I am studying Morse Theory on finite dimensional and compact manifolds using homology groups and relative homology groups on $\mathbb{Z}$. I want to show that this theory could be developed using De Rham cohomology and relative cohomology. What I'd like to ask you is if it's true that $ \forall k \in \{0,...,dim (\mathcal{M}) \}$ and $\forall \, A \subset \mathcal{M}, \, H_k(\mathcal{M},A) \cong H^k (\mathcal{M},A)$. Regarding the same relation between 'absolute' homology and cohomology, I know the Poincaré Duality $H_k (M) \cong H_{n-k}(M) \cong (H^{n-k}(M))^*$ and $(H^{n-k}(M))^* \cong H^{n-k}(M)$ (dim $\mathcal{M} < \infty$) (Is the previous sequence of isomorphisms correct? I might be wrong due to my inexperience in this topic). If this is correct, can I generalize it to Topological couples and consequently to relative homology? How can I prove this?

Thank you for your time.

  • $\begingroup$ No, the actual relative PD is more complex. $\endgroup$ Oct 24, 2021 at 12:43
  • $\begingroup$ Can you suggest me a book where I could learn more about this, please? $\endgroup$
    – Dylan
    Oct 24, 2021 at 14:01
  • $\begingroup$ I will add references later on. $\endgroup$ Oct 24, 2021 at 15:18
  • $\begingroup$ Ok, thank you. I'll wait for your indication. $\endgroup$
    – Dylan
    Oct 25, 2021 at 7:47
  • $\begingroup$ @MoisheKohan and what are the references? I also wanna to know about it. $\endgroup$
    – Z. Liu
    Nov 15, 2021 at 1:01

2 Answers 2


You can find a formulation of relative Poincare Duality for instance in Section 7 (pages 291-298) of

Dold, Albrecht, Lectures on algebraic topology., Classics in Mathematics. Berlin: Springer-Verlag. xi, 377 p. (1995). ZBL0872.55001.

For simplicity, I will assume that a manifold $M$ is oriented and $n$-dimensional, $L\subset K$ are compact subsets of $M$. Then the cup-product with the fundamental class of $M$ defines isomorphisms $$ \check{H}^i(K,L)\cong H_{n-i}(M-L, M-K). $$ (You can apply this, for instance, in the situation when $L=\emptyset$.) Here the left hand side is the Chech cohomology. If $K$ is reasonably nice, say, an ANR, then this is the same as the singular cohomology. Dold proves various variations on this result, for instance, one in section 7.12: $$ \check{H}_c^i(K)\cong H_{n-i}(M, M-K), $$ where $K$ is merely a closed subset of $M$. The subscript $c$ refers to the cohomology with compact support. Take a look at other versions proven in the same section.


Relative Poincare duality is the statement that for a closed, orientable n-manifold $M$ and a compact subset $A$ such that $(X,A)$ satisfies excision, $H^k(M,A) \cong H_{n-k}(M-A)$. This follows from noncompact Poincare duality and the observation that $M-A$ is a manifold with one point compactification equal to $M/A$.

  • $\begingroup$ I don't understand when you say '$M−A$ is a manifold with one point compactification equal to $M/A$'. Do you mean the compactification of Alexandroff? So are you telling me this sentence is false $∀A⊂M$ ,$H_k(M,A)≅H^k(M,A)$ ? Thank you. $\endgroup$
    – Dylan
    Oct 25, 2021 at 7:50
  • $\begingroup$ @Dylan The one point compactification of a locally compact space $X$ is the compact space with a distinguished point such that removing the point gets you $X$. Noncompact Poincare duality relates the homology of $M$ with the cohomology of its one point compactification. And yes, your sentence is wrong for multiple reasons. Poincare duality relates $H^k$ and $H_{n-k}$ for one, but even with this correction you can take $A=\{*\}$ and $k=0$. Then the relative (co)homology is reduced (co)homology and $\bar{H}^0(M)=0 \neq \bar{H}_n(M)=\mathbb{Z}$. $\endgroup$ Oct 25, 2021 at 13:11
  • $\begingroup$ Thank you Connor. $\endgroup$
    – Dylan
    Oct 27, 2021 at 7:33

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