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Stream: learning: questions

Topic: Isomorphisms and superfluous detail

David Egolf (Dec 30 2020 at 00:57):

I'm currently reading "Seven Sketches in Compositionality", where it is noted that if two objects are isomorphic in a category, then they are essentially the same while working in that category. The authors make an analogy to moving a chair between different rooms: the lighting might change, but we have the same chair.

Is there any way to process an isomorphism class of objects to remove variation with superfluous details? To continue the analogy, it seems like it might be easier to study the chair if I didn't have to worry about changing lighting. For example, say I want to study vector spaces and maps between them. Then there are many isomorphic vector spaces, that look different based on our choices for the basis elements. Is there any way to move to a setting where there is only one vector space of a given dimension?

John Baez (Dec 30 2020 at 02:18):

One way is to choose a skeleton for your category. This is an equivalent category in which isomorphic objects are equal - so there's just one object in each isomorphism class!

John Baez (Dec 30 2020 at 02:20):

However, the desire to use a skeleton is often a sign that one hasn't fully processed the lessons of category theory. (One lesson is that equivalent categories are equally good, so the skeleton is no better than the original.)

It could be good to use a skeleton if for some reason you have to pay for each object.

John Baez (Dec 30 2020 at 02:24):

Note: the lesson that equivalent categories are equally good is just one level up from the lesson that isomorphic objects are equally good.

David Egolf (Dec 30 2020 at 03:51):

The idea that it can be okay to have a range of isomorphic objects (or equivalent categories) is counterintuitive to me. On first thought, it seems like a problem arising from failing to specify an object of study precisely enough. However, it does make sense that we will care about objects that contain structure of interest plus additional structure. So, we will probably be forced to have to a class of isomorphic objects.

If I understand correctly, the skeleton category corresponds to the "least adorned" representative of all the categories equivalent to a category of interest. In the chair analogy, this might correspond to the simplest/plainest chair. It makes sense that this plain chair is exactly as good at being a chair as any other chair.

Along similar lines, I was wondering why it is considered okay to have horizontal composition being associative only up to isomorphism in pseudo double categories. I see on nLab that there is an equivalent strictly associative double category for every pseudo double category. (Maybe this is analogous to the skeleton category situation above?). However, I suppose that if these categories are equivalent, then the lack of associativity doesn't really matter.

Dan Doel (Dec 30 2020 at 05:23):

The analogy may not be ideal. Another analogy might be that there are two chairs made precisely enough that they appear the same as far as any observations you can make are concerned. They aren't actually the same chair, but you can study one and be confident that you have knowledge of the other. In some ways this sort of situation actually happens in reality all the time. One stick of butter can be substituted with another of the same sort and brand, even though there is more than one stick of butter in existence.

In the case of sets, you might have $\{1,2\}$ and $\{3,4\}$ which are not, according to ZF, say, actually the same set, but there are isomorphisms between them. If you want to describe some construction on $\{1,2\}$ in detail, you might need to talk about the particular elements. But as long as the construction is sufficiently categorical, the isomorphism could be used to give the details of an equally good construction on $\{3,4\}$. But, the details would actually depend on the specific isomorphism used, even though they are the details of the same sort of abstract structure.

Another consequence of this is that given the details of a categorical structure on $\{1,2\}$, you can get different details of an equally good structure of the same sort via the isomorphism that swaps $1$ and $2$. So in that case even though the set involved in the isomorphism is 'actually the same' on both sides, the choice of isomorphism matters. I think this applies to the vector space example, too. I don't think vector spaces have chosen bases in $\mathsf{Vect}$. So although you might describe a categorical construction on a vector space using a basis you happen to know for it, there could be a non-trivial automorphism on the vector space that would give an equally good description using a different (but isomorphic) basis.

John Baez (Dec 30 2020 at 06:13):

David Egolf said:

The idea that it can be okay to have a range of isomorphic objects (or equivalent categories) is counterintuitive to me.

Yes, one of the big realizations in math is that in most categories of interest there are lots of interesting objects. For example, there are lots of 2-dimensional real vector spaces, not just the $\mathbb{R}^3$ we first meet in school. Any plane tangent to a sphere in 3 dimensions is a 2-dimensional vector space, but they're not all equal to $\mathbb{R}^3$, just isomorphic.

This realization came before category theory and it's one of the things category was invented to deal with.

Along similar lines, I was wondering why it is considered okay to have horizontal composition being associative only up to isomorphism in pseudo double categories.

The reason is that this is how it works in many interesting examples. We need concepts that describe the situations we actually meet in practice. In the original work on double categories, composition was required to be strictly associative in both directions, but this leave out many examples of interest - including most of the ones I work on in applications.

David Egolf (Dec 30 2020 at 06:17):

Thanks for clarifying, Dan. Your emphasis on observation in the analogy got me thinking about what "observation" is in a categorical sense. (This is extra interesting to me because I am interested in imaging systems!)

A reflection on "observation" in a categorical setting:
Important to both chair analogies is what "observations" can be made of the chairs. If I have a magnifying glass, I might not say that the two chairs are the same, when otherwise I might. Trying to connect to a categorical setting, I think an observation is like a functor. The collection of functors from a category correspond to all the observations we can make of it. Observing without a magnifying glass corresponds to some subclass of the observations we can make, maybe the functors of the form $F \circ \mathrm{Blur}$, where $\mathrm{Blur}$ is a functor to some intermediate category that hides microscopic detail. It is as if the underlying category $C$ is not accessible to us, but only $\mathrm{Blur}(C)$. We then assess "blurry" equivalence of two categories $C$ and $D$ in terms of the equivalence of $\mathrm{Blur}(C)$ and $\mathrm{Blur}(B)$. For example, two groups viewed as categories on single objects may not be isomorphic, but if "blurring" sends each group to its underlying discrete category with a single object, then the resulting "blurred" categories are equivalent. Similarly, two categories that we call equivalent may not induce equivalent categories when we add a bit more structure to them.

