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Stream: learning: questions

Topic: Should we be surprised that homology exists?

fosco (Sep 23 2024 at 20:11):

These days I am refreshing my algebraic topology a bit for a little side-project, and I finally found the courage to come out with a huge gap in my understanding of the fundamentals.

On one side, there is an "axiomatic" approach to homology theory, cf. for example chapter 13 of May's concise:

image.png

(the axioms go on, but you certainly know them). This is neat, axiomatic, conceptual, we understand it very well since Eilenberg and Steenrod.

In computing homology, one takes a triangulation / subdivision / filtration of sorts that a space $X$ has, and uses the combinatorics of how "simplices" are arranged. Cohomology dualizes this picture: instead of taking a subdivision of $X$ in parts that are easy to understand, one looks at spaces from test spaces that are easy to understand, defines the singular complex, etc etc.

But then, cohomology can be developed axiomatically as well, and it exhibits phenomena that homology doesn't have (for example, the cohomology of a space is a graded ring). The reason why this happens is somewhat clear to me, together with the fact that cohomology being the exact dual of homology is true at first approximation, but is a statement that should be handled with some care in order not to form false beliefs.

At some point of my studies though, I encountered the following sentence (source: Strom's "Modern Classical Homotopy Theory"):

image.png

All of a sudden, cohomology is a natural object (I see why: because the "cohomology" $H^n(-,G)$ is just a fancy word for "homotopy classes of maps into a $K(G,n)$), while homology is sus and I should be surprised that such a monstrosity even exists?!

fosco (Sep 23 2024 at 20:14):

If Strom isn't just trying to be thought-provoking, i.e. if there is a substantial reason why I should be surprised that homology exists, I would love to know why.

Also, in what sense it is counterintuitive that homology exists? In what sense is it surprising? Surprising as in "there are exotic manifold structures on R^4" or more like "there are fractals"?

Madeleine Birchfield (Sep 23 2024 at 20:15):

If the axiom of choice fails, then there are sets which have non-trivial cohomology, and the cohomology measures the failure of the axiom of choice:

https://golem.ph.utexas.edu/category/2013/07/cohomology_detects_failures_of.html

fosco (Sep 23 2024 at 20:19):

Yes, I know that café post, a pleasant reading. I don't see how it connects to my question

Question which is more or less: "although homology and cohomology are not exactly dual to each other, it doesn't seem surprising to me that both exist, because to a large extent they are dual to each other, not only in the definition, but in their construction: am I in some way biased, and I should instead think that homology is a miracle?"

fosco (Sep 23 2024 at 20:20):

"And if yes, what the hell is Strom trying to make me think exactly?"

Madeleine Birchfield (Sep 23 2024 at 20:23):

My bad, I misread your question.

Mike Shulman (Sep 23 2024 at 20:28):

I see Strom's point, but I think a slight shift in perspective can make homology seem almost as natural as cohomology even from his point of view. The spaces $K(G,n)$ form a spectrum $HG$, and the homotopy classes of maps $X \to K(G,n)$ for varying $n$ can equivalently be defined as the homotopy groups of the mapping spectrum $\mathrm{Map}(X, HG)$. This is the internal-hom in the category of spectra, and internal-homs dualize to tensor products. The tensor product of spectra is called the smash product $\wedge$, and homology can be defined as the homotopy groups of the smash product spectrum $X\wedge HG$.

Madeleine Birchfield (Sep 23 2024 at 20:30):

Mike Shulman said:

homology seem almost as natural as homology

The second homology should be cohomology.

Mike Shulman (Sep 23 2024 at 20:32):

Fixed, thanks.

fosco (Sep 23 2024 at 20:44):

Mike Shulman said:

I see Strom's point, but I think a slight shift in perspective can make homology seem almost as natural as cohomology even from his point of view. The spaces $K(G,n)$ form a spectrum $HG$, and the homotopy classes of maps $X \to K(G,n)$ for varying $n$ can equivalently be defined as the homotopy groups of the mapping spectrum $\mathrm{Map}(X, HG)$. This is the internal-hom in the category of spectra, and internal-homs dualize to tensor products. The tensor product of spectra is called the smash product $\wedge$, and homology can be defined as the homotopy groups of the smash product spectrum $X\wedge HG$.

