Twisted coefficients on coarse spaces and their corona

Abstract

To a metric space X we associate a compact topological space $\nu '({X})$ called the corona of X. Then a coarse map $f:X\rightarrow Y$ between metric spaces is mapped to a continuous map $\nu '({f}):\nu '({X})\rightarrow \nu '({Y})$ between coronas. Sheaf cohomology assigned to a coarse metric space is preserved and reflected by the corona functor. This work reveals new tools to analyze the Higson corona.

1 Introduction

The corona of a proper metric space has been introduced in [15]. We show it suggests a duality between the coarse structure of a metric space and the topology of its boundary.

1.1 Background and related theories

There are a number of dualities and natural equivalences between categories that relate different areas of mathematics.

The Gelfand duality relates topological spaces with their algebra of functions. Note that [2] gives a very concise introduction to Gelfand duality. Denote by $\texttt {LocKTop}$ the category of locally compact Hausdorff spaces and proper¹ continuous maps. Likewise we denote by $\texttt {CommCStar}$ the category of commutative $C^*$-algebras and nondegenerate $*-$homomorphisms. Then the gelfand representation

$$\begin{aligned} \gamma :\texttt {CommCStar}^{op}\rightarrow \texttt {LocKTop}\end{aligned}$$

is a fully faithful functor.

According to [13, Chapter 1.3] Hilbert’s Nullstellensatz implies there is an equivalence between the category of affine algebraic varieties over an algebraically closed field F and the the opposite of the category of finitely generated commutative reduced unital F-algebras.

We introduce another duality in the realm of mathematics. Coarse geometry of metric spaces is related to the topology of compact spaces by designing yet another version of boundary. The new corona functor may not be an equivalence of categories, it is the closest we can get though.

In [11, 12] has been designed and studied a new relation on subsets of a coarse space: Two subsets $A,B\subseteq X$ of a metric space are related by $\delta _\lambda$, written $A\delta _\lambda B$ if $A\curlywedge B$ or $A\cap B\not =\emptyset$. We know the close relation $\curlywedge$ from [7]. Now $\delta _\lambda$ is a proximity relation on subsets of a asymptotically normal² coarse spaces. The boundary of the Smirnov compactification associated to the proximity space $(X,\delta _\lambda )$ has been proven in [12] to be homeomorphic to the Higson corona $\nu X$ if X is a proper metric space.

In [18] has been associated to a metric space X a corona ${\check{X}}$ as a quotient of the Stone-Čhech compactification. The boundary has been studied in various papers [1, 15,16,17,18]. Again if X is a proper metric space then ${\check{X}}$ is homeomorphic to the Higson corona $\nu (X)$.

This paper introduces a boundary on coarse metric spaces which is a functor that maps coarse maps to continuous maps between compact spaces. This assignment suggests a duality since sheaf cohomology on coarse spaces is isomorphic to the sheaf cohomology of its boundary. In designing this functor we looked specifically for this property. We obtain a boundary which is yet again homeomorhic to the Higson corona if X is proper.

If X is a coarsely proper metric space all three definitions of corona, the one introduced in [12] called $\gamma X/\approx$, the one in [18] called ${\check{X}}$ and in the one in this paper called $\nu '({X})$ can be proven to be equivalent.

1.2 Main contributions

To every proximity space we can associate a compact space which arises as the boundary of the Smirnov compactification. The close relation $\curlywedge$ on subsets of a metric space is almost but not quite a proximity relation. We can still mirror the construction of the Smirnov compactification as described in [14].

The Definition 14 of coarse ultrafilters and Definition 18 of the relation asymptotically alike on coarse ultrafilters combine to a coarse version of a cluster on $(X,\curlywedge )$. This concept is very similar to that of a cluster actually. In Lemmas 16, 20 we show that coarse maps preserve coarse ultrafilters modulo asymptotically alike.

In Definition 22 we define a topology on coarse ultrafilters modulo asymptotically alike. We call the resulting space the corona $\nu '({X})$ of a metric space X. Note again this mirrors the topology on clusters in a Smirnov compactification.

Then Lemma 24 shows that $\nu '$ is a functor:

Theorem A

Denote by $\texttt {mCoarse}$ the category with metric spaces as objects and coarse maps modulo close as morphisms. By $\texttt {Top}$ denote the category of topological spaces and continuous maps. Then

$$\begin{aligned} \nu ':\texttt {mCoarse}\rightarrow \texttt {Top}\end{aligned}$$

is a functor. If X is a metric space then $\nu '({X})$ is compact.

Note we are not able to show that $\nu '({X})$ is metrizable in general. In fact Remark 29 and Proposition 30 suggest the opposite.

The Lemma 32 studies the topology of the corona $\nu '({X})$ of a metric space X and shows that open covers of $\nu '({X})$ can be refined by open covers that are induced by coarse covers on X. See [8] for the first definition of coarse covers. This paper also introduces sheaf cohomology on the Grothendieck topology of open covers. Then Corollaries 33, 34, 35, 36, 37 prove this is a useful observation. Given a sheaf on a coarse space, we can assign a sheaf on $\nu '(X)$ which reflects it. The assignment of the value of a sheaf on an open is a bit more sophisticated. In [9] we show that sheaf cohomology with values in the constant sheaf is reflected by the corona functor.

Theorem B

If X is a proper metric space and A an abelian group then

$$\begin{aligned} {\check{H}}_{ct}^q(X;A) ={\check{H}}^{q}({\nu '({X})},{A}). \end{aligned}$$

Here the left side denotes sheaf cohomology on the coarse space X with values in the constant sheaf A and the right side denotes sheaf cohomology on the topological space $\nu '(X)$ with values in the constant sheaf A.

The Lemma 38 shows that every coarsely injective map induces a closed embedding between coronas. Conversely Theorem 39 shows that $\nu '$ reflects epimorphisms.

