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Stream: theory: algebraic topology

Topic: formal properties of the bar construction

Patrick Nicodemus (Dec 24 2021 at 11:57):

There's an interesting discussion here with a few unresolved questions. https://nforum.ncatlab.org/discussion/6644/bar-construction-as-a-cofibrant-replacement/
I would like to resurface this topic and ask if anyone has good references to these questions.
The bar construction is omnipresent, I am looking for concrete references on its properties in modern language. Here is a more specific question. Let $\mathcal{C}$ be an arbitrary category equipped with a comonad $G$.
The bar construction associates to each object in $\mathcal{C}$ a simplicial $G$-coalgebra.

I would like to know if there is any model category structure on the category of simplicial objects in $\mathcal{C}$ which is somehow derived from $G$, or in which certain simplicial objects defined in terms of $G$ turn out to have good properties (for example the simplicial $G$-coalgebras are cofibrant.) In what sense can we say the bar construction is a cofibrant replacement?

As an example take $\mathcal{C}$ to be groups or abelian groups and $G$ the forgetful-free functor of the adjunction, in this case one replaces a given Abelian group by a free resolution; do the free simplicial Abelian groups play an important role with respect to the usual model structure on SAB? I think they are cofibrant but I don't know a source for this fact.

Patrick Nicodemus (Dec 24 2021 at 12:00):

Opening things up more generally, what are your favorite theoretical characterizations of the bar construction and its properties? In whatever formalism. What context do you think it's best to view it in?

Zhen Lin Low (Dec 24 2021 at 12:16):

I don't think it works in such great generality. Actually, I think it doesn't even always work for the comonad associated with the free–forgetful adjunction of a category of algebras for a finitary algebraic theory – there is a technical condition on the (morphisms that would be) trivial cofibrations, which is implied by the existence of fibrant replacements and fibrant path objects, but is not automatic.

Patrick Nicodemus (Jan 10 2022 at 02:57):

Say that we have a category $C$ equipped with a monad, $T$. Given any $T$ algebra $A$ we can form the "bar resolution" of $A$, which is a simplicial $T$ algebra. This is a resolution of $A$ in a sense described in this article. http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/bar+construction
When $T$ is of the form "left tensor with the monoid $M$", and $A$ is a module of the monoid, we might call this $B(M,M,A)$. And it's contractible in the underlying category (not contractible in the category of $M$ modules, so we might call it $EM$, evoking the space $EG$ for a group. I have been trying to get intuition for this construction and one place i'm starting to get lost is how we relate the interesting properties of $EM = B(M,M,1)$ to the interesting properties of $BM =B(1,M,1)$. They play very different roles. $EM$ is a kind of acyclic resolution, $BM$ may not in general be acyclic. What theoretical properties relate $EM$ and $BM$? Why does the bar construction matter here, what kind of interesting properties does $B(1,M,1)$ have?

I am looking for references which would help me to understand this.

Patrick Nicodemus (Jan 10 2022 at 02:58):

As @John Baez said the other day,

No, I'm just saying that there exist people who could figure this out 100 times faster than you, so talking to them - repeatedly, not just once - could greatly accelerate your project.

I am looking for people who think about the bar construction, about $EG$ and $BG$, about cofibrant replacement, who I can talk to repeatedly, not just once. :D

John Baez (Jan 10 2022 at 03:21):

I understand the bar construction reasonably well, Todd Trimble understands it better and has written about it here, I have a pretty good intuition $EG$ and $BG$, and I've taught a course about all this stuff, with notes online, but I don't enjoy the technicalities of model categories.

Mike Shulman (Jan 10 2022 at 05:27):

I generally think of $B(1,M,1)$ as a sort of "delooping" of $M$. This is more true when $M$ is a group, but some part of the intuition works in general. Depending on how you like to think, you can think of $B(1,M,1)$ as a "classifying space" of $M$ because it's a "quotient" of the acyclic $EM$ by the action of $M$. (I don't remember how true that is literally if $M$ is not a group.)

Mike Shulman (Jan 10 2022 at 05:31):

Regarding the general question of when the bar construction gives a cofibrant replacement, there are some newer general results along these lines in section 8 of my paper All (∞,1)-toposes have strict univalent universes.

Patrick Nicodemus (Jan 10 2022 at 19:37):

Thank you John, I've read the letter by Todd Trimble carefully in the past and found it very helpful. Maybe it would help me to ask broader questions so I can figure out where to start. A bar construction is a kind of resolution. What is a resolution, abstractly? I have heard it can be interpreted as a kind of co/fibrant replacement but I don't know where to start learning about that perspective, where can I start learning about this idea of co/fibrant replacement?

Patrick Nicodemus (Jan 10 2022 at 19:39):

Thanks Mike. I am looking at the paper now.

Patrick Nicodemus (Jan 10 2022 at 19:41):

Depending on how you like to think, you can think of B(1,M,1)B(1,M,1)B(1,M,1) as a "classifying space" of MMM because it's a "quotient" of the acyclic EMEMEM by the action of MMM. (I don't remember how true that is literally if MMM is not a group.)

Yes this seems very interesting to me, i'd love to learn more about what is known about this perspective. I have some old papers on classifying spaces by Steenrod and Milgram that I am reading at the moment

Patrick Nicodemus (Jan 10 2022 at 21:33):

by the way i am generally competent in homological algebra, i have studied the textbook by weibel, so all of this is not totally foreign to me. just trying to get a better abstract perspective!

John Baez (Jan 10 2022 at 21:44):

Patrick Nicodemus said:

Thank you John, I've read the letter by Todd Trimble carefully in the past and found it very helpful. Maybe it would help me to ask broader questions so I can figure out where to start. A bar construction is a kind of resolution. What is a resolution, abstractly? I have heard it can be interpreted as a kind of co/fibrant replacement but I don't know where to start learning about that perspective, where can I start learning about this idea of co/fibrant replacement?

I think this example is a great way to get the actual point of cofibrant replacement without drowning in the sea of model categories. (Model categories are good after you get the point, but before you get the point they are just confusing.)

John Baez (Jan 10 2022 at 21:44):

Cofibrant replacement is replacing an object by a similar object that's "less tightly wound", more "loose", so that it's easier to map out of it.

John Baez (Jan 10 2022 at 21:46):

Often cofibrant replacement involves getting rid of equations and replacing them by paths or edges or algebraic things that act like those things. This is what I mean by "less tightly wound".

John Baez (Jan 10 2022 at 21:47):

To really explain this I find it essential to look at examples.

John Baez (Jan 10 2022 at 21:48):

I don't know anyone who can explain this well except James Dolan, so I can't point you to anything to read: I just have to explain it myself.

John Baez (Jan 10 2022 at 21:51):

So: do you know the definition of a resolution of an $R$-module? And have you ever used these to do anything, like calculate an Ext group? I'm just asking because I'd like to go through an example of this sort, and I want to know where to start. It's fine if you don't know anything about this stuff, but if you do know it, I can try to explain what it "really means", by looking at an example or two.

Zhen Lin Low (Jan 10 2022 at 22:37):

I think the model category axioms capture quite well the abstract properties of resolutions. The catch, if you're coming from classical homological algebra, is that you have to first get used to thinking of chain complexes as homological objects in their own right, rather than just a calculational tool or whatever.

Patrick Nicodemus (Jan 10 2022 at 22:46):

yes! thank you, I do know what is meant by a free resolution of module and I can compute Ext and Tor groups in good cases, I've taken a course in homological algebra with lots of homework and I have some computational experience. I know the details of derived functors in the old sense of Cartan-Eilenberg and I have studied the book by Godement where he uses spectral sequences to establish equivalences between sheaf cohomology theories for paracompact Hausdorff spaces.

Patrick Nicodemus (Jan 10 2022 at 22:47):

my question is about organizing the examples I've accumulated into a categorical framework but more examples and data are always helpful. I know some about group cohomology but not much.

Zhen Lin Low (Jan 10 2022 at 22:49):

The bar construction B(G, C, F) can be considered to be a model of the homotopy colimit of F weighted by G. (Whether it actually works or not depends on technical details.) So B(M, M, 1) is supposed to be a model of the homotopy colimit of 1 (a constant diagram) weighted by M (a representable presheaf), therefore is supposed to be homotopy equivalent to 1, which is indeed the case. On the other hand, B(1, M, 1) is supposed to be a model of the homotopy colimit of 1 weighted by 1, i.e. the homotopy coinvariants of the trivial M action. B(C, C, F) and B(1, C, F) are the many-object versions of this.

Zhen Lin Low (Jan 10 2022 at 22:51):

Basically, in short, the bar construction is the ∞-categorical version of the construction of colimits by coproducts and coequalisers.

Patrick Nicodemus (Jan 10 2022 at 23:50):

Ok. Let's say I understand simplicial methods in homotopy theory at the level of a book like the one by May. Where can I get started with infinity categorical methods?

John Baez (Jan 10 2022 at 23:52):

Patrick Nicodemus said:

yes! thank you, I do know what is meant by a free resolution of module and I can compute Ext and Tor groups in good cases, I've taken a course in homological algebra with lots of homework and I have some computational experience. I know the details of derived functors in the old sense of Cartan-Eilenberg and I have studied the book by Godement where he uses spectral sequences to establish equivalences between sheaf cohomology theories for paracompact Hausdorff spaces.