This makes Strom point even less clear to me. I'm sure he knows about smash product of spectra, and yet it seems like he wants to convey the idea that just one between Homology and cohomology is "a natural thing to do". Not to mention I could uno reverse his paragraph saying that we introduced homology in a very concrete combinatorial way, and it turns out to be a family of functors governed by an axiomatic theory. But now we're about to meet cohomology, richer in structure and the functors are even representable !

fosco (Sep 23 2024 at 20:45):

You should be suspicious about the existence of more structure and interesting new phenomena arising from a mere formal dualization process!

fosco (Sep 23 2024 at 20:46):

So, what's going on?

Mike Shulman (Sep 23 2024 at 20:50):

Sorry, I don't think I can guess what was in Strom's head. I agree with you, it's an odd comment. The only thing I can think of is that before one thinks of spectra, representable functors are already a simple and natural thing to consider categorically. Indeed, arguably, cohomology is what leads us to think about spectra in the first place.

John Baez (Sep 23 2024 at 20:54):

Another thing worth pointing out is that homology (at least with coefficients in a field) is naturally a graded co-algebra, where the comultiplication and counit arise by applying the symmetric monoidal functor $H_\ast$ to the maps $X \to X \times X$ and $X \to 1$. This is beautiful stuff: as you know, any object in a cartesian category like $\mathsf{Top}$ is a cocommutative comonoid, and then we're just applying a symmetric monoidal functor $H_\ast$ from $\mathsf{Top}$ to graded vector spaces, and getting a graded coalgebra.

To some extent elementary classes on algebraic topology favor cohomology because people traditionally prefer algebras to coalgebras! To get an algebra out of the comonoid structure that any topological space is born with, we need a contravariant functor.

John Baez (Sep 23 2024 at 20:58):

Here I'm working with coefficients in a field because the Künneth theorem is simpler then: $H_\ast$ is then a strong monoidal functor. But someone who has been thinking about this more recently will know the story over the integers.

Kevin Carlson (Sep 23 2024 at 21:01):

I think the sense of Strom's point depends on what you're working from. But for instance, I could imagine saying that I've gotten some core intuitions and many examples around cohomology just by proving Brown's representability theorem, which has basically no information about algebraic topology in it and requires no inkling about stabilization except the focus on pointed connected spaces. Then if I want whole cohomology theories I just have to find out something about loop spaces. To actually define and construct any homology theories I need to either do the classical thing of singular or simplicial homology, which I need quite a lot of intuition about topology to think to do, or I need to somehow think up basically the entire idea of spectra and their smash product and their homotopy groups (no way am I going to get anywhere with Adams representability starting from zero, I don't think.) That's a lot more.

Mike Shulman (Sep 23 2024 at 21:02):

Along the same lines, we were able to define cohomology in HoTT long before homology.

fosco (Sep 23 2024 at 21:04):

Mike Shulman said:

Along the same lines, we were able to define cohomology in HoTT long before homology.

Ah, interesting. Why? What made it easier to do?

Mike Shulman (Sep 23 2024 at 21:06):

Exactly what we're talking about. Hom-spaces are right there, and homotopy groups are easy, so as soon as we had Eilenberg-Mac Lane spaces we were off with the representable definition of cohomology. But the traditional "hands-on" constructions of homology rely on point-set-level fiddliness that isn't available in HoTT, and smash products of spectra are difficult because you have to work with infinitely many coherent data.

fosco (Sep 24 2024 at 06:20):

Kevin Carlson said:

I think the sense of Strom's point depends on what you're working from. But for instance, I could imagine saying that I've gotten some core intuitions and many examples around cohomology just by proving Brown's representability theorem, which has basically no information about algebraic topology in it and requires no inkling about stabilization except the focus on pointed connected spaces. Then if I want whole cohomology theories I just have to find out something about loop spaces. To actually define and construct any homology theories I need to either do the classical thing of singular or simplicial homology, which I need quite a lot of intuition about topology to think to do, or I need to somehow think up basically the entire idea of spectra and their smash product and their homotopy groups (no way am I going to get anywhere with Adams representability starting from zero, I don't think.) That's a lot more.

I like this answer because I understand what you mean, and I feel it's what Strom might have had in mind, but I find myself very confused by this point of view. It seems to me that you're not really avoiding topology, more than you're just hiding it in another place.

fosco (Sep 24 2024 at 06:25):

Let me explain: right after having defined cohomology functors as $[-,K(G,n)]$ Strom proposes this exercise

image.png

claiming that it can be answered straight from the definition (plus the list of Eilenberg-Steenrod axioms). Although I believe it can be done, I feel like a proper answer can't exist without some (subtle and less subtle) observations on the structure of $[X,K(G,n)]$ that can only come from topology, i.e. from having realized $CP^n, CP^n\land HP^m$, etc. as CW complexes, and studying how cellular maps behave in terms of connectivity and other properties.

fosco (Sep 24 2024 at 06:27):

I'd be happy to see one of these exercises worked out with little to no topology.