2 Metric spaces

Definition 1

Let (X, d) be a metric space. Then the coarse structure associated to d on X consists of those subsets $E\subseteq X^2$ for which

$$\begin{aligned} \sup _{(x,y)\in E}d(x,y)<\infty . \end{aligned}$$

We call an element of the coarse structure entourage. In what follows we assume the metric d to be finite for every $(x,y)\in X^2$.

Definition 2

If X is a metric space a subset $B\subseteq X$ is bounded if the set $B^2$ is an entourage in X.

Definition 3

A map $f:X\rightarrow Y$ between metric spaces is called coarse if

• $E\subseteq X^2$ being an entourage implies that $f^{\times 2}(E)$ is an entourage (coarsely uniform);

• and if $A\subseteq Y$ is bounded then $f^{-1}(A)$ is bounded (coarsely proper).

Two maps $f,g:X\rightarrow Y$ between metric spaces are called close if

$$\begin{aligned} f\times g(\Delta _X) \end{aligned}$$

is an entourage in Y. Here $\Delta _X$ denotes the diagonal in X.

Notation 4

A map $f:X\rightarrow Y$ between metric spaces is called

• coarsely surjective if there is an entourage $E\subseteq Y^2$ such that

  $$\begin{aligned} E[{{\,\textrm{im}\,}}f]=Y; \end{aligned}$$

• coarsely injective if for every entourage $F\subseteq Y^2$ the set $(f^{\times 2})^{-1}(F)$ is an entourage in X.

• two subsets $A,B\subseteq X$ are called coarsely disjoint if for every entourage $E\subseteq X^2$ the set

  $$\begin{aligned} E[A]\cap E[B] \end{aligned}$$

  is bounded.

Remark 5

We study metric spaces up to coarse equivalence. A coarse map $f:X\rightarrow Y$ is a coarse equivalence if

• There is a coarse map $g:Y\rightarrow X$ such that $f\circ g$ is close to $id_Y$ and $g\circ f$ is close to $id_X$.

• or equivalently if f is both coarsely injective and coarsely surjective.

This is [3, Definition 3.D.10]:

Definition 6

(Coarsely proper) If X is a metric space we write

$$\begin{aligned} B(p,r)=\{x\in X:d(x,p)\le r\} \end{aligned}$$

for a point $p\in X$ and $r\ge 0$. The space X is called coarsely proper if there is some $R_0\ge 0$ such that for every bounded subset $B\subseteq X$ the cover

$$\begin{aligned} \bigcup _{x\in B}B(x,R_0) \end{aligned}$$

of B has a finite subcover.

A metric space is said to be locally finite if every bounded set contains finitely many points. Every locally finite metric space is coarsely proper. And a coarsely proper metric space is coarsely equivalent to a locally finite metric space. Namely if $R_0\ge 0$ is as in the definition then a maximal $R_0$-disjoint subset S of X is locally finite and the inclusion $S\subseteq X$ is $R_0$-coarsely surjective.

3 The close relation and coarse covers

This is [7, Definition 9].

Definition 7

(Close relation) Let X be a coarse space. Two subsets $A,B\subseteq X$ are called close if they are not coarsely disjoint. We write

$$\begin{aligned} A\curlywedge B. \end{aligned}$$

Then $\curlywedge$ is a relation on the subsets of X.

Lemma 8

In every metric space X:

• 1. if B is bounded, $B\not \!\!\curlywedge A$ for every $A\subseteq X$

• 2. $U\curlywedge V$ implies $V\curlywedge U$

• 3. $U\curlywedge (V\cup W)$ if and only if $U\curlywedge V$ or $U\curlywedge W$

• 4. for every subspaces $A,B\subseteq X$ with $A\not \!\!\curlywedge B$ there are subsets $C,D\subseteq X$ such that $C\cap D=\emptyset$ and $A\not \!\!\curlywedge (X\setminus C)$, $B\not \!\!\curlywedge X\setminus D$.

Proof

This is [7, Lemma 10, Proposition 11]. $\square$

Remark 9

If $f:X\rightarrow Y$ is a coarse map then whenever $A\curlywedge B$ in X then $f(A)\curlywedge f(B)$ in Y.

We recall [8, Definition 45]:

Definition 10

(Coarse cover) If X is a metric space and $U\subseteq X$ a subset a finite family of subsets $U_1,\ldots ,U_n\subseteq U$ is said to coarsely cover U if for every entourage $E\subseteq X^2$ there exists a bounded set $B\subseteq X$ such that

$$\begin{aligned} U^2\cap (\bigcup _i U_i^2)^c\cap E\subseteq B^2. \end{aligned}$$

Remark 11

Note that coarse covers determine a Grothendieck topology on X. If $f:X\rightarrow Y$ is a coarse map between metric spaces and $(V_i)_i$ a coarse cover of $V\subseteq Y$ then $(f^{-1}(V_i))_i$ is a coarse cover of $f^{-1}(V)\subseteq X$.

Lemma 12

Let X be a metric space. A finite family ${\mathcal {U}}=\{U_\alpha :\alpha \in A\}$ is a coarse cover if and only if there is a finite cover ${\mathcal {V}}=\{V_\alpha :\alpha \in A\}$ of X as a set such that $V_\alpha \not \!\!\curlywedge U_\alpha ^c$ for every $\alpha$.

Proof

This is [7, Lemma 16]. $\square$

Recall [12, Definition 2.1]:

Definition 13

Two subsets of a metric space $S,T\subseteq X$ are called asymptotically alike if there is an entourage $E\subseteq X^2$ such that $E[S]\supseteq T$ and $E[T]\supseteq S$. We write $S\lambda T$ in this case.

4 Coarse ultrafilters

Definition 14

If X is a metric space a system ${\mathcal {F}}$ of subsets of X is called a coarse ultrafilter if

• 1. $A,B\in {\mathcal {F}}$ then $A\curlywedge B$.

• 2. $A,B\subseteq X$ are subsets with $A\cup B\in {\mathcal {F}}$ then $A\in {\mathcal {F}}$ or $B\in {\mathcal {F}}$.

• 3. $X\in {\mathcal {F}}$.

