# Why coarse maps have to be proper?

A map $$f: X \to Y$$ between metric spaces is said to be coarse, if the following two conditions hold:

1. $$f$$ is bornologous, i.e. $$\forall_{R>0} \; \exists_{S>0} \; d(x,y) < R \Rightarrow d(f(x),f(y)) < S$$
2. $$f$$ is (metrically) proper, i.e. preimages of bounded sets are bounded

My question is, why do we need the "2." condition in this definition? Why coarse maps have to be proper?

Some background:

Coarse map is a basic concept of coarse geometry and this definition is given in the book Lectures on Coarse Geometry by John Roe, one of the founding fathers of the subject. I don't know who formulated this definition first, but all references that I've seen point there.

As I understand, the central notion of interest here are coarse equivalences, which are the isomorhpisms in the coarse category, which are those maps that have coarse inverses: for coarse map $$f$$ there must be a coarse map $$g:Y\to X$$ such that $$g\circ f$$ and $$f\circ g$$ are bornotopic to the identity, i.e. $$\sup_{x\in X}\; d( \ g(f(x)) , \ x) < \infty \qquad \sup_{y\in Y}\; d( \ f(g(y)) , \ y) < \infty$$

This gives intuitive results like $$\mathbb{Z}$$ being coarsely equivalent to $$\mathbb{R}$$ - which we imagine as looking at $$\mathbb{Z}$$ from far away, when line of discrete points merge to continuous line. Or that the Cayley diagram of a group $$\pi_1(X)$$ is coarsely equivalent to the universal cover of $$X$$ - no matter what presentation of $$\pi_1(X)$$ we use. In fact, coarse maps have great applications in geometric group theory and I suspect the answer may be hidden there.

But if we exclude properness from the definition, then if $$f$$ is a coarse equivalence, then it is proper. So removing this property does not add any new coarse equivalences. I don't see other reasons why properness should stay.

If we look at a close cousin of coarse spaces, namely uniform spaces (indeed we can look at bornologous/coarse maps as "uniformly bounded" maps), we see that an uniform map is just a uniformly continuous map, i.e. satisfying $$\forall_{\varepsilon > 0} \; \exists_{\delta>0} \; d(x,y)<\delta \Rightarrow d(f(x),f(y)) < \varepsilon$$ while it doesn't have any proper-like condition. (Would it be something along the lines - "images of small sets are small"? Maybe it already holds?)

Moreover, if we consider the product of two metric spaces $$X\times Y$$, then the usual projection maps $$\pi_X,\pi_Y$$ are not coarse, so they don't form a product in the coarse category. This is really nonappealing for me. So I suppose there must be a good reason to do this.

From the terminology, I can expect that equating the concepts of coarse and bornologous maps is not really fruitful, but I want to see the reason why.

Edit: I found a paper[1] which defines the $$\mathbf{Coarse}$$ category with bornological maps, not coarse maps. (It keep the names the same, though.) I suppose naming categories is just a convention, but nevertheless coarse map is still a thing, so my question still holds.

[1] Some categorical aspects of coarse spaces and balleans; D.Dikranjan, N.Zava

• Properness is very natural if you consider the setting of finitely generated subgroups of finitely generated groups: The inclusion $H < G$ is a proper map. Apr 19, 2019 at 23:59
• I don't see why that is anticipated. On the other hand, the projection G×G -> G is not (usually) proper. Examples don't help me here. What I would want to see, is some theorems that work only if the map is coarse (so they need both bornology and properness). For example, for continuous maps we have intermediate value theorem, which illustrates what continuous is all about.
– mz71
Apr 20, 2019 at 0:40
• A homomorphism between finitely generated groups is proper if and only if it has finite kernel. Try to prove this. Apr 20, 2019 at 0:45
• I can prove it. Sadly, I don't see any intuition about coarse maps steaming from it.
– mz71
Apr 20, 2019 at 0:56
• Do you have an intuition for what restricting to bornologous maps would get you? What sort of questions you would answer? Would they be geometric or maybe more analysis or something else? What does having products get you? In your mind what should a coarse category be? What sort of questions should it help answer? Does having the arrows be bornology answer these questions? Even whether or not a coarse map exists between two different spaces is an interesting invariant (between the euclidean plane and hyperbolic plane).
– user29123
Apr 20, 2019 at 1:29

The main point is that I don't really think bornologous maps are really geometric generally. Consider a surjective map on the real numbers which snakes/bounces back and forth going further each time. This does not seem like it should be a coarse maps but it bornologous if you snake back and forth at a constant rate. It isn't metrically proper though since the preimage of 0 will be unbounded. If you are trying to do geometry you probably don't want to deal with maps like.