Great! Thanks for letting me know your background. This is more than enough technical experience for us to do some fun things.

my question is about organizing the examples I've accumulated into a categorical framework but more examples and data are always helpful.

Okay, great. What I want to do is help you understand what it all "really means" - a lot of treatments make this mathematics seem quite technical and a bit dry, but in fact there's something very nice and conceptual about it. Fitting it into a categorical framework is part of what we need to do this, but also we need to talk about things like whether objects are "tightly wound" or not.

John Baez (Jan 10 2022 at 23:55):

Let me think of a good example to get started with.

John Baez (Jan 11 2022 at 00:11):

Okay. This will take a while, but I think it's fun. We'll start with some warmup material from the textbook approach to resolutions, and then we'll try to think more about what it "really means". I may actually get confused at certain points, but I hope that's fun too - that's called "research".

Let's take the abelian group $\mathbb{Z}/2$. Do you know a nice free resolution of this, thinking of it as an object in $\mathsf{AbGp}$, or $\mathbb{Z}\mathrm{Mod}$ if you prefer?

John Baez (Jan 11 2022 at 00:28):

(It'll probably work best if only Patrick answers these questions - and also if Patrick always feels completely free to say "I don't know".)

Patrick Nicodemus (Jan 11 2022 at 00:44):

How lucky I am to have such experts around and willing to spend time discussing this! I appreciate your time.

I'll be comfortable to admit I don't know things, I hope. I've always preferred to work with the more classical stuff for intuition so my knowledge is pretty solid for like, the old stuff people were working on back in the 60s and then drops like a rock for the modern categorical formalisms like model categories, infinity categories and so on. It might be easiest to just assume I understand things before 1970 and nothing after.

But in answer to your question, I'd use the short exact sequence $0\to \mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/2\to 0$ to resolve this, where the map $\mathbb{Z}\to\mathbb{Z}$ is multiplication by two. Lucky us in the Abelian group case we can always do it in two steps, but the higher syzygies for modules not over a PID are a bit mysterious to me in terms of what we are trying to capture.

Zhen Lin Low (Jan 11 2022 at 09:19):

Hmmm. Triangulated categories, which are transitional between classical Cartan–Eilenberg homological algebra and (relatively more) modern model categories, are actually from the 1960s too. Would you say you understand those too? (Half-serious question)

John Baez (Jan 11 2022 at 18:39):

Patrick Nicodemus said:

I've always preferred to work with the more classical stuff for intuition so my knowledge is pretty solid for like, the old stuff people were working on back in the 60s and then drops like a rock for the modern categorical formalisms like model categories, infinity categories and so on. It might be easiest to just assume I understand things before 1970 and nothing after.

Okay, that's fine. It's actually a lot better, for what I'm planning, that you know the classical stuff. You may know it better than me, which would be great.

But in answer to your question, I'd use the short exact sequence $0\to \mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/2\to 0$ to resolve this, where the map $\mathbb{Z}\to\mathbb{Z}$ is multiplication by two. Lucky us in the Abelian group case we can always do it in two steps, but the higher syzygies for modules not over a PID are a bit mysterious to me in terms of what we are trying to capture.

Right, I wanted to start with an example where the resolution is nice and short. Later on, once we know what we're doing, we could try a longer one.

So yes:

$0\to \mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/2\to 0$

is our friend here!

Right now I want to focus on the top part, and think of it as a chain complex that goes on up forever:

$\cdots \to 0 \to 0 \to \mathbb{Z} \stackrel{2}{\to} \mathbb{Z}$

Do you know how this chain complex is related to the chain complex built directly from our original abelian group:

$\cdots \to 0 \to 0 \to 0 \to \mathbb{Z}/2$ ?

Right now I'm looking for any answer that comes to mind, maybe an answer with a classical homological algebra flavor to it. Then I'll talk about some other ways of thinking about this question.

Mike Shulman (Jan 12 2022 at 16:18):

Zhen Lin Low said:

The bar construction B(G, C, F) can be considered to be a model of the homotopy colimit of F weighted by G. (Whether it actually works or not depends on technical details.)

I tried to explain this perspective in my old paper Homotopy limits and colimits and enriched homotopy theory. Section 10, in particular, is about what the bar construction has to do with "homotopy colimits" in an intuitive sense.

Sorry to just pop in and drop links, unfortunately I don't have time to participate more deeply right now...

Patrick Nicodemus (Jan 12 2022 at 16:19):

Zhen Lin Low said:

Hmmm. Triangulated categories, which are transitional between classical Cartan–Eilenberg homological algebra and (relatively more) modern model categories, are actually from the 1960s too. Would you say you understand those too? (Half-serious question)

Uh no not yet. I had a course that discussed them but i kinda bombed through that part of it. I have been wanting to learn more about poincare duality and it seems that triangulated categories were originally conceived as a way of giving a very general setting in which to explore poincare duality, so it's definitely something i'm interested in, I have a copy of Cohomology of Sheaves by Iversen which seems like it gets into this POV a bit, especially in the appendix. I tried to read Chapter 10 of Weibel but it wasn't well-motivated (speaking exclusively for my perspective)

Patrick Nicodemus (Jan 12 2022 at 16:35):

John Baez said:

Patrick Nicodemus said:

I've always preferred to work with the more classical stuff for intuition so my knowledge is pretty solid for like, the old stuff people were working on back in the 60s and then drops like a rock for the modern categorical formalisms like model categories, infinity categories and so on. It might be easiest to just assume I understand things before 1970 and nothing after.

Okay, that's fine. It's actually a lot better, for what I'm planning, that you know the classical stuff. You may know it better than me, which would be great.

But in answer to your question, I'd use the short exact sequence $0\to \mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/2\to 0$ to resolve this, where the map $\mathbb{Z}\to\mathbb{Z}$ is multiplication by two. Lucky us in the Abelian group case we can always do it in two steps, but the higher syzygies for modules not over a PID are a bit mysterious to me in terms of what we are trying to capture.

Right, I wanted to start with an example where the resolution is nice and short. Later on, once we know what we're doing, we could try a longer one.

So yes:

$0\to \mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/2\to 0$

is our friend here!

Right now I want to focus on the top part, and think of it as a chain complex that goes on up forever:

$\cdots \to 0 \to 0 \to \mathbb{Z} \stackrel{2}{\to} \mathbb{Z}$

Do you know how this chain complex is related to the chain complex built directly from our original abelian group:

$\cdots \to 0 \to 0 \to 0 \to \mathbb{Z}/2$ ?

Right now I'm looking for any answer that comes to mind, maybe an answer with a classical homological algebra flavor to it. Then I'll talk about some other ways of thinking about this question.

hmm in the category of chain complexes this should be quasi-isomorphic to the original by means of the augmentation map, but not homotopy equivalent.

I have been thinking about this question on my own recently, so my asking about this stuff is not totally ex nihilo, so let me say some of what i've been able to piece together. I read the first few pages of the book by Gabriel and Zisman on the calculus of fractions and homotopy theory and their abstract perspective was very helpful. From what they point out, it's natural to conclude that ordinary free resolution can be seen as a special case of a Cartan-Eilenberg resolution in the case where the chain complex consists of a single object. Working in the homotopy category of chain complexes concentrated in positive degree, a C-E resolution gives a left adjoint to the forgetful functor from the category of complexes whose objects are all injective, which is in turn equivalent to the category of fractions given by inverting the quasi-isomorphisms. This all is very cool and seems super crucial in understanding things. But to me that's just a beginning, I'm not completely satisfied there, I don't really know why we would want to identify quasi-isomorphic things, or replace an object with a chain complex, or why we would define derived functors. I mean honestly the origin of chain complexes itself is still perplexing. This is not to ask anyone to answer these questions immediately, just to give an idea of what I'm thinking, let us continue on the current conversational path.

Patrick Nicodemus (Jan 12 2022 at 16:42):

By the way, my background is in logic and I have some knowledge of homotopy type theory. Certainly it would take us very far afield to start talking about this right now but I just want to bring it up to share what is on my mind. It seems that homological algebra and homotopy theory have sort of discovered a new notion of equality or identification which is more subtle than the standard one, instead of saying "x = y" we say "p is an identification between x and y". We do this in presenting a gadget by generators and relations (whenever we focus on the relations in their own right as mathematical objects instead of simply worrying about the quotient object as a 'truncated' entity), we do this in homotopy theory with homotopy colimits (which i still only have the barest intuitive feeling for), we do it in Martin-Lof type theory with the inhabitants of the equality type and so on. I am in some sense trying to understand what is going on with this notion of equality, like, what are the kind of problems which require us to deal with identifications as objects in their own right, what does this way of thinking get us. Somehow I just don't understand how dealing with the $R$-module $M$ leads us to the higher syzygies between relations of generators in $M$ and how this comes back to bear on $M$ itself.

Patrick Nicodemus (Jan 12 2022 at 16:42):

I don't expect anyone to respond to everything I have written here.

Patrick Nicodemus (Jan 12 2022 at 16:46):

I guess I would just ask @John Baez, not to ignore this entirely, but to take it into account and continue with the examples and the "tightly/loosely wound" metaphor.

John Baez (Jan 12 2022 at 17:06):

It seems that homological algebra and homotopy theory have sort of discovered a new notion of equality or identification which is more subtle than the standard one, instead of saying "x = y" we say "p is an identification between x and y".

Yes, exactly.