Lemma 15

If X is a metric space and ${\mathcal {F}}$ a coarse ultrafilter on X and if $A\not \in {\mathcal {F}}$ then $A^c\in {\mathcal {F}}$.

Proof

Assume the opposite, both $A,A^c\not \in {\mathcal {F}}$. Then $X=A\cup A^c\not \in {\mathcal {F}}$ which is a contradiction to axiom 3. $\square$

Lemma 16

If $f:X\rightarrow Y$ is a coarse map between metric spaces and ${\mathcal {F}}$ is a coarse ultrafilter on X then

$$\begin{aligned} f_*{\mathcal {F}}:=\{A\subseteq Y:f^{-1}(A)\in {\mathcal {F}}\} \end{aligned}$$

is a coarse ultrafilter on Y.

Proof

• 1. If $A,B\in f_*{\mathcal {F}}$ then $f^{-1}(A),f^{-1}(B)\in {\mathcal {F}}$. This implies $f^{-1}(A)\curlywedge f^{-1}(B)$ which implies $A\curlywedge B$.

• 2. If $A,B\subseteq Y$ are subsets with $A, B\not \in f_*{\mathcal {F}}$ then $f^{-1}(A), f^{-1}(B)\not \in {\mathcal {F}}$. Which implies $f^{-1}(A)\cup f^{-1}(B)\not \in {\mathcal {F}}$. Thus $A\cup B\not \in {\mathcal {F}}$.

• 3. $f^{-1}(Y)=X$. Thus $Y\in f_*{\mathcal {F}}$.

$\square$

Theorem 17

If X is a coarsely proper metric space and $Z\subseteq X$ an unbounded subset then there is a coarse ultrafilter ${\mathcal {F}}$ on X with $Z\in {\mathcal {F}}$.

Proof

We just need to prove there is a coarse ultrafilter ${\mathcal {F}}$ on Z. Then $i_*{\mathcal {F}}$ where $i:Z\rightarrow X$ is the inclusion has the required properties.

We can assume that Z is locally finite. Then the bounded sets are exactly the finite sets. The rest of the proof is very similar to the proof of [14, Theorem 5.8]. Let $\sigma$ be a non-principal ultrafilter on Z (Thus every $A\in \sigma$ is not finite). We define

$$\begin{aligned} {\mathcal {F}}:=\{A\subseteq Z:A\curlywedge C \text{ for } \text{ each } C\in \sigma \} \end{aligned}$$

We check that ${\mathcal {F}}$ is a coarse ultrafilter on Z:

• 1. If $A,B\in {\mathcal {F}}$ let $C\subseteq X$ be a subset. Then either $C\in \sigma$ or $C^c\in \sigma$. This implies both $A\curlywedge C,B\curlywedge C$ or both $A\curlywedge C^c,B\curlywedge C^c$. Thus for every $C\subseteq Z$ we have $C\curlywedge A$ or $C^c\curlywedge B$ this implies $A\curlywedge B$.

• 2. If $A,B\subseteq Z$ are subsets with $A,B\not \in {\mathcal {F}}$ then there are $C_1,C_2\in \sigma$ with $A\not \!\!\curlywedge C_1,B\not \!\!\curlywedge C_2$. Then

  $$\begin{aligned} A\cup B\not \!\!\curlywedge C_1\cap C_2. \end{aligned}$$

  Since $C_1\cap C_2\in \sigma$ we have $A\cup B\not \in {\mathcal {F}}$.

• 3. $Z\in {\mathcal {F}}$ since $Z\curlywedge A$ for every nonbounded subset $A\subseteq Z$.

$\square$

Definition 18

We define a relation on coarse ultrafilters on X: two coarse ultrafilters ${\mathcal {F}},{\mathcal {G}}$ are asymptotically alike, written $A\lambda B$ if for every $A\in {\mathcal {F}},B\in {\mathcal {G}}$:

$$\begin{aligned} A\curlywedge B \end{aligned}$$

Lemma 19

The relation asymptotically alike is an equivalence relation on coarse ultrafilters on X.

Proof

The relation is obviously symmetric and reflexiv. We show transitivity. Let ${\mathcal {F}}_1,{\mathcal {F}}_2,{\mathcal {F}}_3$ be coarse ultrafilters on X such that ${\mathcal {F}}_1\lambda {\mathcal {F}}_2$ and ${\mathcal {F}}_2\lambda {\mathcal {F}}_3$. We show ${\mathcal {F}}_1\lambda {\mathcal {F}}_3$. Assume the opposite. There are $A\in {\mathcal {F}}_1$ and $B\in {\mathcal {F}}_3$ such that $A\not \!\!\curlywedge B$. Then there are subsets $C,D\subseteq X$ with $C^c\cup D^c=X$ and $C^c\not \!\!\curlywedge A,D^c\not \!\!\curlywedge B$. Now one of $C^c,D^c$ is in ${\mathcal {F}}_2$. This contradicts ${\mathcal {F}}_1\lambda {\mathcal {F}}_2$ and ${\mathcal {F}}_2\lambda {\mathcal {F}}_3$. Thus transitivity follows. $\square$

Lemma 20

If two coarse maps $f,g:X\rightarrow Y$ between metric spaces are close and ${\mathcal {F}},{\mathcal {G}}$ are asymptotically alike coarse ultrafilters on X then $f_*{\mathcal {F}}\lambda g_*{\mathcal {G}}$ in Y.

Proof

If $A\in f_*{\mathcal {F}},B\in g_*{\mathcal {G}}$ then $f^{-1}(A)\in {\mathcal {F}},g^{-1}(B)\in {\mathcal {G}}$. This implies $f^{-1}(A) \curlywedge g^{-1}(B)$. Thus there are subsets $S\subseteq f^{-1}(A),T\subseteq g^{-1}(B)$ which are not bounded such that $S\lambda T$. Since f, g are close we have $f(S)\lambda g(T)$. Now $f(S)\subseteq A,g(T)\subseteq B$ are not bounded since f, g are coarsely proper. This implies $A\curlywedge B$. Thus $f_*{\mathcal {F}}\lambda g_*{\mathcal {F}}$ in Y. $\square$

Proposition 21

Let ${\mathcal {F}},{\mathcal {G}}$ be two coarse ultrafilters on a metric space X. Then ${\mathcal {F}}\lambda {\mathcal {G}}$ if and only if for every $A\in {\mathcal {F}}$ there is an element $B\in {\mathcal {G}}$ with $A\lambda B$.