David Michael Roberts (Sep 24 2024 at 06:29):

I think that starting from the representable view of cohomology, and then defining homology to be the homotopy groups of smashing with the appropriate spectrum, why this has anything to do with the usual naive singular/simplicial/cellular homology is indeed a bit mysterious to me.

Mike Shulman (Sep 24 2024 at 06:30):

@David Michael Roberts Are you saying that that's more mysterious than why the representable view of cohomology has anything to do with singular/simplicial/cellular cohomology? Or just equally mysterious?

Mike Shulman (Sep 24 2024 at 06:35):

I don't think the connection to cellular (co)homology is all that mysterious, since a cell complex is really just a way of building a space out of (homotopy) colimits, and both functors $[-,HG]$ and $- \wedge HG$ on spectra preserve colimits, so "of course" homology and cohomology of a cell complex can be decomposed along the cell structure.

David Michael Roberts (Sep 24 2024 at 06:36):

I think that homology as defined in a way that would have been recognisable by people 100 years ago feels reasonably natural, but starting from the point of view that cohomology is maps to a spectrum, it's very very different. Certainly the link between traditional cohomology and the representable view is highly nontrivial, too.

But my intuition is very much not like that of a homotopy theorist!

David Michael Roberts (Sep 24 2024 at 06:36):

(messages overlapped, but I can't spend more time thinking about my vague guess any more, sorry!)

fosco (Sep 24 2024 at 07:20):

$[-,HG]$ preserves limits, not colimits -or does it?

David Michael Roberts (Sep 24 2024 at 07:25):

well, preservation of colimits at the infty-category level, right? But then as a functor out of an opposite category, one might be fussing over what is preserved under contravariant functors

fosco (Sep 24 2024 at 07:26):

A contravariant hom(-,X) sends colimits (i.e. limits in the domain) to limits (i.e. limits in the codomain)

fosco (Sep 24 2024 at 07:28):

now, for hom(X,-) to preserve colimits, X has to be a retract of a representable; for hom(-,X) to preserve colimits (i.e. send limits to colimits), X has to be...?

Mike Shulman (Sep 24 2024 at 08:02):

$[-,HG] : \rm Top \to \rm Top^{\rm op}$ preserves colimits!

Mike Shulman (Sep 24 2024 at 08:04):

When talking about a contravariant functor, I'd prefer to be able to say that it "preserves colimits" and mean that it takes colimits in its domain to limits in its codomain, and dually. It seems more natural to me to use the word that we normally use for the thing in the domain being preserved, rather than turning it around.

Mike Shulman (Sep 24 2024 at 08:05):

The conventional representation of a contravariant functor from $C$ to $D$ as a covariant functor from $C^{\rm op}$ to $D$ is arbitrary, after all -- we could just as well represent it as a covariant functor from $C$ to $D^{\rm op}$.

Daniel Gratzer (Sep 24 2024 at 15:40):

Here is a story that makes homology seem quite natural/inevitable to me, at least if one is willing to work with $\infty$-categories and perhaps at the expense of the original geometric motivations.

To start with, we regard homology as just one functor from spaces to the derived $\infty$-category $\mathcal{D}(\mathbb{Z})$ (chain complexes up to quasi-equivalence). This bundles up all of the $H_i$ s into a single functor $H : \mathcal{S} \to \mathcal{D}(\mathbb{Z})$ and one can post-compose with the homology group functors $\mathcal{D}(\mathbb{Z}) \to \mathrm{Ab}$ to obtain the original family of functors.

We can translate the Eilenberg-Steenrod axioms to this setting.
In particular, colimits in $\mathcal{D}(\mathbb{Z})$ induce long exact sequences of homology groups, one can replace the various axioms concerning long-exact sequences and additivity with the single property that $H$ sends colimits to colimits. The axiom that homotopic maps of spaces induce equal maps on homology groups is an automatic consequence of working with things $\infty$-categorically. Finally, the dimension axiom tells us that $H(\ast) = \mathbb{Z}$.

Any cocontinuous functor out of $\mathcal{S}$, however, is determined by its value on the point and so we can replace the entire set of axioms with just "homology is the unique cocontinuous functor $\mathcal{S} \to \mathcal{D}(\mathbb{Z})$ sending $\ast$ to $\mathbb{Z}$". This is also a perfectly valid (and, to my mind, somewhat elegant) definition of homology. This same argument will work just fine for cohomology, provided one inserts some $\mathrm{op}$ s where needed.

