I am dealing with a compact topological space $X$, such that $X\setminus S $ is an $n$-manifold, here $S\subset X$ is a compact subset of $X$ ($S$ stands for singular locus).
An example could be the Hawaiian earring where $n=1$ and the singular locus can be taken to be the origin.
Now, I would like to understand the relation between the top Cech cohomology and singular homology of the pair $\check {H^n}(X, S)$, $H^n(X, S)$.
In particular I expect that $\check {H^n}(X, S)= H_c^n(X\setminus S)$ (cohomology with compact support) and that the latter is generated by the connected components of $X\setminus S$, is this correct? How to prove it?
What about the singular cohomology $H^n(X,S)$ ? Is it isomorphic to $\check {H^n}(X, S)$?
Best Answer
Cech cohomology groups can be defined for very general pairs $(X,A)$, but best results are obtained for compact (including Hausdorff!) pairs. This completely suffices for your question.
Cech cohomology has a number of interesting properties.
For compact polyhedral pairs (or more generally compact CW pairs) it agrees with singular cohomology.
Continuity: If $(X,A)$ is the inverse limit of an inverse system $(\mathbf X, \mathbf A)$ of compact pairs $(X_\alpha,A_\alpha)$, i.e. $(X,A) = \varprojlim (\mathbf X, \mathbf A)$, then $\check H^*(X,A)$ is the direct limit of Cech cohomology groups $\varinjlim \check H^*(\mathbf X, \mathbf A)$.
Strong excision: $\check H^*(X,A) = \check H^*(X/A,*)$. The latter is the reduced Cech cohomology of the quotient $X/A$.
The Hawaiian earring is the inverse limit of an inverse sequence $X_n= $ wedge of $n$ copies $S^1_i$ of the circle. The bonding maps $p_n : X_{n} \to X_{n-1}$ are the retractions mapping $S^1_n$ to the basepoint. Using 1. and 2. we get
$$\check H^1(X,S) = \check H^1(X) = \Sigma = \bigoplus_{i=1}^\infty \mathbb Z .$$
The first singular cohomology group of the Hawaiian earring is also $\bigoplus_{i=1}^\infty \mathbb Z$. To see this, we can use the exact universal cooeficient sequence $$0 \to Ext(H_{n−1}(X),\mathbb Z) \to H^n(X) \to Hom(H_n(X),\mathbb Z) \to 0$$ For $n =1$ we have $H_0(X) = 0$, thus $$H^1(X) \approx Hom(H_1(X),\mathbb Z).$$
The first singular homology group of the Hawaiian earring has the form $\Pi \oplus (\Pi/\Sigma)$, where $\Pi = \prod_{i=1}^\infty \mathbb Z$ is the Baer-Specker-group (see here). By the way, the first Cech homology group is $\Pi$. We therefore get $$H^1(X) \approx Hom(\Pi,\mathbb Z) \oplus Hom(\Pi/\Sigma,\mathbb Z) .$$ The first summand is isomorphic to $\Sigma$ (see here). The second summand can be identified with the subgroup of $Hom(\Pi,\mathbb Z)$ annihalating $\Sigma$. It is therefore a free Abelian group of (at most) countably infinite rank.
Here is an example where you can see that in general you cannot expect $\check H^n(X,S) = H^n(X,S)$ in the top dimension.
Let $X = S^2$ and $S = W$ be a copy of the Warsaw circle. The singular cohomology of $W$ is trivial in all positive dimensions, its Cech cohomology agrees with that of $S^1$. Considering the long exact cohomology sequences of the pair $(S^2,W)$ we get $$ 0 = H^1(W) \to H^2(S^2,W) \to H^2(S^2) = \mathbb Z \to H^2(W) = 0 ,$$ i.e. $H^2(S^2,W) = \mathbb Z$, and $$ 0 = \check H^1(S^2) \to \check H^1(W) = \mathbb Z \to \check H^2(S^2,W) \to \check H^2(S^2) = \mathbb Z \to \check H^2(W) = 0 .$$ This short exact sequence splits and we get $\check H^2(S^2,W) = \mathbb Z \oplus \mathbb Z$.