Proof

This has already been proved in [15, Lemma 4.2]. $\square$

5 Topological properties

Definition 22

Let X be a metric space. Denote by ${{\hat{\nu }}} (X)$ the set of coarse ultrafilters on X. We define a relation $\curlywedge$ on subsets of ${{\hat{\nu }}}(X)$ as follows. Define for a subset $A\subseteq X$:

$$\begin{aligned} \textsf{cl}({A})=\{{\mathcal {F}}\in {{\hat{\nu }}}(X):A\in {\mathcal {F}}\} \end{aligned}$$

Then $\pi _1\not \!\!\curlywedge \pi _2$ if and only if there exist subsets $A,B\subseteq X$ such that $A\not \!\!\curlywedge B$ and $\pi _1\subseteq \textsf{cl}({A}),\pi _2\subseteq \textsf{cl}({B})$.

Theorem 23

The relation $\curlywedge$ on ${{\hat{\nu }}}( X)$ is a proximity relation. If ${\mathcal {F}}\in {{\hat{\nu }}}( X)$ Then its closure is

$$\begin{aligned} {{\bar{{\mathcal {F}}}}}=\{{\mathcal {G}}\in {{\hat{\nu }}}( X):{\mathcal {G}}\lambda {\mathcal {F}}\} \end{aligned}$$

Thus $\curlywedge$ is a separated proximity relation on the quotient $\nu '({X}):={{\hat{\nu }}}( X)/\lambda$. We call $\nu '({X})$ with the induced topology of $\curlywedge$ the corona of X.

Proof

We check the axioms of a proximity relation:

• 1. if $\pi _1\curlywedge \pi _2$ then $\pi _2\curlywedge \pi _1$ since the definition is symmetric in $\pi _1,\pi _2$.

• 2. If $\pi _1=\emptyset$ then $\pi _1\subseteq \textsf{cl}({\emptyset })$. Now for every $\pi _2\subseteq \nu '({X})$ we have $\pi _2\subseteq \textsf{cl}({X})$ and $\emptyset \not \!\!\curlywedge X$. Thus $\pi _1\not \!\!\curlywedge \pi _2$.

• 3. If $\pi _1\cap \pi _2\not =\emptyset$ and if $\pi _1\subseteq \textsf{cl}({A}), \pi _2\subseteq \textsf{cl}({B})$ then there is some coarse ultrafilter ${\mathcal {F}}\in \pi _1\cap \pi _2$, thus $A\in {\mathcal {F}}$ and $B\in {\mathcal {F}}$. This implies $A\curlywedge B$. Thus we have shown that $\pi _1\curlywedge \pi _2$.

• 4. Let $\pi _1,\pi _2\subseteq \nu '({X})$ be subsets such that for every $\rho \subseteq \nu '({X})$ we have $\pi _1\curlywedge \rho$ or $\pi _2\curlywedge \rho ^c$. Let $A,B\subseteq X$ be subsets such that $\pi _1\subseteq \textsf{cl}({A}),\pi _2\subseteq \textsf{cl}({B})$. Let $C\subseteq X$ be a subset. Now one of $\pi _1\curlywedge \textsf{cl}({C})$ or $\pi _2\curlywedge \textsf{cl}({C})^c$ holds. We have $\pi _1\subseteq \textsf{cl}({A}),\textsf{cl}({C})\subseteq \textsf{cl}({C})$ and $\pi _2\subseteq \textsf{cl}({B}),\textsf{cl}({C})^c\subseteq \textsf{cl}({C^c})$. Thus $A\curlywedge C$ or $B\curlywedge C^c$. Now $C\subseteq X$ was arbitrary thus $A\curlywedge B$. We have shown $\pi _1\curlywedge \pi _2$.

• 5. Let ${\mathcal {F}},{\mathcal {G}}\in \nu '({X})$ be two coarse ultrafilters such that ${\mathcal {F}}{{\bar{\lambda }}}{\mathcal {G}}$. Then there exist subsets $A,B\subseteq X$ such that $A\not \!\!\curlywedge B$ and $A\in {\mathcal {F}},B\in {\mathcal {G}}$. Thus ${\mathcal {F}}\not \!\!\curlywedge {\mathcal {G}}$.

• 6. if ${\mathcal {F}},{\mathcal {G}}\in \nu '({X})$ are two coarse ultrafilters such that ${\mathcal {F}}\lambda {\mathcal {G}}$ then for every $A\in {\mathcal {F}},B\in {\mathcal {G}}$ we have $A\curlywedge B$. This implies ${\mathcal {F}}\curlywedge {\mathcal {G}}$.

$\square$

Lemma 24

If $f:X\rightarrow Y$ is a coarse map between metric spaces then f induces a proximity map $\nu '({f}):\nu '({X})\rightarrow \nu '({Y})$.

Proof

Define for a subset $S\subseteq \nu '({X})$:

$$\begin{aligned} f_*S=\{[f_*{\mathcal {F}}]:[{\mathcal {F}}]\in S\} \end{aligned}$$

Let $\pi _1,\pi _2\subseteq \nu '({X})$ be two subsets with $f_*\pi _1\not \!\!\curlywedge f_*\pi _2$ in $\nu '({Y})$. Then there are subsets $A,B\subseteq Y$ such that $A\not \!\!\curlywedge B$ and $f_*\pi _1\subseteq \textsf{cl}({A}),f_*\pi _2\subseteq \textsf{cl}({B})$. Then $f^{-1}(A)\not \!\!\curlywedge f^{-1}(B)$ and $\pi _1\subseteq \textsf{cl}({f^{-1}(A)}),\pi _2\subseteq \textsf{cl}({f^{-1}(B)})$. This implies $\pi _1\not \!\!\curlywedge \pi _2$. $\square$

Corollary 25

Denote by $\texttt {mCoarse}$ the category of metric spaces and coarse maps modulo close. By $\texttt {Proximity}$ denote the category of proximity spaces and p-maps. Then

$$\begin{aligned} \nu ':\texttt {mCoarse}\rightarrow \texttt {Proximity}\end{aligned}$$

is a functor.