John Baez (Jan 12 2022 at 17:07):

Somehow I just don't understand how dealing with the $R$-module $M$ leads us to the higher syzygies between relations of generators in $M$ and how this comes back to bear on $M$ itself.

John Baez (Jan 12 2022 at 17:08):

Yes, that's exactly what I want to get into, but I'd like to work my way toward it slowly with the help of some examples so we can understand it very concretely and vividly.

John Baez (Jan 12 2022 at 17:13):

So, I'd like to keep chewing away on this example:

Right now I want to focus on the top part, and think of it as a chain complex that goes on up forever:

$\cdots \to 0 \to 0 \to \mathbb{Z} \stackrel{2}{\to} \mathbb{Z}$

Do you know how this chain complex is related to the chain complex built directly from our original abelian group:

$\cdots \to 0 \to 0 \to 0 \to \mathbb{Z}/2$ ?

@Patrick Nicodemus replied:

hmm in the category of chain complexes this should be quasi-isomorphic to the original by means of the augmentation map, but not homotopy equivalent.

Okay, that's exactly right. In simple terms I think you're saying there's a map from

$C' = \big\{ \cdots \to 0 \to 0 \to \mathbb{Z} \stackrel{2}{\to} \mathbb{Z} \big\}$

to our original

$C = \big\{ \cdots \to 0 \to 0 \to 0 \to \mathbb{Z}/2 \big\}$

which induces isomorphisms on homology groups, but not one the other way around.

John Baez (Jan 12 2022 at 17:14):

The map from $C'$ to $C$ is the obvious one, namely the only one that maps the $\mathbb{Z}$ term to the $\mathbb{Z}/2$ term by modding out by 2.

John Baez (Jan 12 2022 at 17:15):

There's no map of chain complexes going the other way except for the zero map.

John Baez (Jan 12 2022 at 17:16):

So these two chain complexes are "the same in a way" (which we dub "quasi-isomorphic") yet also not isomorphic nor even equivalent (that is, homotopy equivalent).

Patrick Nicodemus (Jan 12 2022 at 17:18):

Yeah.

John Baez (Jan 12 2022 at 17:18):

I hope there's sort of an obvious vague intuitive sense in which $C'$ is like an "unrolled" or "loosened-up" version of $C$. But we want to gradually make that more and more precise, and also understand more and more why it pays to to "unroll" objects in this way, and how we can do it using the bar construction.

John Baez (Jan 12 2022 at 17:19):

Okay, so let's try something here.

John Baez (Jan 12 2022 at 17:19):

Since right now we're dealing with chain complexes (of abelian groups) that have at most 2 terms, let's just think about 2-term chain complexes for now.

John Baez (Jan 12 2022 at 17:20):

Later we can add more terms.

Patrick Nicodemus (Jan 12 2022 at 17:20):

Yeah, I mean free objects are easy to manipulate, easy to build maps out of. They're the simplest groups, you can do linear algebra with them. So a complex of free objects seems loose to me.

John Baez (Jan 12 2022 at 17:20):

So we've got

$C' = \big\{ \mathbb{Z} \stackrel{2}{\to} \mathbb{Z} \big\}$

and

$C = \big\{ 0 \to \mathbb{Z}/2 \big\}$

John Baez (Jan 12 2022 at 17:21):

Now, there's a nice way to turn a 2-term chain complex of abelian groups into a category, where the objects are the 0-chains.

John Baez (Jan 12 2022 at 17:21):

Can you guess how this trick works?

John Baez (Jan 12 2022 at 17:25):

I guess I'm mainly asking what the morphisms in this category should be.

Patrick Nicodemus (Jan 12 2022 at 17:29):

Well, by Dold-Kan a chain complex is equivalent to a simplicial Abelian group and categories can also be regarded as certain 2-truncated simplicial gadgets, so it seems plausible to me for sure. Using Dold-Kan as a guide here, the boundary map of $C'$, because it is a differential, should be interpreted as sending a $1$ cell to the difference of its $0$-cells, so I guess we could design a category where elements of $C'_1$ are morphisms, elements of $C'_0$ are objects, and for two cells $x,y$ in $C'_0$ you would have a morphism from $x$ to $y$ for each element in $C'_1$ which lives in the fiber over $y-x$. I guess the identity morphism would be given by zero and composition would be given by addition. This makes sense with the connection to topology, I think, I mean if $H_1$ is the Abelianized version of $\pi_1$ then this is like an abelianized version of the fundamental groupoid. Am I on the right track here?

Patrick Nicodemus (Jan 12 2022 at 17:32):

This doesn't work exactly because the same element of $C_1'$ would be a morphism between different objects, the homsets aren't disjoint, but it seems close

John Baez (Jan 12 2022 at 17:44):

Umm, all that sounds promising, but I'm not sure I got your answer. Suppose you have a 2-term chain complex $D$ - I'll call it $D$ because it's just any 2-term chain complex, not the two we're specially interested in. And suppose you have two 0-chains in $D$, say $x$ and $y$. What are the morphisms from $x$ to $y$?

John Baez (Jan 12 2022 at 17:44):

The more concretely and simply you can say this, the happier I'll be.

John Baez (Jan 12 2022 at 17:48):

To set up the notation, say

$D = \{D_1 \stackrel{d}{\to} D_0 \}$

where I'm calling the boundary map $d$ instead of $\partial$ because it's easier to type.

Patrick Nicodemus (Jan 12 2022 at 18:04):

$\operatorname{Hom}(x,y):=d^{-1}(\{y-x\})$, where $\{ y -x \}$ is the singleton and $d^{-1}$ is the preimage.

John Baez (Jan 12 2022 at 18:06):

Okay, great! Here's how I'd say the same thing: morphism from $x$ to $y$ is a 1-chain $f$ with $df = y - x$.

Patrick Nicodemus (Jan 12 2022 at 18:06):

Yes, then we're on the same page.

John Baez (Jan 12 2022 at 18:07):

So the picture of a morphism is this:

$\bullet \stackrel{f}{\longrightarrow} \bullet$

John Baez (Jan 12 2022 at 18:08):

A 1-chain $f$ is a "movable arrow", but when we nail down its source $x$ we know its target $y$.

John Baez (Jan 12 2022 at 18:10):

If you've ever taught vectors to kids you've had to clarify the difference between a vector that starts at the origin, and a "movable arrow" that can start anywhere. They find this confusing. I see this online:

Why make the distinction between points and vectors? A vector need not start at the origin: it can be located anywhere!

John Baez (Jan 12 2022 at 18:16):

So a 1-chain is a "movable arrow", while an actual morphism can be seen as a pair consisting of a 0-chain and a 1-chain.

John Baez (Jan 12 2022 at 18:18):

And just to be completely detailed: how do you compose morphisms in the category built from a 2-term chain complex?

Patrick Nicodemus (Jan 12 2022 at 19:17):

It should be given by addition. The fact that the homomorphism $d$ is additive gives us that this is well-defined. Also, one takes the zero element in $D_1$ to be the identity morphism for $x$; since $x-x$ is zero, $0$ indeed lies in $d^{-1}(x-x)$.

John Baez (Jan 12 2022 at 20:56):

Right! A morphism from $x$ to $y$ is just a 1-chain $f$ with $df = y - x$, and given a morphism $g$ from $y$ to $z$ we just add them to get a morphism from $x$ to $z$

John Baez (Jan 12 2022 at 20:56):

So here's another puzzle: are the categories coming from our chain complexes $C$ and $C'$ equivalent, or not?

John Baez (Jan 12 2022 at 20:57):

For anyone lurking and watching, I mean the 2-term chain complexes

$C' = \big\{ \mathbb{Z} \stackrel{2}{\to} \mathbb{Z} \big\}$

and

$C = \big\{ 0 \to \mathbb{Z}/2 \big\}$

Patrick Nicodemus (Jan 13 2022 at 17:11):

sorry, i got distracted yesterday. The two categories should be equivalent, yes. The category associated to $\mathcal{C}$ is the discrete category with two objects, and the category associated to $\mathcal{C}'$ is the coproduct of two "contractible" categories, each one has countably many objects all isomorphic to each other by a unique isomorphism.

Patrick Nicodemus (Jan 13 2022 at 17:11):

I do find this a bit counterintuitive/surprising, so it feels "wrong". I am sure there is no equivalence which respects the additive structure of the objects or morphisms, as there is no nonzero homomorphism $\mathbb{Z}/2\to\mathbb{Z}$

John Baez (Jan 13 2022 at 22:29):

Patrick Nicodemus said:

sorry, i got distracted yesterday.

Don't worry - there's a limit to the rate at which I can do this myself.

The two categories should be equivalent, yes. The category associated to $C$ is the discrete category with two objects, and the category associated to $C'$ is the coproduct of two "contractible" categories, each one has countably many objects all isomorphic to each other by a unique isomorphism.

Great! Yes, in the jargon of category theory, the category associated to $C$ is a skeletal version of the category associated to $C'$ - that is, an equivalent category where isomorphic objects are equal.

John Baez (Jan 13 2022 at 22:31):

Category theorists often start with a category and then choose a skeletal version of it, for convenience of calculation.

John Baez (Jan 13 2022 at 22:32):

But notice here we are going in the other direction: we started with a skeletal category and 'deskeletonized' it, or 'resolved' it.

