Lefschetz duality for non-compact relative manifolds 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.
 A: 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:


*

*$M$ a closed orientable manifold, then $H^i(M)=H_{n-i}(M)$.

*$M$ a compact orientable manifold with boundary, then $H^i(M)=H_{n-i}(M,\partial M)$ and $H^i(M,\partial M)=H_{n-i}(M)$.

*$M$ 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).

A: 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.
