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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?

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  • $\begingroup$ 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. $\endgroup$ Dec 11, 2018 at 13:12
  • $\begingroup$ @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. $\endgroup$
    – user45765
    Dec 11, 2018 at 14:43

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$H_\star(X,X \setminus O)\cong H_\star(X,X \setminus \{x\})$ 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 $\text{dirlim}_{U \in \mathcal{U}'} H_*(X, U) \cong H_\star(X,X \setminus \{x\})$.

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