2

Let $C$ be a compact set in Euclidean space and $i$ an integer. Is the inverse limit $$\underset{C\subset U}{\lim_\leftarrow}H_i(U,C)$$ over all open sets $U$ containing $C$ of relative homology groups always zero?

  • Hi Alexander, are you sure this is an inverse limit and not a direct limit? Anyway, your question is more suitable for MathStackExchange, so I've voted to move it there. – Mark Grant Nov 27 '18 at 14:24
  • U containing C, so should be inverse limit. –  Nov 27 '18 at 20:17
  • This is confusing. I assume that open subsets form a directed poset via standard inclusion ordering. I also assume that bonding maps are induced from inclusions. But $H_i$ is a covariant functor. How is that an inverse system? – freakish Nov 28 '18 at 10:26
  • @freakish The system $\mathfrak{U}(C)$ of open neigborhoods of $C$ is usually ordered by $U \le V$ if $U \supset V$ (reverse inclusion). $\mathfrak{U}(C)$ is directed and may be understood as an inverse system of topological spaces: The bonding map $f^V_U : V \to U$ is the inclusion. – Paul Frost Nov 28 '18 at 11:15
  • @freakish The main reason why one doesn't consider the order $U \le V$ if $U \subset V$ is that $\mathfrak{U}(C)$ would have $X = \mathbb{R}^n$ as a maximum. That is, the direct system $\mathfrak{U}(C)$ would be isomorphic to the trivial system ${ X }$. – Paul Frost Nov 28 '18 at 14:26
  • @PaulFrost So basically if a poset $P$ is directed both "upwards" and "downwards" then the direct limit over $P$ is the same as the inverse limit over $P^{op}$ and vice versa? – freakish Nov 28 '18 at 14:43
  • @freakish A poset $(P,\le)$ may be regarded as a (small) category by taking as the set of morphisms from $a$ to $b$ either ${(a,b)}$ if $a \le b$ or $\emptyset$ if $a \not\le b$. Let $\mathcal{C}$ be a category and $P$ be directed. A direct (inverse) system in $\mathcal{C}$ over $P$ is a covariant functor $F : P \to \mathcal{C}$ ($G : P^{op} \to \mathcal{C}$). Obviously the following types of systems can be identified: (1) direct (inverse) systems in $\mathcal{C}$ over $P$. (2) inverse (direct) systems in $\mathcal{C}^{op}$ over $P$. – Paul Frost Nov 28 '18 at 16:13
  • Thus, in the opposite categories $\mathcal{C}$ and $\mathcal{C}^{op}$ the direct limit of $F : P \to \mathcal{C}$ (inverse limit of $G : P^{op} \to \mathcal{C}$) corresponds to the inverse limit of $F^{op} : P^{op} \to \mathcal{C}^{op}$ (direct limit of $G^{op} : P \to \mathcal{C}^{op}$). Note that if both $P$ and $P^{op}$ are directed, then we may consider direct (inverse) systems in $\mathcal{C}$ over $P$ and over $P^{op}$. However, there is no relation between these systems and their limits. – Paul Frost Nov 28 '18 at 16:14
  • For example, the inverse limit of $\mathfrak{U}(C)$ with "reverse" order is $C$ whereas the direct limít $\mathfrak{U}(C)$ with "normal" order is always $\mathbb{R}^n$. – Paul Frost Nov 28 '18 at 16:14

1 Answers1

1

If you work with singular homology, then the answer is "no".

Let $C$ be the topologist's sine curve in $\mathbb{R}^2$. There is a cofinal inverse sequence $(U_n)$ in $\mathfrak{U}(C)$ such that each $U_n$ is homeomorphic to an open disk. Let $i_n : U_{n+1} \to U_n$ denote inclusion. Consider the long exact reduced homology sequence of the pair $(U_n,C)$. We have $H_1(U_n) = \tilde{H}_0(U_n) = 0$, hence we get an isomorphism $$\partial(n) =\partial : H_1(U_n,C) \to \tilde{H}_0(C) = \mathbb{Z} .$$ Concerning $\tilde{H}_0(C)$ recall that $C$ has two path components. By naturality we get $\partial(n) (i_n)_* = \partial(n+1)$ which shows $$\underset{C\subset U}{\lim_\leftarrow}H_1(U,C) \approx \underset{n\in \mathbb{N}}{\lim_\leftarrow}H_1(U_n,C) \approx \mathbb{Z} .$$

Paul Frost
  • 76,394
  • 12
  • 43
  • 125