NB: Mike's definition using spectra and $H\mathbb{Z}$ lurks within this construction: one can further factor $\mathcal{S} \to \mathcal{D}(\mathbb{Z})$ through the functor $\mathrm{Sp} \to \mathcal{D}(\mathbb{Z})$ given by smashing with $H\mathbb{Z}$

Mike Shulman (Sep 24 2024 at 16:05):

That's a very nice story! More generally, the point is that since the $\infty$-category $\mathcal{S}$ of spaces is the free cocompletion of a point, any object of any cocontinuous $\infty$-category uniquely determines an "invariant of spaces" valued in that $\infty$-category.

Fancier versions of this get more press, like the fact that any commutative Frobenius monoid in any category determines an invariant of 2-dimensional manifolds (a "TQFT"), or the similar statement about fully dualizable objects and the cobordism hypothesis. But in general whenever you have a category that's freely generated by something-or-other, any particular something-or-other somewhere else determines an invariant of objects of that category.

Mike Shulman (Sep 24 2024 at 16:06):

And it's also true in 1 dimension that any object of any cocomplete 1-category determines an "invariant of sets" valued in that category. But of course that's not usually considered very interesting...

Mike Shulman (Sep 24 2024 at 16:07):

And this is getting a bit digressive, but anyone who doesn't find it sufficiently well-motivated to pass to $\infty$-categories from the get-go can make the same statement about derivators instead.

Kevin Carlson (Sep 24 2024 at 17:40):

That's a great story for ordinary homology although I think it's still a bit tough for extraordinary homology theories; basically you've got to somehow come up with spectra as the universal stable $\infty$-category with a cocontinuous functor out of spaces first, and only then you can be like, alright, every spectrum then induces a new cocontinuous functor by the universal property and get off to the races.

Mike Shulman (Sep 24 2024 at 17:40):

I think you only need to know that spectra is a cocomplete $\infty$-category; its universality or stability doesn't matter.

Kevin Carlson (Sep 24 2024 at 17:42):

fosco said:

i.e. from having realized $CP^n, CP^n\land HP^m$, etc. as CW complexes, and studying how cellular maps behave in terms of connectivity and other properties.

Well, wait, surely Strom has given $CP^n$ etc as cell complexes and is expecting you to use those constructions with the Eilenberg-Steenrod axioms?

Mike Shulman (Sep 24 2024 at 17:45):

That was my assumption too, but I wasn't as sure what he expected you to do with $\Omega S^{n+1}$.

fosco (Sep 24 2024 at 17:48):

Kevin Carlson said:

fosco said:

i.e. from having realized $CP^n, CP^n\land HP^m$, etc. as CW complexes, and studying how cellular maps behave in terms of connectivity and other properties.

Well, wait, surely Strom has given $CP^n$ etc as cell complexes and is expecting you to use those constructions with the Eilenberg-Steenrod axioms?

Yes, CW complexes are introduced very early on; I'd be happy to see how one should reason!

Kevin Carlson (Sep 24 2024 at 21:11):

Oh, so, it's probably important that he's not asking about real projective space. Suppose I glue an $n+2$-dimensional cell to an $n$-dimensional complex $X$, producing $Y$ and that I already know the cohomologies of spheres and that $n$-dimensional complexes have no cohomology above dimension $n.$ Then the long exact sequence in cohomology looks like $0\to H^{n+2}(Y)\to H^{n+1}(S^{n+1})\to 0\to H^{n+1}(Y)\to 0\to H^n(X)\to H^n(Y)\to 0 \to \cdots.$ So you don't have to know how the degree of a map relates to the map on cohomology to calculate this one.

Daniel Gratzer (Sep 25 2024 at 12:48):

Kevin Carlson said:

That's a great story for ordinary homology although I think it's still a bit tough for extraordinary homology theories

Indeed, for ordinary (co)homology, we benefit from the fact that $H\mathbb{Z}$-modules are already familiar objects with the derived category. If one wanted to explain general theory one could try and pitch this in terms of reduced excisive functors. However, I guess the nice part of this simpler story is that it isolates that (co)homology is an invariant counting something about a space (colimit preservation) and that the flex is whether we choose to count using the multiplicative or additive structure of $\mathcal{D}(\mathbb{Z})$. And, returning to the original question, that both are equal natural at this level of abstraction