This reverse process, of systematically 'adding extra flab', is what 'taking a resolution' or 'doing cofibrant replacement' is all about. We want to understand it better and understand why it's a good thing to do.

Zhen Lin Low (Jan 13 2022 at 22:33):

Patrick Nicodemus said:

I do find this a bit counterintuitive/surprising, so it feels "wrong". I am sure there is no equivalence which respects the additive structure of the objects or morphisms, as there is no nonzero homomorphism $\mathbb{Z}/2\to\mathbb{Z}$

Hmmm. I feel it shouldn't be surprising. Maybe you just haven't thought about it before. Given a general chain complex of abelian groups (concentrated in non-negative degrees), as you say, the Dold–Kan theorem gives you a simplicial abelian group that "contains the same information", in the sense that the two categories are equivalent. So you get a forgetful functor from chain complexes to simplicial sets. Why do we care about this? Well, this forgetful functor is homotopical, in the sense that a morphism of chain complexes is a quasi-isomorphism if and only if the forgetful functor sends it to a weak homotopy equivalence of simplicial sets. This is because the Dold–Kan theorem also gives you a natural isomorphism between homology groups on one side and homotopy groups on the other side. Furthermore the underlying simplicial set of a simplicial abelian group is a Kan complex, so weak homotopy equivalences between them are actually homotopy equivalences! But the quasi-inverse is merely a morphism of simplicial sets.

For chain complexes concentrated in degrees 0 and 1, you only need to worry about things happening in homotopy dimension < 2, so you can work with 1-groupoids instead of simplicial sets, and this is what John's exercise is about.

John Baez (Jan 13 2022 at 22:49):

Patrick Nicodemus said:

I do find this a bit counterintuitive/surprising, so it feels "wrong".

I think we'll see it's a quite general thing.

I am sure there is no equivalence which respects the additive structure of the objects or morphisms, as there is no nonzero homomorphism $\mathbb{Z}/2\to \mathbb{Z}$

Yes, that's the really interesting part!

I think the pattern we're seeing is like this:

Suppose $A$ is an abelian group,

$C = \{ 0 \to A \}$

is the corresponding 2-term chain complex of abelian groups, and

$C' = \{ F_1 \stackrel{d}{\longrightarrow} F_0 \}$

is a free resolution of $A$, or more precisely a 2-term chain complex of abelian groups where $F_0,F_1$ are free abelian,

$\mathrm{ker} d = 0$

and

$F_0/ \mathrm{im} d \cong A$

Then the hopefully obvious map

$p: C' \to C$

is a quasi-isomorphism but is typically not a chain homotopy equivalence.

John Baez (Jan 13 2022 at 22:57):

We can turn $C$ and $C'$ into categories which I'll call $\mathrm{Cat}(C)$ and $\mathrm{Cat}(C')$ using the method we discussed. Then $p$ gives a functor which I'll call

$\mathrm{Cat}(C): \mathrm{Cat}(C') \to \mathrm{Cat}(C)$

This is an equivalence of categories. $\mathrm{Cat}(C)$ is skeletal, i.e. isomorphic objects are equal. But $\mathrm{Cat}(C')$ is usually not.

John Baez (Jan 13 2022 at 22:59):

Here's one big moral: if we think of $A$ as presented by generators and relations, the relations become isomorphisms in $\mathrm{Cat}(C')$.

John Baez (Jan 13 2022 at 23:01):

$A$ was a mere set, which you can think of as a discrete category, but we've puffed it up to an equivalent category where the relations have become isomorphisms!

John Baez (Jan 13 2022 at 23:01):

This is the essence of 'resolution'.

John Baez (Jan 13 2022 at 23:04):

But I haven't yet gotten to what I called the "interesting part", namely this observation of yours:

I am sure there is no equivalence which respects the additive structure of the objects or morphisms, as there is no nonzero homomorphism $\mathbb{Z}/2\to\mathbb{Z}$

John Baez (Jan 13 2022 at 23:04):

You see, I was lying just now when I said $A$ was a mere set: of course it's an abelian group.

John Baez (Jan 13 2022 at 23:06):

Similarly $\mathrm{Cat}(C)$ and $\mathrm{Cat}(C')$ are not mere categories: they are abelian group objects in Cat. This expresses the fact that we can add and subtract objects, and also morphisms.

John Baez (Jan 13 2022 at 23:07):

I guess we should dig into that and see why it's true. But anyway, the "interesting part" is that while $\mathrm{Cat}(C)$ and $\mathrm{Cat}(C')$ are equivalent as categories, they're not equivalent as abelian group objects in Cat.

John Baez (Jan 13 2022 at 23:09):

The point is that while we can find an equivalence going back from $\mathrm{Cat}(C)$ back to $\mathrm{Cat}(C')$, we can't find one which respects the additive structure of the objects or morphisms, as there is no nonzero homomorphisms $\mathbb{Z}/2 \to \mathbb{Z}$...

... as you said.

John Baez (Jan 13 2022 at 23:12):

Well, I've said a lot of stuff. You can prove these things I alluded to, or ask questions about them, if you want:

1) There's a map sending 2-term chain complexes to categories, chain maps to functors, and chain homotopies to natural isomorphisms.

2) This map $\mathrm{Cat}$ actually sends 2-term chain complexes to abelian group objects in Cat.

3) It does similar things to chain maps and chain homotopies.

John Baez (Jan 13 2022 at 23:14):

4) Any abelian group $A$ automatically gives an abelian group object in Cat, which I've been calling $\mathrm{Cat}(C)$.

5) Any 2-term free resolution of $A$ gives an abelian group object in Cat called $\mathrm{Cat}(C')$.

6) $\mathrm{Cat}(C)$ and $\mathrm{Cat}(C')$ are equivalent as categories but usually not equivalent as abelian group objects in Cat.

Patrick Nicodemus (Jan 14 2022 at 00:18):

Zhen Lin Low said:

Furthermore the underlying simplicial set of a simplicial abelian group is a Kan complex, so weak homotopy equivalences between them are actually homotopy equivalences!

Yes, this was the key observation I needed, thank you.

John Baez (Jan 15 2022 at 23:50):

Was the stuff I wrote too complicated and/or mysterious? For example, should we talk more about "abelian group objects in Cat"?

Patrick Nicodemus (Jan 18 2022 at 10:51):

Hi John; I just didn't get a chance to sit down and think through it, I hope it isn't overwhelming but I often procrastinate a bit on the task of making sense of complex matrerial. I will read it carefully now.

Patrick Nicodemus (Jan 18 2022 at 12:05):

I do understand what is meant by an Abelian group object in $\mathbf{Cat}$, it is the same definition as in any category with products.

Patrick Nicodemus (Jan 18 2022 at 12:07):

So a quasi-isomorphism is an equivalence of categories. Hmm.

Ok, here's what I'll note.

The homology classes in $H_0(C')$ are the connected components of the groupoid $\mathbf{Cat}(C')$.

Let $\alpha : C'\to C$ be a chain map, and write $\mathbf{Cat}(\alpha)$ for the associated functor between categories, $\alpha_\ast : H_0(C')\to H_0(C)$ for the induced map between homology groups.

I claim that $\mathbf{Cat}(\alpha)$ is fully faithful iff $\alpha_\ast$ is injective. To see this, note that between two objects in the category $C'$ there is at most one morphism, as the boundary map $d : C_1\to C_0$ is injective, so $Hom(x,y)$ is empty or a singleton. So $\alpha$ is fully faithful iff $\alpha(x)\cong \alpha(y)$ in $\mathbf{Cat}(C)$, already we have $x\cong y$. But because homology classes correspond to connected components of the groupoid, this is precisely to say that $\alpha_\ast$ is injective on homology classes.

I claim that $\mathbf{Cat}(\alpha)$ is essentially surjective if $$\alpha_\ast$ is surjective; this is clear from the observation that homology classes correspond to groupoid components, and a functor between groupoids is essentially surjective if the induced map between the two sets of connected components is surjective.

Thus if a map $\alpha : C'\to C$ is a quasi-isomorphism, then $\mathbf{Cat}(\alpha)$ is an equivalence, and conversely.

Patrick Nicodemus (Jan 18 2022 at 12:21):

John Baez said:

Here's one big moral: if we think of $A$ as presented by generators and relations, the relations become isomorphisms in $\mathrm{Cat}(C')$.

Yes, I can see this now. For $x\in A$, each different object in the isomorphism class which is the "fiber" of $C'$ over $x$ corresponds to a formal linear combination of generators which returns $x$, and the isomorphisms link them together as being equivalent.

Patrick Nicodemus (Jan 18 2022 at 12:23):

John Baez said:

Similarly $\mathrm{Cat}(C)$ and $\mathrm{Cat}(C')$ are not mere categories: they are abelian group objects in Cat. This expresses the fact that we can add and subtract objects, and also morphisms.

By the way, is this equivalent to them being internal category objects in $\mathbf{Ab}$?

Patrick Nicodemus (Jan 18 2022 at 12:23):

I think I understand what's going on so far, if you wish to continue.

Zhen Lin Low (Jan 18 2022 at 12:24):

Yes. Ab(Cat(X)) is equivalent to Cat(Ab(X)) for any category X with finite limits. It basically boils down to the fact that limits commute with limits.

Patrick Nicodemus (Jan 18 2022 at 12:25):

Thank you!

