I'd like to use the formulation of Lefschetz duality stated here, but I can't seem to find a reference for this particular version of it, and it doesn't seem quite right to me.

The exact statement in question is: Let $X$ be a Hausdorff topological space, and let $A \subset X$ be a subspace such that the complement $X - A$ is an orientable topological $n$-manifold. Then, for any abelian group $G$ and any $i$, $$H_i(X, A; G) \cong H_c^{n - i}(X - A; G),$$ where $H_i$ and $H_c^{n - i}$ denote singular homology and compactly supported singular cohomology, respectively.

The problem is, by Poincaré duality, $H_c^{n - i}(X - A; G) \cong H_i(X - A; G)$, so if the above statement is true, then $H_i(X - A; G) \cong H_i(X, A; G)$. But that definitely doesn't seem right; for example, if $X = \mathbb{R}^2$ and $A = \{(0, 0)\}$, then $H_1(X - A; G) \cong G$ because $X - A$ is homotopic to a circle, but $H_1(X, A; G) \cong \tilde{H}_1(X; G) = 0$.

Is the above statement true in this generality? If so, where can I find a proof, and where is my attempt at a counterexample mistaken? If not, what's the correct formulation?

(The main case I'm interested in is where $X$ is a complex projective variety and $A$ is a closed subvariety containing the singular locus of $X$.)

I asked this question on Math.SE already, but there are still no answers after 10 days.

Also, I'm aware of the similar formulation of Lefschetz duality proved on p. 297 of Spanier's Algebraic Topology, but it's not quite the same, and I don't see how one follows from the other.

  • $\begingroup$ I remember being confused about this as well. Maybe some assumption on $A$ is missing? Or we need to take the closure of $X-A$ (and this needs to be a manifold with boundary?)? Back when I needed a statement like this, I found exercise 35 on page 260 of Hatcher's book sufficient. $\endgroup$ – Thomas Rot Apr 26 '14 at 1:09
  • $\begingroup$ Now that Mingcong gave a proof for the smooth case, I feel that for the general case($A$ can be singular and contain the singular locus of $X$), if one takes the log resolution of the pair $(X,A)$, the inverse image of $A$ is a SNC divisor which should have something similar to "the unit disc bundle of tubular nbh"(as locally the equation is nice). The image of the "tubular nbh" in $X$ deformation retracts to $A$ and might serve the purpose as well. $\endgroup$ – Honglu Apr 26 '14 at 22:32

I have a slightly modified version of answer which might help you.

Assume $X$ is a closed(compact + without boundary) manifold and $A$ is a closed submanifold. Then let $M = X - D(N(A))$ where $D(N(A))$(sorry for the silly notation) is the open disc bundle of tublar neighborhood of $A$. Then $M$ is a compact manifold with boundary diffeomorphic to the sphere bundle of $N(A)$. So consider the pair $(M,\partial M)$ and its cohomology. By Poincare duality we have $H^i(M,\partial M;\mathbb{Z}) = H_{n-i}(M;\mathbb{Z})$(Which comes from the version that boundary is the disjoint union of 2 manifold and take the second one to be empty, I think it can be found on Hatcher or Kosinski). (And I omit the coefficient $\mathbb{Z}$ from now on)Then $H_{n-i}(M) = H_{n-i}(X - A)$ since as topological spaces they are homotopy equivalent. Now $H^i(M,\partial M) = H^i(M/\partial M)$ by excision. But $H^i(M/\partial M) = H^i(X/A) = H^i(X,A)$. The first equation is because they are homeomorphic as topological spaces(They are both $X-A$ joining a point). So after all $H^i(X,A) = H_{n-i}(X - A)$ assuming $X$ is closed manifold and $A$ a closed submanifold.

You can also do the same by consider $H_i(M,\partial M;\mathbb{Z})$ at the beginning and you will end at $H_i(X,A) = H^{n-i}(X - A)$. Hopefully my argument makes sense and this works for you.

For more general case I do not know because here I require $A$ to be a submanifold so I can benefit from taking the sphere bundle of tublar neighborhood. I also assume $X$ is compact to avoid anything about compact supported cohomology. This version looks working for your example if you replace $X$ by $S^2$ since we require compactness.

I am sorry that this should be a comment since it does not answer your question precisely but I do not have enough reputation for leaving a comment. And I am dumb at $\LaTeX$ so I use $=$ for isomorphism between groups.

  • $\begingroup$ Indeed, the statement in the question might need $X$ to be compact, which is okay for projective varieties. For singular varieties it should still be true, at least people are saying that on the Grothendieck ring of varieties, the topological euler characteristic(with compact support?) is additive. $\endgroup$ – Honglu Apr 26 '14 at 4:16
  • $\begingroup$ Yes, this is the original classical version of the duality, assuming $X$ compact (relative) manifold. One can have it the other way around: $H_i(X,A)=H^{n-i}(X-A)$. One can play around and say that $H^*(X,A)=H^*_c(X-A)$ or $H_*(X,A)=H_*^c(X-A)$, where the upper $c$ stands for closed supports. But one shouldn't mix the two. In all reasonable cases, one can add a boundary to the manifold, making it compact, write down the "right" duality, and see what can be done (excision and such) to get rid of the boundary. $\endgroup$ – Alex Degtyarev Apr 26 '14 at 5:13
  • $\begingroup$ I think we can be careful here because if $M$ is a compact manifold with boundary, then $H^*_c(M)$ is different from $H^*_c(M - \partial M)$ in general. For example let $M$ be the standard closed ball in $\mathbb{R}^3$, then $H^i_c(M - \partial M) = H^i_c(\mathbb{R}^3)$ which is non-zero only when $i = 3$. But $M$ itself is compact so $H^i_c(M) = H^i(M)$ thus $i = 0$ is the only nontrivial case. This tells me that when you want to get compact supported cohomology on the right hand side you may end up with $H^{n-i}_c(M)$ which may not be $H^{n-i}_c(X-A)$. $\endgroup$ – Mingcong Zeng Apr 26 '14 at 21:54
  • $\begingroup$ Thanks for the answer; this makes sense at a first glance, though I'll need some time to think it over and make sure I really understand what's going on. $\endgroup$ – Daniel Hast Apr 27 '14 at 7:11
  • $\begingroup$ @Mingcong Zeng, Why is $X-A$ homotopic to $M=X-N(A)$ ? I am sure that $A$ is homotopic to $N(A)$, is it true for the complement? $\endgroup$ – H-H Dec 5 '19 at 7:28

Your counterexample looks correct to me. I don't see an obvious way to add hypotheses on $X$ or $A$ so that the statement called "Lefschetz duality" in the link you give becomes correct.

Some correct versions of Poincar\'e(--Lefschetz) duality are the following:

  1. $M$ a closed orientable manifold, then $H^i(M)=H_{n-i}(M)$.
  2. $M$ a compact orientable manifold with boundary, then $H^i(M)=H_{n-i}(M,\partial M)$ and $H^i(M,\partial M)=H_{n-i}(M)$.
  3. $M$ a (not necessarily compact) orientable manifold, then $H^i_c(M)=H_{n-i}(M)$, and $H^i(M)=H_{n-i}^\mathrm{lf}(M)$ ("locally finite" homology, also known as Borel--Moore homology).

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