Theorem 26

If X is a metric space then the space $\nu '({X})$ is compact.

Proof

Let $(A_i)_i$ be a family of arbitrary small closed sets in $\nu '({X})$ with the finite intersection property. We need to show $\bigcap _i A_i\not =\emptyset$. It is sufficient to show this property for $A_i=\textsf{cl}({B_i})$ for every i where $B_i\subseteq X$ are subsets for every i. We can assume $(\textsf{cl}({B_i}))_i$ is maximal among families of closed subsets of $\nu '({X})$ of the form $(\textsf{cl}({B_i}))_i$ with the finite intersection property. Define

$$\begin{aligned} {\mathcal {F}}:=\{B_i:i\}. \end{aligned}$$

Then ${\mathcal {F}}$ is a coarse ultrafilter:

• 1. if $B_i,B_j\in {\mathcal {F}}$ and $B_i\not \!\!\curlywedge B_j$ then $\textsf{cl}({B_i})\cap \textsf{cl}({B_j})=\emptyset$ which is a contradiction to the finite intersection property.

• 2. If $A,B\not \in {\mathcal {F}}$ then there are $B_1,\ldots ,B_n\in {\mathcal {F}}$ with

  $$\begin{aligned} \textsf{cl}({B_1})\cap \cdots \cap \textsf{cl}({B_n})\cap \textsf{cl}({A})=\emptyset . \end{aligned}$$

  And there are $C_1,\ldots ,C_n\in {\mathcal {F}}$ with

  $$\begin{aligned} \textsf{cl}({C_1})\cap \cdots \cap \textsf{cl}({C_m})\cap \textsf{cl}({B})=\emptyset . \end{aligned}$$

  Assume the opposite, $A\cup B\in {\mathcal {F}}$. Since $\textsf{cl}({A})\cup \textsf{cl}({B})=\textsf{cl}({A\cup B})$ we get

  $$\begin{aligned}&\textsf{cl}({A\cup B})\cap \textsf{cl}({B_1})\cap \cdots \cap \textsf{cl}({B_n})\cap \textsf{cl}({C_1})\cap \cdots \cap \textsf{cl}({C_m})\\&\quad =\textsf{cl}({A\cup B})\cap ( (\textsf{cl}({B_1})\cap \cdots \cap \textsf{cl}({B_n})\cap \textsf{cl}({C_1})\cap \cdots \cap \textsf{cl}({C_m})\cap \textsf{cl}({A}))\\&\qquad \cup (\textsf{cl}({B_1})\cap \cdots \cap \textsf{cl}({B_n})\cap \textsf{cl}({C_1})\cap \cdots \cap \textsf{cl}({C_m})\cap \textsf{cl}({B})))\\&\qquad \subseteq (\textsf{cl}({B_1})\cap \cdots \cap \textsf{cl}({B_n})\cap \textsf{cl}({A})) \cup (\cap \textsf{cl}({C_1})\cap \cdots \cap \textsf{cl}({C_m})\cap \textsf{cl}({B}))\\&\quad =\emptyset . \end{aligned}$$

  This is a contradiction to the finite intersection property.

• 3. $X\in {\mathcal {F}}$ by maximality.

Then

$$\begin{aligned} {\mathcal {F}}\in \bigcap _i\textsf{cl}({B_i}) \end{aligned}$$

Thus the intersection of the $\textsf{cl}({B_i})$ is nonempty. $\square$

Proposition 27

If X is a proper metric space then $\nu '({X})$ is homeomorphic to the Higson corona $\nu (X)$ of X.

Proof

We prove that $\nu '({X})$ is an equivalent definition to ${\check{X}}$, the corona of X as defined in [16, Chapter 2] if X is a coarsely proper metric space. Then [16, Proposition 1] implies that $\nu '({X})$ and $\nu (X)$ are homeomorphic if X is a proper metric space.

Without loss of generality we can assume that X is locally finite. Then the bounded sets are exactly the finite sets. In this case the non-principal ultrafilters are exactly the cobounded ultrafilters. By Theorem 17 we can associate to every non-principal ultrafilter a coarse ultrafilter and likewise every coarse ultrafilter on X is induced by a non-principal ultrafilter on X by [14, Theorem 5.8]. By Proposition 21 two non-principal ultrafilters are parallel if and only if their induced coarse ultrafilters are asymptotically alike. Finally the topology on $\nu '({X})$ and ${\check{X}}$ has been defined in the same way. $\square$

Notation 28

If $A,B\subseteq X$ are two subsets of a metric space and $x_0\in X$ a point then define

$$\begin{aligned} \chi _{A,B}:{\mathbb {N}}&\rightarrow {\mathbb {R}}_+\\ i&\mapsto d(A\setminus B(x_0,i),B\setminus B(x_0,i)) \end{aligned}$$

Let ${\mathcal {F}},{\mathcal {G}}$ be two coarse ultrafilters on X. The distance of ${\mathcal {F}}$ to ${\mathcal {G}}$ is at least $f\in {\mathbb {R}}_+^{\mathbb {N}}$, written $d({\mathcal {F}},{\mathcal {G}})\ge f$ if there are $F\in {\mathcal {F}},G\in {\mathcal {G}}$ with $\chi _{F,G}+c\ge f$ for some $c\ge 0$.

Remark 29

Note that by [5, Proposition 8.1] there is no countable subset $S\subseteq {\mathbb {R}}_+^{\mathbb {N}}$ with the property that for every $f\in {\mathbb {R}}_+^{\mathbb {N}}$ there is an element $s\in S$ and $c\ge 0$ with $f+c\ge s$.