Zhen Lin Low (Jan 18 2022 at 12:32):

Something less easy to explain is that every internal category in Ab(X) (or, for that matter, Grp(X)) is automatically an internal groupoid. This is a consequence of the Mal'cev property of Ab(X) (or Grp(X)), but the significance of this fact is not clear to me. Personally I would prefer to think of internal abelian groups in the category of groupoids (thereby connecting with homotopy theory) rather than internal categories in the category of abelian groups...

Patrick Nicodemus (Jan 18 2022 at 12:38):

not to interrupt the current train of thought but i have a closely related question that maybe you can help me with, and it's about something that the Gabriel-Zisman theory of localization somehow fails to capture. Gabriel and Zisman set up this scenario where you can take a category of, say, chain complexes. You can quotient out by the relation of homotopy equivalence and then everything after that is 1-categorical. So working in this homotopy category you can identify a class of distinguished morphisms (the quasi isomorphisms) and this distinguished subcategory of complexes of injective objects. The Cartan-Eilenberg resolution is a kind of reflection functor into this subcategory, and it has the property that every map from a chain complex $C$ into an complex of injectives $A$ extends uniquely up to homotopy along a C-E resolution $C\to I$.

Patrick Nicodemus (Jan 18 2022 at 12:39):

The problem is that there's something infinity-categorical going on here which the 1-categorical formulation doesn't capture, namely that I think the extension of the map $C\to A$ along the C-E resolution $C\to I$ should be actually unique up to higher homotopy, or maybe there's a homotopy equivalence of mapping spaces $A^C\simeq A^I$, or something like this

Patrick Nicodemus (Jan 18 2022 at 12:40):

I am looking for like, some kind of infinity categorical generalization of the Gabriel-Zisman localization theory which would help me to understand what is going on here with this uniqueness of the extension up to higher homotopy, I want to adapt this theory to the bar construction in something that I am working on and I want to know what the right language is

Patrick Nicodemus (Jan 18 2022 at 12:41):

I don't know much about higher category theory so if you can recommend an introduction to higher cats as a whole I will gladly dive in, I embrace technical detail if this helps

Zhen Lin Low (Jan 18 2022 at 12:45):

I'm not sure exactly what you mean by GZ-localisation. You mean their general construction of localisations using arbitrary zigzags, or their specific construction of hom-sets as filtered colimits of sets of spans?

Patrick Nicodemus (Jan 18 2022 at 12:47):

Um, I think I mean the "ordinary" localization which has a 1-categorical universal property in $\mathbf{Cat}$ with respect to a class of morphisms $\Sigma$, I once heard someone refer to this as GZ-localization to distinguish it from some kind of higher-categorical localization.

Zhen Lin Low (Jan 18 2022 at 12:53):

Well, you mentioned the homotopy category K and the derived category D, and both of these are 1-categorical localisations of the chain complex category Ch, wrt chain homotopy equivalences and quasi-isomorphisms, respectively. But GZ also show that D is a localisation of K in a very controlled way. The former is easy to generalise to (∞, 1)-categories, in the sense that you can just define things by universal properties and appeal to general nonsense to get existence, but the latter is an explicit construction and would require some work to generalise.

Zhen Lin Low (Jan 18 2022 at 12:54):

Aside from that point, as far as I know the homotopy category K is not much studied using (∞, 1)-categorical techniques – the emphasis is on the derived category D. The same can be said for the classical homotopy category of topological spaces or simplicial sets vs the homotopy category of CW-complexes or ∞-groupoids.

Patrick Nicodemus (Jan 18 2022 at 12:56):

Let me rephrase it this way: whenever you have a category $C$ with reflective subcategory $D$, say with left adjoint $G : C\to D$, there is an equivalence of categories between $D$ and the localization of $C$ at a class of morphisms $\Sigma$, namely the class of maps $f$ with $G(f)$ an isomorphism.

I am looking for some kind of cousin of this result which is higher-categorical in the following sense - if we have $C$ a category and $D$ a reflexive subcategory, with $G : C\to D$ a kind of (higher categorical?) right adjoint, characterized by the fact that for $A$ in $C$ and $X$ in $D$, there is a homotopy equivalence of mapping spaces $X^A \simeq X^{G(A)}$

Patrick Nicodemus (Jan 18 2022 at 12:56):

because it seems like something like this is what is happening with C-E resolutions.

Zhen Lin Low (Jan 18 2022 at 12:57):

The equivalence of mapping spaces is just the statement that you have a reflective adjunction, no...?

Patrick Nicodemus (Jan 18 2022 at 12:58):

I have no idea. I'm sorry. I don't know higher category theory, I don't know what the definition of a higher categorical right/left adjoint is or would be. That's why i'm asking you for references.

Zhen Lin Low (Jan 18 2022 at 12:58):

OK, never mind higher category theory then. How about enriched category theory? You know how enriched adjunctions work?

Patrick Nicodemus (Jan 18 2022 at 13:00):

Hmm, yes, I can't remember the definition off the top of my head but it would just be a natural isomorphism of $\mathcal{V}$-presheaves $Nat(F(-),-)\cong Nat(-,G(-))$ i guess?

Patrick Nicodemus (Jan 18 2022 at 13:00):

so we could take everything to be enriched over $SSet$ or $Ch(Ab)$ i guess, and get a reasonable theory there

Patrick Nicodemus (Jan 18 2022 at 13:01):

is there a reasonable notion of localization for this SSet-enriched setting which this kind of reflexive subcategory would be equivalent to a "localization" of $\mathcal{C}$ at a distinguished class of morphisms

Patrick Nicodemus (Jan 18 2022 at 13:02):

I don't know what it means to formally invert an element of a simplicial set, I guess.

Zhen Lin Low (Jan 18 2022 at 13:07):

sSet-enriched categories ("simplicial categories") and Ch-enriched categories ("dg-categories") can be thought of as (∞, 1)-categories, yes, but you don't get (∞, 1)-category theory just by specialising enriched category theory. You have to tweak definitions here and there, replacing isomorphisms by equivalences in the right places and adding coherence conditions etc. But, the big picture is more or less the same.

Zhen Lin Low (Jan 18 2022 at 13:10):

There is a theory of localisations of (∞, 1)-categories, developed in the language of quasicategories. My memory is poor so I don't remember if there is a result along the lines of what you want but my intuition tells me that it is true. There is a fairly high-level proof for ordinary categories that should translate.

John Baez (Jan 19 2022 at 18:59):

Patrick Nicodemus said:

Thus if a map $\alpha : C'\to C$ is a quasi-isomorphism, then $\mathbf{Cat}(\alpha)$ is an equivalence, and conversely.

Nice!

John Baez (Jan 19 2022 at 19:11):

So let me summarize where we are, and the annoying situation we find ourselves in.

We've seen that 2-term chain complexes (of abelian groups) are "the same" as categorical abelian groups - which is my desperate attempt to find something quicker to say than "abelian group objects in Cat".

Furthermore, abelian groups can be seen as a special case of categorical abelian groups. They are just the discrete ones: the ones with only identity morphisms.

We've seen that a free resolution of an abelian group $A$ amounts to taking a presentation of that group and replacing relations by isomorphisms to get a categorical abelian group, which we've been calling $A'$.

$A'$ is "just a puffed up version" of $A$ in the following sense: there's a map of categorical abelian groups

$p: A' \to A$

which is an equivalence of categories. I'll call $p$ a resolution or cofibrant replacement.

But the annoying thing is that while there's an equivalence of categories going back, say

$q: A \to A'$

we usually can't choose this to be a map of abelian categorical groups!

We've seen this happening already in our favorite example, where

$C = \big\{ 0 \to \mathbb{Z}/2 \big\}$

and

$C' = \big\{ \mathbb{Z} \stackrel{2}{\to} \mathbb{Z} \big\}$

John Baez (Jan 19 2022 at 19:13):

There's no map of categorical abelian groups from $C$ back to $C'$ except the zero map, since such a map amounts to a group homomorphism from $\mathbb{Z}/2$ to $\mathbb{Z}$.

John Baez (Jan 19 2022 at 19:18):

Intuitively speaking: we can easily "curl up" $\mathbb{Z}$ to get $\mathbb{Z}/2$, but we can't "uncurl" $\mathbb{Z}/2$ and get $\mathbb{Z}$.

John Baez (Jan 19 2022 at 19:20):

So the annoying thing is this: $C$ and $C'$ are abelian categorical groups, and we know (because it follows from things you proved) that there's an equivalence of categories

$q : C \to C'$

but there's no way to choose this equivalence to be a map of abelian categorical groups, even though

$p: C' \to C$

is an equivalence and a map of abelian categorical groups.

John Baez (Jan 19 2022 at 19:22):

All this hints that our concept of "map of abelian categorical groups" is too strict - at least too strict for some purposes.

John Baez (Jan 19 2022 at 19:25):

For example, this annoying phenomenon doesn't happen with monoidal categories - at least not if you do things right. If you have an equivalence of monoidal categories

$p: X \to Y$

you can always find an equivalence of monoidal categories

$q : Y \to X$

going back. But if you do things wrong, the problem happens: I think if you have an equivalence of monoidal categories that's a strict monoidal functor

$p: X \to Y$

you can't always find an equivalence of categories that's a strict monoidal functor going back:

$q: Y \to X$.

John Baez (Jan 19 2022 at 19:27):

Okay, so here's my puzzle: what's a good non-strict notion of "map between abelian categorical groups", that could help us out of our annoying situation?