John Baez (Dec 30 2020 at 06:25):

The idea of "blurring" is to some extent captured by the concept of a 'forgetful functor'. Interestingly, there's no mathematical definition of a forgetful functor; it's more of an attitude we can have toward a functor: namely, that it forgets details of some sort.

David Egolf (Dec 30 2020 at 06:26):

Dan Doel said:

Another consequence of this is that given the details of a categorical structure on $\{1,2\}$, you can get different details of an equally good structure of the same sort via the isomorphism that swaps $1$ and $2$. So in that case even though the set involved in the isomorphism is 'actually the same' on both sides, the choice of isomorphism matters.

This is intriguing to me, but I don't quite follow you. I think you are saying that applying two different isomorphisms to an object can yield objects that expose different details of the same structure. I would have thought that all these objects would be isomorphic, and so have all the same structure. A simple example might help me understand.

Dan Doel (Dec 30 2020 at 06:45):

I don't think that's what I meant.

An example might be this. $\{3,4\}$ is a coproduct $1 + 1$, where $1$ is some designated singleton set, say $\{\bullet\}$. That means there are functions $ι_1, ι_2 : 1 → \{3,4\}$, and given functions $f, g : 1 → Z$, there is a unique function $[f,g] : \{3,4\} → Z$, such that $[f,g] \circ ι_1 = f$ and $[f,g] \circ ι_2 = g$

Now, you can describe particular details for all these things.

$$ι_1(\bullet) = 3 \\ ι_2(\bullet) = 4 \\ [f,g](3) = f(\bullet) \\ [f,g](4) = g(\bullet)$$

But, there is an isomorphism $h : \{3,4\} → \{3,4\}$

$$h(3) = 4 ; h(4) = 3$$

and it's possible to use that isomorphism to get a different coproduct structure on $\{3,4\}$:

$$ι_1(\bullet) = 4 \\ ι_2(\bullet) = 3 \\ [f,g](4) = f(\bullet) \\ [f,g](3) = g(\bullet)$$

And for any other construction that needed only the fact that $\{3,4\}$ is a coproduct $1 + 1$ in some way, each of these would be equally good, just differing on particular details that don't matter for the overall structure.

David Egolf (Dec 30 2020 at 06:48):

Okay, I misunderstood you. I think you are just pointing out that isomorphic objects can have different "particular details". Thanks for the clarification!

Dan Doel (Dec 30 2020 at 06:50):

Not just that, but a single object can be isomorphic to itself in a non-trivial way, so that it can be an example of a structure in two ways that differ on the exact details.

Dan Doel (Dec 30 2020 at 06:50):

But it doesn't really matter which details you choose.

Dan Doel (Dec 30 2020 at 06:51):

You can technically plug in a usb-c cable in two ways, but it doesn't matter which.

David Egolf (Dec 30 2020 at 06:53):

A related question... Does this mean that if I need a map from an object defined by a universal property, I can choose any representative as the source of the map from the resulting isomorphism class of objects? (There are some constructions on nLab that specify maps from objects like these.)

David Egolf (Dec 30 2020 at 06:55):

For example, to define a map from the product of two sets, it should be okay (I think!) to define this in terms of what the map does on a particular product.

Dan Doel (Dec 30 2020 at 06:56):

Yes, I think so.

Dan Doel (Dec 30 2020 at 07:13):

Incidentally, set theory kind of works this way, too, if you think about it. There are potentially many ways of specifying a set with the same elements, and the only thing that tells you that those different ways of specifying them actually give you 'the same' set is an axiom, which works because nothing in set theory can tell apart two sets with exactly the same elements. So even if there were two distinct sets with the same elements, they could be freely substituted for one another as far as set theory is concerned.

Eric Forgy (Dec 30 2020 at 07:20):

Maybe unsurprisingly relevant (it seems to resonate at least), I was just (re)reading Section 3 Natural Numbers and Finite Sets of From Finite Sets to Feynman Diagrams and it gives some intuition about universal properties and how it doesn't matter what choice you make.

Eric Forgy (Dec 30 2020 at 07:26):

The wonderful thing about this definition is that now, if you have your disjoint union and I have mine, we automatically get maps going both ways between the two, and a little fiddling shows that these maps are inverses. (If you have never seen this done, you should do it yourself right now.) Thus, while there is not a unique disjoint union of sets, any two disjoint unions are canonically isomorphic. Experience has shown that knowing a set up to canonical isomorphism is just as good as knowing it exactly. For this reason, we may actually speak of ‘the’ disjoint union of sets, so long as we bear in mind that we are using the word ‘the’ in a generalized sense here.

Eric Forgy (Dec 30 2020 at 07:29):

(I think that reasoning applies to any universal property, not just disjoint union.)

John Baez (Dec 30 2020 at 07:30):

Yes.

Fawzi Hreiki (Dec 30 2020 at 07:40):

If you work in a meta-theory without equality on objects, you literally cannot say anything which is not isomorphism invariant (at least when working within one category)

Fawzi Hreiki (Dec 30 2020 at 07:40):

E.g. Makkai’s FOLDS