Proposition 30

If X is a metric space then the topology on $\nu '({X})$ is coarser than the topology $\tau _d$ induced by d.

Proof

Every point ${\mathcal {F}}\in \nu '({X})$ has a base of neighborhoods $(\textsf{cl}({A}) ^c)_A$ where the index A runs over subsets of X with $A\not \!\!\curlywedge B$ for some $B\in {\mathcal {F}}$. Then the $\chi _{A,B}$-neighborhood of ${\mathcal {F}}$ lies in $\textsf{cl}({A})^c$. $\square$

6 Coarse cohomology with twisted coefficients

Coarse cohomology with twisted coefficients has been introduced in [8]. Coarse covers determine a Grothendieck topology on a coarse space and a coarse map gives rise to a morphism of Grothendieck topologies. The resulting sheaf cohomology on coarse spaces is called coarse cohomology with twisted coefficients.

Definition 31

Let $A\subseteq X$ be a subset of a metric space. Define $\textsf{cl}({A})\subseteq \nu '({X})$ as those classes of coarse ultrafilters for which there is one representative ${\mathcal {F}}\in \textsf{cl}({A})$.

Lemma 32

If X is a metric space

• then $(\textsf{cl}({A})^c)_{A\subseteq X}$ are a base for the topology induced by $\curlywedge$ on $\nu '({X})$.

• If $(U_i)_i$ is a coarse cover of X then $(\textsf{cl}({U_i^c})^c)_i$ is an open cover of $\nu '({X})$.

• For every open cover $(V_i)_i$ of $\nu '({X})$ there is a coarse cover $(U_i)_i$ of X such that $(\textsf{cl}({U_i^c})^c)_i$ refines $(V_i)_i$.

• If $(U_i)_i$ is a coarse cover of X then $\textsf{cl}({U_i^c})^c\cap \textsf{cl}({U_j^c})^c=\textsf{cl}({(U_i\cap U_j)^c})^c$ for every ij.

Proof

• First we show that if $A\subseteq X$ is a subset then $\textsf{cl}({A})$ is closed. Now the topology induced the proximity relation $\curlywedge$ on $\nu '({X})$ is defined by the Kuratowski closure operator

  $$\begin{aligned} \pi \mapsto \{{\mathcal {F}}:{\mathcal {F}}\curlywedge \pi \}. \end{aligned}$$

  Let ${\mathcal {F}}\in \nu '({X})$ be an element with ${\mathcal {F}}\curlywedge \textsf{cl}({A})$. Then $B\in {\mathcal {F}}$ implies $B\curlywedge A$. Thus $A\in {\mathcal {G}}$ with ${\mathcal {G}}\lambda {\mathcal {F}}$. Thus ${\mathcal {F}}\in \textsf{cl}({A})$. We show $(\textsf{cl}({S})^c)_{S\subseteq X}$ are a base for the topology of $\nu '({X})$. If $A\subseteq \nu '({X})$ is closed and ${\mathcal {F}}\in A^c$ then ${\mathcal {F}}\not \!\!\curlywedge A$. Thus there are $S,T\subseteq X$ with $S\in {\mathcal {F}}, A\subseteq \textsf{cl}({T})$ such that $S\not \!\!\curlywedge T$. But then ${\mathcal {F}}\not \!\!\curlywedge \textsf{cl}({T})$ thus ${\mathcal {F}}\in \textsf{cl}({T})^c\subseteq A^c$.

• We proceed by induction on the number of components of $(U_i)_i$.

  • 1. If U coarsely covers X then $U^c$ is bounded. Thus $\textsf{cl}({U^c})=\emptyset$ which implies $\textsf{cl}({U^c})^c=\nu '({X})$.

  • 2. If $U_1^c\not \!\!\curlywedge U_2^c$ then $\textsf{cl}({U_1^c})\not \!\!\curlywedge \textsf{cl}({U_2^c})$ which implies $\textsf{cl}({U_1^c})\cap \textsf{cl}({U_2^c})=\emptyset$. Thus $\textsf{cl}({U_1^c})^c\cup \textsf{cl}({U_2^c})^c=\nu '({X})$.

  • 3. If $U,V,U_1,\ldots ,U_n$ coarsely covers X then $U\cup V,U_1,\ldots ,U_n$ coarsely cover X and U, V coarsely cover $U\cup V$. By induction hypothesis $\textsf{cl}({(U\cup V)^c})^c,\textsf{cl}({U_1^c})^c,\ldots ,\textsf{cl}({U_n^c})^c$ is an open cover of $\nu '({X})$. Now $\textsf{cl}({(U\cup V)^c})^c\subseteq \textsf{cl}({U\cup V})$ and $\textsf{cl}({U^c})^c,\textsf{cl}({V^c})^c$ are an open cover of $\textsf{cl}({U\cup V})$. As a result $\textsf{cl}({U^c})^c,\textsf{cl}({V^c})^c,\textsf{cl}({U_1^c})^c,\ldots ,\textsf{cl}({U_n^c})^c$ are an open cover of $\nu '({X})$.

• We can refine $(V_i)_i$ by $(\textsf{cl}({U_i^c})^c)_i$ for some $U_i\subseteq X$. By Thereom 26 we can refine $(\textsf{cl}({U_i^c})^c)_i$ by a finite subcover $\textsf{cl}({U_1^c})^c,\ldots ,\textsf{cl}({U_n^c})^c$. Now

  $$\begin{aligned} \textsf{cl}({U_1^c})\cap \ldots \cap \textsf{cl}({U_n^c})=\emptyset \end{aligned}$$

  implies $\bigcap _i E[U_i^c]$ is bounded for every entourage $E\subseteq X^2$. This is equivalent to $(U_i)_i$ being a coarse cover.