Graham Manuell (Jan 19 2022 at 20:58):

Wow! Why have I never seen anyone explain chain complexes this way before?!
(I have a guess for the answer to the puzzle, but I'll keep quiet for now so as to not ruin it for the people this was actually intended for.)

Zhen Lin Low (Jan 19 2022 at 22:25):

I think if you look into the literature about non-abelian cohomology you'll see discussion about "crossed modules" or "cat^1-groups". I remember being thoroughly unable to understand why these were the objects of study. Some years later I understood Dold–Kan better but had forgotten about cat^1-groups so I failed to make the connection until just the other day, through this discussion.

Graham Manuell (Jan 19 2022 at 22:39):

Yeah, I know a little about non-abelian cohomology and so I have at least a vague understanding of crossed modules, but I didn't know the link between (weak) equivalences of internal categories in Ab and quasi-isomorphisms of chain complexes. When I learnt about quasi-isomorphisms they were defined in a completely arbitrary way and I never understood the motivation before.

Zhen Lin Low (Jan 19 2022 at 22:43):

I didn't imagine I would hear someone say quasi-isomorphisms are defined in a completely arbitrary way... what would not have been arbitrary, then? Chain homotopy equivalences?

Mike Shulman (Jan 19 2022 at 23:31):

I think if you don't know why to care about homology groups, quasi-isomorphisms can look pretty arbitrary. Just like if you don't know why to care about homotopy groups, weak homotopy equivalences look pretty arbitrary.

Zhen Lin Low (Jan 19 2022 at 23:44):

Of course. I was taking it as given that that homology/homotopy groups were not considered to be defined arbitrarily, though I suppose they too might be considered unmotivated.

Graham Manuell (Jan 20 2022 at 11:45):

Yes. I also have trouble with homology and homotopy groups and weak homotopy equivalences. But even granting homology groups are important, do you not agree that equivalences of categories at least seem more fundamental than inducing isomorphisms on the homology groups?

Zhen Lin Low (Jan 20 2022 at 13:19):

I don't disagree, but I also believe in conservation of complexity (or arbitrariness or inexplicability) and I think you're taking for granted the translation of chain complexes – not even general chain complexes! only chain complexes of abelian groups concentrated in degrees 0 and 1 – into cat^1-groups. I'm not convinced this translation is a natural thing to do. To me, it feels like an arbitrary construction designed to make true the theorem that a morphism of chain complexes in degrees 0...1 is a quasi-isomorphism if and only if it corresponds to a weak equivalence of groupoids. I do not know of any independent justification of this construction.

Graham Manuell (Jan 20 2022 at 14:51):

I see. Well, presumably quasi-isomorphisms of more general chain complexes (at least ones concentrated in nonnegative degree) can be understood in terms of equivalences of internal higher categories in a similar way (and we could work internal to abelian categories other than Ab). I am more happy with internal categories than I am with chain complexes, so I guess from my perspective if the translation isn't natural, then so much the worse for chain complexes.

John Baez (Jan 20 2022 at 18:09):

Graham Manuell said:

Wow! Why have I never seen anyone explain chain complexes this way before?!

Because they learned about chain complexes, grew to accept them, and didn't realize that people learning about them might enjoy a new viewpoint on them? This material is known; people just don't deploy it early enough in the process of teaching homological algebra (in my opinion).

John Baez (Jan 20 2022 at 18:14):

Zhen Lin Low said:

I don't disagree, but I also believe in conservation of complexity (or arbitrariness or inexplicability) and I think you're taking for granted the translation of chain complexes – not even general chain complexes! only chain complexes of abelian groups concentrated in degrees 0 and 1 into cat^1-groups.

I'm starting with chain complexes of abelian groups concentrated in degrees 0 and 1 because I'm trying to explain things in a way that's easy to follow. Later on, if Patrick stays interested, we can generalize the story to arbitrary chain complexes. But I'm mainly trying to provide intuition, not state a bunch of theorems.

Patrick Nicodemus (Jan 20 2022 at 18:26):

I see what you mean, the example of strict vs nonstrict monoidal categories is helpful; indeed a categorical Abelian group should be a special type of strict symmetric monoidal category, one equipped with an "inversion" functor $(-)^{-1}$

Any functor $F$ from $\mathbf{Cat}(C) \to \mathbf{Cat}(C')$ which is a right inverse to $\mathbf{Cat}(\alpha)$ will send $0$ to an even number and $1$ to an odd number. Such a functor is then additive up to isomorphism in the sense that we have $F(n)+F(m) \cong F(n+m)$ for $n, m\in \{0,1\}$, as any two natural numbers of the same parity are isomorphic. This isomorphism will always be natural as $\mathbb{Z}/2\mathbb{Z}$ is a discrete category, and so the naturality condition is trivial or degenerate.

Patrick Nicodemus (Jan 20 2022 at 18:30):

The other equational group laws should hold up to isomorphism, I think, for example inverses should be preserved up to isomorphism. I won't try and state what it means for all these isomorphisms to be coherent but surely in such simple categories all these complex coherence conditions are trivially satisfied as there is at most one isomorphism between objects

John Baez (Jan 20 2022 at 18:32):

Okay, let me try to think about this.

Yes, a categorical abelian group is a strict symmetric monoidal category with an inversion functor obeying the axioms an inverse should, on the nose (as equations)... but by the way, a categorical abelian group is a bit better than just this!

Can you see what extra axiom holds?

(This is not the most important thing you said, but it's good to get this straightened out, and this extra axiom is sort of fun.)

Patrick Nicodemus (Jan 20 2022 at 18:51):

Hmm I may be missing your point here, I wrote down the diagrams that characterize an Abelian group object and I am able to categorize them into the ones that describe a symmetric monoid with respect to the Cartesian product and some others governing the inverse but I am not sure what the substantial difference is - perhaps you are asking me to say that the identity map $m(x,e) \cong x$ should be a natural transformation $m(-,e) \cong \operatorname{id}_{\mathcal{C}}$ or something like this

Patrick Nicodemus (Jan 20 2022 at 18:52):

but this is perhaps already guaranteed by saying that $m(-,e) = \operatorname{id}$ as functors.

Patrick Nicodemus (Jan 20 2022 at 18:54):

Do you have a hint as to where I should be looking for the discrepancy?

Patrick Nicodemus (Jan 21 2022 at 17:20):

John Baez said:

Because they learned about chain complexes, grew to accept them, and didn't realize that people learning about them might enjoy a new viewpoint on them? This material is known; people just don't deploy it early enough in the process of teaching homological algebra (in my opinion).

I am a big fan of returning to the foundations often to reexamine them if only for pedagogical reasons. I feel that homological algebra courses could be much improved if we were more able to explain to students what the meaning is behind, say, Hilbert's Syzygy theorem. For me there is an "easy" answer - homological dimension is some kind of measure of internal complexity, as if the relations are themselves bound by nontrivial higher relations and so on then there has to be a deal of "unfurling" that goes on before one reaches full internal independence; Hilbert's syzygy theorem caps this unfurling and so puts an upper bound on complexity. But this does not feel like an exhaustive telling of the story, there is more to be said about the meaning of Hilbert's syzygy theorem than those two sentences. I don't mean to say that the easy answer is inaccurate, more that it is the beginning.

John Baez (Jan 21 2022 at 17:27):

Right, I like to keep digging at the basics of things. There's little chance that the first version of an important idea is the only interesting version or even the deepest version.

John Baez (Jan 21 2022 at 17:33):

Just to list the two questions that are floating around in our conversation:

1. An abelian categorical group is the same as a symmetric strict monoidal groupoid $C$ with an 'inverse' functor ${}^{-1} : C \to C$ making it into a group object in Cat... which also has one other property. What's that property?

2. The obvious morphisms of abelian categorical groups are 'too strict', in the sense that not every such morphism that's an equivalence of categories has an equivalence going back that's also such a morphism. This is the reason people 'formally invert' such morphisms. But an alternative is to find a less strict kind of map between abelian categorical groups - what's that?

John Baez (Feb 04 2022 at 06:13):

Since this conversation seems to have ended, let me answer my questions above.

1. The other property that any abelian categorical group has is that the symmetry $S_{x,y} : x \otimes y \to y \otimes x$ is the identity natural transformation. So an abelian group object in Cat is not merely symmetric but commutative.

2. The less strict sort of map I was alluding to is a symmetric monoidal functor.

Patrick Nicodemus (Feb 07 2022 at 22:22):

John Baez said:

Since this conversation seems to have ended, let me answer my questions above.

1. The other property that any abelian categorical group has is that the symmetry $S_{x,y} : x \otimes y \to y \otimes x$ is the identity natural transformation. So an abelian group object in Cat is not merely symmetric but commutative.

2. The less strict sort of map I was alluding to is a symmetric monoidal functor.

Hi John,
My apologies, I have a bad habit of procrastinating on digesting complex material or intimidating questions. I am interested in continuing this conversation if you are. I will work through these and try to understand (1.) and why (2.) is appropriate, I did not think about the symmetry condition the first time I sat down to think through this.

John Baez (Feb 07 2022 at 22:29):

Okay - I was afraid my questions may have seemed intimidating. Sorry! I hope you see the answers are not extremely complicated. But you should also feel free to ask questions or just say "hmm, I don't get it."

