$H^n(X, S)$ and $\check {H^n}(X,S)$ when $X\setminus S$ is an $n$-manifold (Hawaiian earring related)

Question

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 $\check {H^n}(X, S)$?

Answer 1

Here is the worked out example of the Hawaiian earring. I'm afraid I don't have much to say about the general case, because the relative Cech cohomology groups seems very badly behaved formally. However, this is too long to leave as another comment.

To set this up, let $C_n\subseteq\mathbb{R}^2$ be the circle of radius $1/n$ about $(0,1/n)$, so that $X=\bigcup_{n\ge1}C_n\subseteq\mathbb{R}^2$ is the earring space and $S=\{(0,0)\}\subseteq X$ is the bad point. The complement $X\setminus S$ is homeomorphic to $\coprod_{n\ge1}\mathbb{R}$, a $1$-manifold.

The relative cohomology group $H^1(X,S)$ is canonically isomorphic to $H^1(X)$ via pullback $H^1(X,S)\stackrel{\sim}{\rightarrow}H^1(X)$, since $S$ is a singleton (this follows from the LES of the pair $(X,S)$), which is isomorphic to $\bigoplus_{n\ge1}\mathbb{Z}$, see here.

In general, if $Y=\coprod_{i\in I}Y_i$ is a coproduct of topological spaces, there is a natural isomorphism between the cochain complexes $C^{\bullet}(Y)\stackrel{\sim}{\rightarrow}\prod_{i\in I}C^{\bullet}(Y_i)$, which is given by pullback along $Y_i\subseteq Y$ in each factor. In this way, the support of a cochain on $Y$ corresponds to the disjoint union of the supports of the restricted cochains on the $Y_i,\,i\in I$. A disjoint union of sets is compact if and only it is finite and each summand is compact, so this isomorphism identifies the compactly supported cochain complexes $C_c^{\bullet}(Y)\stackrel{\sim}{\rightarrow}\bigoplus_{i\in I}C_c^{\bullet}(Y_i)$. On cohomology, this yields canonical isomorphisms $H_c^k(Y)\stackrel{\sim}{\rightarrow}\bigoplus_{i\in I}H_c^k(Y_i)$ for all $k\ge0$. In the example, applying this observation to $X\setminus S\cong\coprod_{n\ge1}\mathbb{R}$, we obtain $H_c^1(X\setminus S)\cong\bigoplus_{n\ge1}H_c^1(\mathbb{R})\cong\bigoplus_{n\ge1}\mathbb{Z}$.

Let $X$ be any topological space and $S\subseteq X$ a singleton. For each open cover $\mathcal{U}$ of $X$, there is a natural pullback $H^k(N(\mathcal{U}),N(\mathcal{U}\vert_S))\rightarrow H^k(N(\mathcal{U}))$ for all $k\ge0$, which is an isomorphism for $k>0$ by the corresponding LES in cohomology. Taking the limit, we obtain an isomorphism $\check{H}^k(X,S)\stackrel{\sim}{\rightarrow}\check{H}^k(X)$.

Now let $X$ be the earring space again. For each $i\ge1$, let $\mathcal{U}_i$ be the cover of $X$ that consists of the following open sets:

• the union of $\bigcup_{n\ge i+1}C_n$ and arcs of diameter $<1/i$ about $(0,0)$ in each $C_n$ for $n=1,\dotsc,i$,
• arcs of diameter $<1/i$ in each $C_n$ for $n=1,\dotsc,i$, such that for each $n\in\{1,\dotsc,i\}$, the collection of all arcs in $C_n$ considered here together with the single arc contained in the previously mentioned open set form an open cover of $C_n$, where each arc intersects exactly two adjacent ones.

This part is a bit handwavy, but I don't wanna be more explicit. It's not hard to see pictorially that these choices are easily arrangeable in a way such that $\mathcal{U}_{i+1}$ refines $\mathcal{U}_i$ for all $i\ge1$. Furthermore, each open cover $\mathcal{U}$ of $X$ is refined by $\mathcal{U}_i$ for $i\gg0$, as a consequence of the Lebesgue number lemma (note that the unions $\bigcup_{n\ge i+1}C_n$ have diameter shrinking to $0$ as $i\rightarrow\infty$). This implies the directed system of the $\mathcal{U}_i,\,i\ge0$ is cofinal in the directed system of all open covers of $X$, whence we can compute $\check{H}^1(X)\cong\varinjlim H^1(N(\mathcal{U}_i))$.

However, and I defer to the pictorial intuition again, it is by construction that $N(\mathcal{U}_i)\cong\bigvee_{n=1}^iS^1$ (here, it is relevant that we chose to avoid triple and higher intersections), and the refinments induce simplicial maps which realize the canonical projections $\bigvee_{n=1}^{i+1}S^1\rightarrow\bigvee_{n=1}^iS^1$ collapsing the $i+1$-th $S^1$ to the wedge point for all $i\ge1$. On cohomology, this corresponds to the canonical inclusion in the $i+1$-th summand $\bigoplus_{n=1}^i\mathbb{Z}\rightarrow\bigoplus_{n=1}^{i+1}\mathbb{Z}$. In total, $\check{H}^1(X)\cong\varinjlim H^1(\bigvee_{n=1}^iS^1)\cong\varinjlim\bigoplus_{n=1}^i\mathbb{Z}=\bigoplus_{n\ge1}\mathbb{Z}$.

Answer 2

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.

1. For compact polyhedral pairs (or more generally compact CW pairs) it agrees with singular cohomology.

2. 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 $\varinjlim \check H^*(\mathbf X, \mathbf A)$.

3. 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 $\check H^2(S^2,W) = \mathbb Z \oplus \mathbb Z$.