• Since $U_i^c\cup U_j^c\supseteq U_i^c,U_j^c$ the inclusion $\textsf{cl}({(U_i\cap U_j)^c})^c\subseteq \textsf{cl}({U_i^c})^c\cap \textsf{cl}({U_j^c})^c$ is obvious. For the reverse inclusion note if ${\mathcal {F}}$ is a coarse ultrafilter and $U_i^c,U_j^c\not \in {\mathcal {F}}$ then $U_i^c\cup U_j^c\not \in {\mathcal {F}}$.

$\square$

Corollary 33

The functor $\nu '$ preserves and reflects finite coproducts.

Proof

In case X is a proper metric space there is a homeomorphism $\nu '({X})=\nu (X)$. Then [19, Proposition 4.5, Theorem 4.6] already states this result. Now Lemma 32 serves an alternative proof: A coarse disjoint union³$(U_i)_i$ of X is mapped to a union of topological connection components $(\nu '({U_i}))_i$. If for two subsets $A,B\subseteq X$ the closed sets $\textsf{cl}({A})\not \!\!\curlywedge \textsf{cl}({B})$ are disjoint then A, B are coarsely disjoint. $\square$

There is a close connection between sheaves on the Grothendieck topology of coarse covers and sheaves on the topological space $\nu '(X)$. In [9] we assign to every sheaf ${\mathcal {F}}$ on coarse covers of X a sheaf ${\mathcal {F}}^\nu$ on $\nu '(X)$. Likewise a sheaf ${\mathcal {G}}$ on $\nu '(X)$ is mapped to a sheaf $\hat{{\mathcal {G}}}$ on coarse covers of X. Then sheaf cohomology is reflected and preserved by this functoriality:

Corollary 34

If X is a metric space and A an abelian group then

$$\begin{aligned} {\check{H}}^q_{ct}(X;A_X) = {\check{H}}^{q}({\nu '({X})},{A_{\nu '(X)}}) \end{aligned}$$

here the left side denotes sheaf cohomology on the coarse space X with values in the constant sheaf A and the right side denotes sheaf cohomology on the topological space $\nu '(X)$ with values in the constant sheaf A.

A proof can be found in [9].

Corollary 35

If two coarse maps between metric spaces $f,g:X\rightarrow Y$ are close then the induced maps $f_*:{\check{H}}^q_{ct}(X;{\mathcal {F}})\rightarrow {\check{H}}^q_{ct}(Y;f_*{\mathcal {F}})$ and $g_*:{\check{H}}^q_{ct}(X;{\mathcal {F}})\rightarrow {\check{H}}^q_{ct}(Y,{\mathcal {Y}},g_*{\mathcal {F}})$ are isomorphic.

Proof

This has already been proved in [8, Theorem 72]. Now we provide another proof: A coarse map $f:X\rightarrow Y$ between coarse spaces gives rise to a morphism of Grothendieck topologies $Y_{ct},X_{ct}$ in the same way as $\nu '({f})$ gives rise to a morphism of Grothendieck topologies of the topological spaces $\nu '({Y}),\nu '({X})$. Thus the direct image functors $f_*$ and $\nu '({f})_*$ are basically the same maps on sheaf level. If f, g are close coarse maps then $\nu '({f})=\nu '({g})$ by Lemma 20, Lemma 24. Thus the result follows. $\square$

Corollary 36

Let X be a metric space. If ${\mathcal {F}}$ is a sheaf on X and $A,B\subseteq X$ are two subsets that coarsely cover X there is a Mayer-Vietoris long exact sequence in cohomology:

$$\begin{aligned} \cdots&\rightarrow {\check{H}}^{i-1}({A\cap B},{{\mathcal {F}}})\rightarrow {\check{H}}^{i}({A\cup B},{{\mathcal {F}}})\rightarrow {\check{H}}^{i}({A},{{\mathcal {F}}})\times {\check{H}}^{i}({B},{{\mathcal {F}}})\\&\rightarrow {\check{H}}^{i}({A\cap B},{{\mathcal {F}}})\rightarrow \cdots \end{aligned}$$

Proof

This is already [8, Theorem 74]. Now we provide an alternative proof: If $A,B\subseteq X$ are two subsets of a metric space then $\nu '({A}),\nu '({B})$ can be realized as closed subsets of $\nu '({X})$ by Lemma 38. By [10, II.Mayer-Vietoris sequence 5.6] there is a long exact Mayer-Vietoris sequence for every sheaf ${\mathcal {F}}$ on X. $\square$

Corollary 37

If X is a proper metric space and the asymptotic dimension $n={{\,\textrm{asdim}\,}}(X)$ of X is finite then

$$\begin{aligned} {\check{H}}^{q}({X},{{\mathcal {F}}})=0 \end{aligned}$$

for $q>n$ and ${\mathcal {F}}$ a sheaf on X.

Proof

Note that $\nu '({X})$ is paracompact since $\nu '({X})$ is compact. By [6, Chapitre II.5.12] it is sufficient to show that the covering dimension of $\nu '({X})$ does not exceed n. Since [4, Theorem 1.1] showed that $\dim (\nu X)\le {{\,\textrm{asdim}\,}}(X)$ and Proposition 27 showed $\nu '({X})=\nu X$ are homeomorphic the result follows. $\square$

7 On morphisms

Lemma 38

If $f:X\rightarrow Y$ is a coarsely injective coarse map between metric spaces then $\nu '({f})$ is a closed embedding.

Proof

We show if $i:Z\rightarrow X$ is an inclusion of metric spaces then $\nu '({i})$ is a closed embedding.

First we prove $\nu '({i})$ is injective: Let ${\mathcal {F}},{\mathcal {G}}$ be two coarse ultrafilters on Z. If $i_*{\mathcal {F}}\lambda i_*{\mathcal {G}}$ then ${\mathcal {F}}\lambda {\mathcal {G}}$ obviously.

Now we prove $\nu '({i}) (\nu '({Z}))=\textsf{cl}({Z})$. If ${\mathcal {F}}$ is a coarse ultrafilter on X with $Z\in {\mathcal {F}}$ then define

$$\begin{aligned} {\mathcal {F}}|_Z:=\{A\subseteq Z: A\in {\mathcal {F}}\}. \end{aligned}$$

Then ${\mathcal {F}}|_Z$ is obviously a coarse ultrafilter on Z.