One thing you mention is that your understanding is that being coarsely equivalent and "bornology equivalence" give the same equivalences and that you think this is the main point of interest. Certainly that equivalence is interesting but it is not necessarily the point, you want interesting invariants. Even whether or not coarse maps exist between different spaces is interesting. For example consider the spaces: regular 4-valent tree(standard Cayley graph of free group on two generators), hyperbolic 2-space, hyperbolic 3-space, Euclidean 2-space, and Eulcidean 3-space. What are the bornologous maps between pairs of these spaces(there are tons)? What are the coarse maps between spaces? Are there any? I think that by thinking about this you will realize that existence of coarse maps(and non existence) actually tell you something about the spaces geometry.

Related to the above is asymptotic dimension, which coarse maps play nicely with. If you have a coarse map $$X \to Y$$ then $$\operatorname{asdim} X \leq \operatorname{asdim} Y$$. The same is not true at all for bornologous maps. A particularly important theorem along these lines is that if a finitely generated group has finite asymptotic dimension then it satisfies the Novikov conjeture(a big open problem in topology having to do with whether or not something is invariant under homotopy).

In geometry one is often interested in boundaries and/or compactifications of a space $$X$$, and particularly how these interact with maps. Now under some constraints(see Lectures on Coarse Geometry by Roe) you can get coarse maps $$X \to Y$$ extend to maps on the boundary. I am not to familiar with Higson corona and Higson compactification but these are in some sense universal boundaries which you would be interested in. Certainly bornologous maps do not in general extend to maps on boundaries.

I am not familiar with the theory but there is a theory of coarse cohomology, and this is a (contravariant)functor on coarse maps.

I haven't really worked in the generality of coarse structure(not necessarily just looking at metric spaces) but another reason is it has to do with that definition, which intuitively is where you define all your negligible sets, and for metric spaces it is natural that negligible sets are the bounded ones. Just like in topology, or uniform continuity, preimage of open is open and it is natural to say preimage of negligible is negligible. From this perspective you might even consider bornology to be the added on condition in order to add uniformity.

What is added by having projections coarse maps or products in the coarse category? You can still have metric products, but I don't really see what is of interest in making sure that projections are coarse maps.

Coarse maps is not the end all be all of coarse geometry though, there are interesting maps which are not proper, and geometric information can be pulled from them. For example to study mapping class groups of surfaces Masur and Minsky developed a theory of "hierarchies" where geometry of the mapping class group is studied by looking at (not proper) projections to something called curve graphs of subsurfaces.

• Or you can instead talk about the bornology category
– user29123
Apr 19, 2019 at 3:33
• If projections were coarse, $(X\times Y,\pi_X ,\pi_Y )$ would be a product in coarse category. Currently it isn't. I want coarse category to have products.
– mz71
Apr 19, 2019 at 9:17
• @mzg147 Is there a reason you want products other than you just want them? Not all categories have products(fields don't have products for example). If you want to just work with bornology do that but if you are interested in geometry/coarse geometry bornology collapse and don't necessarily have much to do with geometry. There are lots of bornologous maps so you won't be able to make statements like there is no coarse map from hyperbolic space to hyperbolic plane. Also the coarse category is more general than just metric spaces.
– user29123
Apr 19, 2019 at 12:56
• I just say then there must be a good reason why we take this definition. I understand why fields are what they are. But I don't know why the snake map that you mentioned in your answer shouldn't be coarse. Why bornology don't have to do with geometry but coarseness do?
– mz71
Apr 19, 2019 at 15:15
• Could you also explain what properness give us when studying asymptotic dimension?
– mz71
Apr 19, 2019 at 15:17