(1) is a interesting because only in certain rather specialized branches of math do people study "commutative strict monoidal categories", rather than the more flexible "symmetric strict monoidal categories".

An abelian group object in Cat has to be commutative, not merely symmetric, because there's no room for commutativity up to isomorphism in concept of "abelian group object" - abelian means the multiplication commutes exactly. So a 2-term chain complex is going to give a commutative strict monoidal category of a certain sort.

John Baez (Feb 07 2022 at 22:31):

I have also bumped into commutative monoidal categories in two other branches of math (Petri nets and categories of line bundles).

Patrick Nicodemus (Feb 07 2022 at 22:35):

Ah, yes. That makes sense. I have not heard of a symmetric monoidal functor but I am able to make an educated quess and write down the commutative square which, I assume, characterizes when a monoidal functor is symmetric monoidal. In our case where we consider everything strictly, do we get this for free as a consequence of the fact that functors send identities to identities?

Patrick Nicodemus (Feb 07 2022 at 22:37):

This question may be incoherent as a consequence of the fact that I haven't yet worked out what you're saying on pencil and paper

John Baez (Feb 07 2022 at 23:24):

Patrick Nicodemus said:

Ah, yes. That makes sense. I have not heard of a symmetric monoidal functor but I am able to make an educated quess and write down the commutative square which, I assume, characterizes when a monoidal functor is symmetric monoidal. In our case where we consider everything strictly, do we get this for free as a consequence of the fact that functors send identities to identities?

Yes, exactly. Another way to put it is this: any homomorphism between groups that happen to be commutative is automatically "an abelian group homomorphism", no extra properties are required, because being abelian is just a property of a group, not an extra structure.

Similarly, being an abelian group object in Cat is just an extra property of a group object in Cat, not an extra structure. So every map between group objects in Cat that happen to be abelian is a "morphism of abelian group objects".

(And an abelian group object in Cat is a 2-term chain complex of abelian groups, and a morhpism between these is a chain map.)

John Baez (Feb 07 2022 at 23:26):

On the other hand, in general being symmetric is an extra structure on a monoidal category, so not every monoidal functor between symmetric monoidal categories is a symmetric monoidal functor.

Mike Shulman (Feb 07 2022 at 23:56):

Huh. That's an aspect of commutative monoidal categories that I never thought about before. Does it apply only to strict monoidal functors between commutative monoidal categories, or also to strong/pseudo monoidal functors?

Zhen Lin Low (Feb 08 2022 at 00:00):

I was expecting the answer to John's question to be anafunctors, but maybe I've been doing model categories too long...

John Baez (Feb 08 2022 at 00:07):

Mike Shulman said:

Huh. That's an aspect of commutative monoidal categories that I never thought about before. Does it apply only to strict monoidal functors between commutative monoidal categories, or also to strong/pseudo monoidal functors?

I'm only claiming it for strict monoidal functors; that's what is relevant to my story here.

I think it's not true for strong monoidal functors. The condition for a strong monoidal functor to be symmetric is this commutative square Patrick alluded to:

John Baez (Feb 08 2022 at 00:08):

When the symmetric monoidal categories are commutative the braidings $\gamma$ are the identity, and when the monoidal functor is strict the laxators $\phi$ are identities too, so the square automatically commutes.

Mike Shulman (Feb 08 2022 at 00:09):

Ok, yes, that makes sense. But I still don't think it's something that ever occurred to me before, that every strict monoidal functor between commutative monoidal catgories is automatically a symmetric monoidal functor.

John Baez (Feb 08 2022 at 00:12):

I noticed it when Jade and I were writing Open Petri nets, and talking about how Petri nets present free commutative monoidal categories. We described commutative monoidal categories and maps between them in a few different ways, and the quickest way is that they're just strict monoidal functors.

Patrick Nicodemus (Feb 09 2022 at 09:16):

Ok, I have thought through this and it makes sense to me. So for these Abelian group objects in Cat, we might consider them as strict symmetric monoidal categories. But since the right notion of sameness of two categories is an equivalence rather than an isomorphism, in spite of the fact that the monoidal product itself is strict, in general we shouldn't demand that morphisms between these categories be strict monoidal functors.

Patrick Nicodemus (Feb 09 2022 at 09:25):

You are trying to lay a strict product on top of a nonstrict notion of sameness and you shouldn't expect a strict monoidal product to transfer nicely along a non-strict equivalence (rather than an isomorphism of categories)

John Baez (Feb 09 2022 at 16:41):

Yes, this is great! And while it causes some problems, this is what we're doing when we treat abelian group actions in Cat as 2-term chain complexes and study chain maps between them!

John Baez (Feb 09 2022 at 17:25):

I know two ways out of this problem:

• work with symmetric monoidal functors $f: C \to D$ that aren't strict

• cofibrant replacement: replace the source of $f$ by a 'cofibrant' chain complex $C'$ that's quasi-isomorphic to $C$; then we get a chain map $f' : C' \to D$ that serves as substitute for $f: C \to D$.

John Baez (Feb 09 2022 at 17:27):

Traditionally in homological algebra people always take the second route!

John Baez (Feb 09 2022 at 17:29):

Notice that in the second route, even though $C$ is a 2-term chain complex, $C'$ may have more nonzero terms. Since we're talking about chain complex of abelian groups right now, I think we can always choose $C'$ to be a 3-term chain complex. This is the same as an abelian group object in 2Cat.

John Baez (Feb 09 2022 at 17:30):

So far in this discussion we've only talked about a simpler situation, where $C$ is just a 1-term chain complex, and $C'$ is a 2-term chain complex.

John Baez (Feb 09 2022 at 17:32):

Then $C$ is called an abelian group and $C'$ is called its resolution.

Mike Shulman (Feb 09 2022 at 17:36):

Here's another question that hasn't occurred to me before. What exactly is it that forces $C'$ to have more nonzero terms than $C$? If we stay in the world of monoidal categories, it is true that any non-strict symmetric monoidal functor $C\to D$ can be represented, up to isomorphism, as a span $C \leftarrow C' \to D$ of symmetric strict monoidal functors, where $C'$ is an ordinary symmetric monoidal category, not a symmetric monoidal 2-category. Is the point that this $C'$ might not be commutative even if $C$ and $D$ are?

John Baez (Feb 09 2022 at 20:22):

Hmm, since I like examples you're making me want to think about this in our favorite example, where in the language of chain complexes we take

$C = \big\{ 0 \to \mathbb{Z}/2 \big\}$

and

$C' = \big\{ \mathbb{Z} \stackrel{2}{\to} \mathbb{Z} \big\}$

with $p: C' \to C$ being the obvious quasi-isomorphism. Then we can take $D = C'$ and look at the span of quasi-isomorphisms

$C \stackrel{p}{\leftarrow} C' \stackrel{1}{\to} C'$

John Baez (Feb 09 2022 at 20:24):

So why can't we use a 1-term chain complex in the middle here?

John Baez (Feb 09 2022 at 20:25):

Maybe this example is too "lowly" to shed light on your question, since we're resolving an abelian group in 0Cat and getting an abelian group object in 1Cat, while your question starts at the 1-caegory level.

Mike Shulman (Feb 09 2022 at 21:25):

Yeah, I'm not sure we can see this at the 0-category level. At least, not if the codomain $D$ is also a 1-term complex, since then there's no room for any non-strict functors $C\to D$ in the first place. Certainly if $D$ has more than 1 term, we might need to resolve $C$, and I would expect that to be true in the 1-category case too: to represent all non-strict symmetric monoidal 2-functors $C\to D$ where $D$ can be a 2-category, we would need to consider spans $C \leftarrow C'\to D$ where $C'$ is a 2-category, even if $C$ were only a 1-category. But I'm asking about the case when $D$ is a 1-category too.

Patrick Nicodemus (Feb 10 2022 at 07:44):

John Baez said:

I know two ways out of this problem:

• work with symmetric monoidal functors $f: C \to D$ that aren't strict

• cofibrant replacement: replace the source of $f$ by a 'cofibrant' chain complex $C'$ that's quasi-isomorphic to $C$; then we get a chain map $f' : C' \to D$ that serves as substitute for $f: C \to D$.

Ok, yes, this makes sense to me. So $C'$ and $D'$ have the property that any map $f'$ between them which is an equivalence in the underlying 2-category $\mathbf{Cat}$ is already an equivalence in the 2-category of Abelian group objects in $\mathbf{Cat}$, (strict) homomorphisms between them, and natural transformations respecting the strict monoidal product. This property fails for categories which aren't cofibrant, we only get an equivalence in the larger 2-category whose 1-cells are (nonstrict) symmetric monoidal functors. (But, if we wanted, we could primarily work in this larger 2-category and this would somewhat circumvent the need for cofibrant replacement to begin with.)
Am I right here?

John Baez (Feb 10 2022 at 16:11):

I haven't worked out all this stuff very carefully, so I need to think a bit. I noticed you're mentioning $D'$, which I didn't mention. I'm pretty sure the whole point of cofibrancy is that we only need to do cofibrant replacement for the source, not the target. So I want to take the paragraph you wrote and change it to this:

John Baez (Feb 10 2022 at 16:12):

So $C'$ has the property that any map $f'$ out of it that is an equivalence in the underlying 2-category $\mathbf{Cat}$ is already an equivalence in the 2-category of Abelian group objects in $\mathbf{Cat}$, (strict) homomorphisms between them, and natural transformations respecting the strict monoidal product. This property fails for categories which aren't cofibrant, we only get an equivalence in the larger 2-category whose 1-cells are (nonstrict) symmetric monoidal functors. (But, if we wanted, we could primarily work in this larger 2-category and this would somewhat circumvent the need for cofibrant replacement to begin with.)