Now we show $i_*{\mathcal {F}}|_Z\lambda {\mathcal {F}}$: if $A\in i_*{\mathcal {F}}|_Z,B\in {\mathcal {F}}$ then $A\cap Z\in {\mathcal {F}}$. Thus $A\curlywedge B$. $\square$

Theorem 39

If $f:X\rightarrow Y$ is a coarse map between coarsely proper metric spaces and if $\nu '({f})$ is surjective then f is coarsely surjective.

Proof

If $f:X\rightarrow Y$ is not coarsely surjective then there exists a nonbounded subspace $Z\subseteq Y$ such that $f(X)\not \!\!\curlywedge Z$. By Theorem 17 there exists a coarse ultrafilter ${\mathcal {F}}$ on Y with $Z\in {\mathcal {F}}$. This implies $f(X)\not \in {\mathcal {F}}$. Thus ${\mathcal {F}}\not \in {{\,\textrm{im}\,}}\nu '({f})$. $\square$

Lemma 40

A coarse map $f:X\rightarrow Y$ between metric spaces is coarsely injective if for every two subsets $A,B\subseteq X$ the relation implies $f(A)\not \!\!\curlywedge f(B)$.

Proof

The proof is similar to the proof of [12, Theorem 2.3]. Let f have the above property. Then for every two subsets $A,B\subseteq X$ the relation $A{{\bar{\lambda }}} B$ in X implies $f(A){{\bar{\lambda }}} f(B)$ in Y. Assume the opposite, f is not coarsely injective. Then there is some $r\ge 0$ and a sequence $(x_n,y_n)_n\subseteq X^2$ such that $d(f(x_n),f(y_n))<r$ and $d(x_n,y_n)>n$. Now the sequences $(x_n)_n,(y_n)_n$ satisfy the hypothesis of [12, Lemma 2.2] which leads to a contradiction. $\square$

8 Side notes

Remark 41

(Space of Ends) Let X be a metric space. The relation $\sim$ on $\nu '({X})$ which is defined by belonging to the same topological connection component is an equivalence relation. Similarly as in [7, Side Notes] we obtain the space of ends by Freudenthal of X, if we assume X to be proper geodesic.

Recall from [12] that the relation $\delta _\lambda$ on subsets of a proper metric space is defined by $A\delta _\lambda B$ if $A\curlywedge B$ or $A\cap B\not =\emptyset$. It is a proximity relation on X.

Remark 42

It has been shown in [12] that the Higson corona $\nu X$ of a proper metric space X arises as the boundary of the Smirnov compactification of the proximity space $(X,\delta _\lambda )$. Then [14, Theorem 7.7] implies that $A\delta _\lambda B$ if and only if $A\cap B\not =\emptyset$ or $({{\bar{A}}}\cap \nu X)\cap ({{\bar{B}}}\cap \nu X)\not =\emptyset$ for every subsets $A,B\subseteq X$. Thus if $A\cap B=\emptyset$ then $A\not \!\!\curlywedge B$ if and only if $({{\bar{A}}}\cap \nu X)\cap ({{\bar{B}}}\cap \nu X)=\emptyset$.

Recall the following theorem from [7]:

Theorem 43

Let X be a metric space. A finite family $U_1,\ldots , U_n$ is a coarse cover if and only if there is a finite cover $V_1,\ldots , V_n$ of X as a set such that $V_i\not \!\!\curlywedge U_i^c$ for every $i\in \{1,\ldots ,n\}$.

A cover $U_1,\ldots ,U_n$ is called a $\delta _\lambda$-cover of X if $U_1,\ldots ,U_n$ is a coarse cover and $U_1\cup \cdots \cup U_n=X$.

Lemma 44

Let X be a metric space. A finite family $U_1,\ldots ,U_n$ is a $\delta _\lambda$-cover if and only if there exists a cover $V_1,\ldots ,V_n$ of X as a set such that $V_i{{\bar{\delta }}}_\lambda U_i^c$ for every i.

Proof

We first show the forward direction. Suppose $U_1,\ldots ,U_n$ is a $\delta _\lambda$-cover. For every $i\in \{1,\ldots ,n\}$ we define

$$\begin{aligned} V_i:=\{x\in X\mid d(x,U_j^c)\le d(x,U_i^c)\forall j\in \{1,\ldots ,{{\hat{i}}},\ldots ,n\}\}. \end{aligned}$$

If $x\in X$ then $\{d(x,U_1^c),\ldots ,d(x,U_n^c)\}$ contains a maximal element. Thus there is some i with $x\in V_i$. This implies the $V_i$ cover X. Then $V_1,\ldots ,V_n$ is a cover of X and

$$\begin{aligned} V_i\cap U_i^c&=\{x\in X|d(x,U_j^c)\le d(x,U_i^c)\forall j\in \{1,\ldots ,{{\hat{i}}},\ldots ,n\},x\in U_i^c\}\\&=U_1^c\cap \cdots \cap U_n^c\\&=(U_1\cup \cdots \cup U_n)^c\\&=\emptyset \end{aligned}$$

Together with Theorem 43 we obtain $V_i{{\bar{\delta }}} U_i^c$.

Now we show the reverse direction. Suppose $U_1,\ldots ,U_n$ are subsets and $V_1,\ldots , V_n$ a cover of X with $V_i{{\bar{\delta }}} U_i^c$. Then in particular $V_i\cap U_i^c=\emptyset$. Then

$$\begin{aligned}&U_1\cup \cdots \cup U_n \supseteq V_1\cup \cdots \cup V_n\\&\quad =X. \end{aligned}$$

Now in particular $V_i\not \!\!\curlywedge U_i^c$ for every i and Thereom 43 implies that $U_1,\ldots ,U_n$ is a coarse cover. Thus $U_1,\ldots ,U_n$ are a $\curlywedge '$-cover. $\square$