manifold. Then also $X\setminus B$ is an open orientable $d$-manifold. We have two interesting maps $$p:(X,A)\to (X,B)~~~\text{ and }~~~i:X\setminus B\to X\setminus A.$$ as well as two Poincaré–Lefschetz isomorphisms $$\mathrm{pl}_1:H^{d-*}(X,A)\to H_*(X\setminus A)~~~\text{ and }~~~\mathrm{pl}_2:H^{d-*}(X,B)\to H_*(X\setminus B).$$ Is the following true: $$\mathrm{pl}_1\circ p^* = i_*\circ \mathrm{pl}_2.$$
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2$\begingroup$ Surely the answer will be yes once the question is understood, but what kind of cohomology do you mean? I want to replace cohomology of $(X,A)$ by compactly supported cohomology of $X\backslash A $. $\endgroup$– Tom GoodwillieCommented Jun 17, 2019 at 23:37
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$\begingroup$ Doesn't this follow from $i_*(\alpha) \cap \sigma = i_*(\alpha \cap i^*(\alpha))$? (I haven't much time to think about this so this may be completely wrong.) $\endgroup$– Najib IdrissiCommented Jun 18, 2019 at 7:34
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$\begingroup$ @TomGoodwillie: Yes, I was a bit unprecise: I mean the Čech cohomology $\check{H}^{d-*}(X,A)$ and $\check{H}^{d-*}(X,B)$. Then the isomorphisms are exactly the ones from [Spanier: Algebraic Topology, Thm. 6.2.19]. In my case, $(X,A)$ and $(X,B)$ are relative cell complexes, so we may replace $\check{H}^{d-*}$ by $H^{d-*}$. $\endgroup$– FKranholdCommented Jun 18, 2019 at 8:16
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$\begingroup$ Okay, and this Cech cohomology is canonically isomorphic to the compactly supported cohomology of $X\backslash A$, right? And the duality isomorphism (for the manifold $U=X\backslash A$) is then given by capping compactly supported cocycles of $U$ with a locally finite orientation class. So it seems that the question becomes: is this isomorphism from compactly supported cohomology to homology (for oriented manifolds without boundary) natural for inclusion of an open subset of an oriented manifold? Which it clearly is. $\endgroup$– Tom GoodwillieCommented Jun 18, 2019 at 12:19
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$\begingroup$ Yes, I think your first sentence is correct and can be easily checked by excision! Let $d_A:\check{H}^{d-*}(X,A)\to H_c^{d-*}(X\setminus A)$ be the iso, and likewise $d_B$. Moreover, let $q_A:H_c^{d-*}(X\setminus A)\to H_*(X\setminus A)$ be the Poincaré isomorphism. Then you claim that the following two statements are clear: $q_A=i_*\circ q_B\circ i^*$ and $i^*\circ d_A\circ p^* = d_B$. From those two, my claim follows. The first statement should be clear, as you said. What about the second? $\endgroup$– FKranholdCommented Jun 18, 2019 at 12:49
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