John Baez (Feb 10 2022 at 16:13):

Is this what you meant? Or does it at least seem plausible?

John Baez (Feb 10 2022 at 16:16):

By the way, I wouldn't say "categories which aren't cofibrant", because I think cofibrancy here is a property of abelian group objects in $\mathbf{Cat}$, not merely of their underlying categories. It might turn out to depend only on the underlying category, but I don't have any evidence for that. (Maybe you were just trying to write quickly instead of pedantically here.)

Mike Shulman (Feb 10 2022 at 16:57):

I don't think I agree with the change -- saying that a map is an equivalence involves referring to a map back in the other direction, and if $D$ isn't also cofibrant then such a map $D\to C'$ might not exist unless we also resolve $D$ to $D'$.

Patrick Nicodemus (Feb 12 2022 at 02:09):

Yes, @Mike Shulman I am thinking of the interesting property of free resolutions that a quasi-isomorphism between complexes of free Abelian groups is a homotopy equivalence automatically, and similarly a weak homotopy equivalence between CW complexes is a homotopy equivalence. but I do not know much about cofibrancy and I am not sure if this aspect of resolutions is part of what cofibrancy is trying to capture.

Patrick Nicodemus (Feb 12 2022 at 02:12):

Sorry for the miscommunication there.

Zhen Lin Low (Feb 12 2022 at 05:10):

Roughly speaking, yes. In a model category, a cofibrant object "sees" every trivial fibration as a split epi homotopy equivalence. (This is in the same sense that a projective object "sees" every epimorphism as a split epimorphism.) Dually, a fibrant object "sees" every trivial cofibration as a split mono homotopy equivalence. (Compare injective objects.) Thus every weak equivalence between cofibrant–fibrant objects is a homotopy equivalence.

Some important model categories have the pleasant property that every object is cofibrant (e.g. simplicial sets qua ∞-groupoids, simplicial sets qua (∞, 1)-categories, the injective model structure on chain complexes...) and some have the dual property of every object is fibrant (e.g. topological spaces qua ∞-groupoids, the projective model structure of chain complexes, ...), but only a small handful have the property that every object is cofibrant–fibrant (e.g. topological spaces, categories, chain complexes of vector spaces), and most have neither property.

Graham Manuell (Feb 12 2022 at 16:28):

Mike Shulman said:

Here's another question that hasn't occurred to me before. What exactly is it that forces $C'$ to have more nonzero terms than $C$? If we stay in the world of monoidal categories, it is true that any non-strict symmetric monoidal functor $C\to D$ can be represented, up to isomorphism, as a span $C \leftarrow C' \to D$ of symmetric strict monoidal functors, where $C'$ is an ordinary symmetric monoidal category, not a symmetric monoidal 2-category. Is the point that this $C'$ might not be commutative even if $C$ and $D$ are?

I don't understand this stuff very well, but I think that there is nothing requiring that $C'$ has more nonzero terms here if you can choose it based on the map $f$. If we take resolutions, then they satisfy the stronger property that we can use a single choice of $C'$ for all maps $f$ and this is where we might pick up more nonzero terms. I'd be happy to be corrected though.

John Baez (Feb 12 2022 at 17:31):

I think you're right. We were mainly talking about the case where you pick the chain complex $C'$ with once and for all, with the desired property for all maps $f$, since then $C'$ must be a 'resolution' or 'cofibrant replacement' of $C$, and we were trying to understand these concepts. Even here of course we don't always need $C'$ to be an $(n+1)$-term chain complex when $C$ is an $n$-term chain complex, but we sometimes do. What people usually consider interesting is that we don't need more than $n+1$ terms. Mike was turning things around and pointing out that from the unorthodox standpoint we're taking now, it's actually interesting that we often need more than $n$ terms!

Patrick Nicodemus (Feb 12 2022 at 23:43):

Zhen Lin Low said:

Roughly speaking, yes. In a model category, a cofibrant object "sees" every trivial fibration as a split epi homotopy equivalence. (This is in the same sense that a projective object "sees" every epimorphism as a split epimorphism.) Dually, a fibrant object "sees" every trivial cofibration as a split mono homotopy equivalence. (Compare injective objects.) Thus every weak equivalence between cofibrant–fibrant objects is a homotopy equivalence.

Some important model categories have the pleasant property that every object is cofibrant (e.g. simplicial sets qua ∞-groupoids, simplicial sets qua (∞, 1)-categories, the injective model structure on chain complexes...) and some have the dual property of every object is fibrant (e.g. topological spaces qua ∞-groupoids, the projective model structure of chain complexes, ...), but only a small handful have the property that every object is cofibrant–fibrant (e.g. topological spaces, categories, chain complexes of vector spaces), and most have neither property.

Interesting observation, thank you.

Patrick Nicodemus (Feb 12 2022 at 23:50):

I guess I should look into cofibrancy in model categories to begin to understand this notion of "unwinding."

John Baez (Feb 13 2022 at 01:53):

To start understanding this notion of "unwinding" I think it actually helps to look at concrete examples, esp. the classic example of chain complexes.

We're just getting started on that project, but we've already seen how, when we are resolving an abelian group by a 2-term chain complex, the "unwinding" turns relations in some presentation of an abelian group into isomorphisms. This "unwinds" our group in the sense that nontrivial equations between group elements

$x = y$

get "unwound" into isomorphisms

$\alpha: x \to y$

(which as we've seen are also edges in a simplicial complex).

John Baez (Feb 13 2022 at 01:57):

It would get more exciting when we consider $R$-modules for general rings $R$, because then besides relations we have to deal with relations-between-relations, relations-between-relations-between-relations, and so on: so-called "syzygies".

John Baez (Feb 13 2022 at 01:57):

And if we ever start talking about the bar construction, we'll see how that gives an "automatic" unwinding procedure not only for $R$-modules but for a vast collection of algebraic gadgets - roughly speaking algebras of any monad.

John Baez (Feb 13 2022 at 02:00):

The model-category-theoretic definition of cofibrancy just axiomatizes some properties we want this unwinding process to have. But the bar construction reveals a lot about what unwinding (= cofibrant replacement) actually looks like.

Patrick Nicodemus (Feb 13 2022 at 02:47):

Ok, I have no desire to rush the conversation.

Patrick Nicodemus (Feb 13 2022 at 02:47):

Let's keep discussing the concrete examples and what unwinding looks like for these, then

Mike Shulman (Feb 13 2022 at 19:45):

In the case of ordinary symmetric monoidal categories, it's also true that we can use a single $C'$ for all maps: it's called the "pseudomorphism classifier". So I still don't understand why for chain complexes we need an extra dimension if we want to use the same $C'$ for all maps.

Patrick Nicodemus (Feb 14 2022 at 04:24):

(I am following the conversation but I don't currently have any comments or immediate questions on what has been said so far.)

Patrick Nicodemus (Feb 23 2022 at 02:06):

Hmm I would like to continue this conversation but I don't exactly know what question to ask to prime the pump. Can we continue the thread of concrete examples for this?

Patrick Nicodemus (Feb 23 2022 at 02:06):

@John Baez if you have any time in the next few days

John Baez (Feb 23 2022 at 02:51):

I guess one direction to go is to look at modules of some ring of global dimension bigger than one.

John Baez (Feb 23 2022 at 02:51):

This would get us into "syzygies".

John Baez (Feb 23 2022 at 02:53):

Maybe we should just stay a bit general and look at how the bar construction provides a resolution of any $R$-module, and think about how this is "unwinding" all the relations in the module, and the relations between relations (syzygies), and relations between relations between relations, and so on.

John Baez (Feb 23 2022 at 02:54):

Still I think an example would make it more vivid!

John Baez (Feb 23 2022 at 02:57):

Let's see: do we both know how to use the bar construction to get a free resolution of an $R$-module? (I say "both" because it will take me a while to remember.)

Patrick Nicodemus (Feb 24 2022 at 14:34):

Yeah an example will be good, I'll get out a hom alg book later and look for an example.

Patrick Nicodemus (Feb 24 2022 at 14:35):

If $M$ is an $R$ module, then write $F_0$ for the free $R$-module on the underlying set of elements of $|M|$, and inductively we should define $F_{n+1} = R^{|F_n|}$.

Patrick Nicodemus (Feb 24 2022 at 14:37):

Then the boundary maps are given by an alternating sum of face maps, where (letting $\varepsilon_M : R^{|M|}\to M$ be the canonical map that sends each formal linear combination to its an evaluation in $R$, and $G$ be the comonad of the adjunction between Sets and R-modules)

$\partial_i = G^i\varepsilon G^{n-i}$

Patrick Nicodemus (Feb 24 2022 at 14:37):

and $d = \sum (-1)^i \partial_i$

Patrick Nicodemus (Feb 24 2022 at 14:38):

One could also normalize this complex by killing off degeneracies, I don't recall off the top of my head whether this gives a free resolution

John Baez (Apr 26 2022 at 18:32):

I will be very busy for about the next month... sorry!

Patrick Nicodemus (May 01 2022 at 15:40):

No problem, thank you!