Relative homology and limit

Let $$X$$ be a smooth manifold and $$O\subset X$$ be an closed set containing a non-trivial neighbourhood of $$x\in X$$. The reason to ask the question is to clarify the relationship between limit and relative homology. Recall that in algebraic geometry, an affine scheme $$Spec(R)$$ has stalk $$R_p$$ with $$p\in Spec(R)$$ where $$R_p$$ can be realized as direct limit over open sets $$p\in D(f)\subset Spec(R)$$. I wish to see whether there is connection between limit and relative homology.

Suppose $$O$$ is small enough say a small ball around $$x$$. Then $$H_\star(X,X-O)\cong H_\star(X,X-\{x\})$$ via deformation retraction. Take any homology class $$\gamma\in H_\star(X,X-O)$$. I can always lift to a $$\gamma'$$ cycle of $$Z_\star(X,X-O)$$ first and then perform subdivision. Note that I am basically invoking subdivision procedure and identifying the chain complex level quasi isomorphism and this does not change the relevant homology class information. Now $$H_\star(X,X-O)$$ will throw away the cycles sitting inside $$X-O$$. So I can keep dividing $$\gamma'$$ and in hope to get a cycle "totally" lying in $$O$$ as all cycles in $$X-O$$ are removed by quotient.

$$\textbf{Q1:}$$ Does this infinite subdivision procedure makes sense? It looks like I am considering open covering of $$X$$ and looking at the cycles supported on those coverings. In other words, eventually, I have deformed original $$\gamma'$$ cycle to a cycle lying on $$O$$ and this $$\gamma'$$ may very likely touch boundary of $$O$$ as $$O$$ is closed.

Now $$H_\star(X,X-O)\to H_\star(X,X-\{x\})$$ is by excision first and then following deformation retraction.

$$\textbf{Q2:}$$ Can I say "$$H_\star(X,X-\{x\})$$ is directed limit of $$H_\star(X,U)$$ with $$x\not\in U$$? It is clear that if $$U$$ is small enough, one can do excision again and deformation retraction argument. Here I wish to draw analogy with algebraic geometry setting.

$$\textbf{Q3:}$$ What is the weakest assumption on $$X,O$$ to keep $$Q1,Q2$$ holding?

• I can't at this early hour of the morning follow what you're saying...but it might be helpful to work through an example in low dimensions, like, say, the torus $S^1 \times S^1$, which you can draw as a square with edges identified, and triangulated by a 3x3 grid of lines with diagonals drawn in. Dec 11, 2018 at 13:12
• @JohnHughes The homology I am talking about is singular homology. Simplicial homology does not have this sort of obvious freedom. Given a covering $U_i$ of $X$, then $C_\star(X;U_i)$ is quasi isomorphic to $C_\star(X)$ where latter denotes singular chains supported on $U_i$. This quasi isomorphism is given by homotopy between identity map and subdivision map. Dec 11, 2018 at 14:43

$$H_\star(X,X \setminus O)\cong H_\star(X,X \setminus \{x\})$$ is a serious requirement on $$O$$. For example, it is wrong if $$O = X$$. An adequate choice for $$O$$ is this: If $$X$$ is $$n$$-dimensional, choose a chart $$\phi : U \to \mathbb{R}^n$$ such that $$\phi(x) = 0$$, where $$U$$ is an open neighborhood of $$x$$. Let $$D(0,r) = \{ x \in \mathbb{R}^n \mid \lVert x \rVert \le r \} \subset \mathbb{R}^n$$ be a closed ball and $$\mathring{D}(0,r)$$ its interior which is an open ball. For $$O = O(\phi,r) = \phi^{-1}(D(0,r))$$ both $$X \setminus O$$ and $$X \setminus \{ x \}$$ have $$X \setminus \phi^{-1}(\mathring{D}(0,2r))$$ as a strong deformation retract, therefore the inclusion $$i : X \setminus O \to X \setminus \{ x \}$$ is a homotopy equivalence. Hence $$i_* : H_*(X \setminus O) \to H_*(X \setminus \{ x \})$$ is an isomorphism. The long exact sequences of the pairs $$(X,X \setminus O)$$ and $$(X,X \setminus \{ x \})$$ and the five lemma gives us then an isomorphism $$H_\star(X,X \setminus O) \to H_\star(X,X \setminus \{x\})$$.
Now it seems that you consider the set $$\mathcal{U}$$ of all open sets $$U \subset X$$ not containing $$x$$. This is set is partially ordered by inclusion (i.e. $$U \le U'$$ if $$U \subset U'$$). It is clearly a directed set and it has $$X \setminus \{x\}$$ as maximum. Hence the directed system $$H_*(X, U)$$ indexed by $$\mathcal{U}$$ trivially has $$H_\star(X,X \setminus \{x\})$$ as its direct limit.
But perhaps you want to admit only open $$U \subset X$$ such that $$O_U = X \setminus U$$ is a closed neighborhood of $$x$$. The set $$\mathcal{U}'$$ of all these $$U$$ is a directed subset of $$\mathcal{U}$$ which does not have a maximum. We know that the $$O(\phi,r)$$ constitute a neigborhood basis for $$x$$. Hence the $$U(\phi,r) = X \setminus O(\phi,r)$$ form a cofinal subset of $$\mathcal{U}'$$, hence the direct limits $$\text{dirlim}_{U \in \mathcal{U}'} H_*(X, U)$$ and $$\text{dirlim}_{r >0} H_*(X,U(\phi,r))$$ are isomorphic. Since all $$i_* : H_*(X,U(\phi,r)) \to H_\star(X,X \setminus \{x\})$$ are isomorphisms, we end again with $$\text{dirlim}_{U \in \mathcal{U}'} H_*(X, U) \cong H_\star(X,X \setminus \{x\})$$.