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Stream: learning: questions

Topic: Joy of Cats

James Wood (Apr 14 2020 at 09:33):

I don't have any insight for the actual question, but you need double-$ to get the MathJax rendering. e.g, $$f$$

user 170039 (Apr 17 2020 at 13:43):

I was wondering whether there already exists a theory of digraphs in the same spirit as category theory. To be precise, let's define a digraph $\bf{D}$ as a triple $(\mathcal{O},\operatorname{Hom}, \mathcal{M})$ where $\mathcal{O}$ is the set/class of nodes and $\operatorname{Hom}:\mathcal{O}\times\mathcal{O}\to\mathcal{M}$, where $\mathcal{M}$ is a set/class. An element of $\operatorname{Hom}(A,B)$ for any two $A,B\in \mathcal{O}$ being called an edge from $A$ to $B$.

user 170039 (Apr 17 2020 at 13:47):

More specifically, the goal of such a theory would be to see how far can we generalize the theory of categories in this setting. An important part of this investigation would be to see whether important theorems of category theory (like, e.g., Yoneda Lemma) could also be generalized in this setting in a reasonable manner.

Morgan Rogers (he/him) (Apr 17 2020 at 14:04):

user 170039 said:

I was wondering whether there already exists a theory of digraphs in the same spirit as category theory. To be precise, let's define a digraph $\bf{D}$ as a pair $(\mathcal{O},\operatorname{Hom})$ where $\mathcal{O}$ is the set/class of nodes and $\operatorname{Hom}:\mathcal{O}\times\mathcal{O}\to\mathcal{O}$, an element of $\operatorname{Hom}(A,B)$ for any two $A,B\in \mathcal{O}$ being called an edge from $A$ to $B$.

Is that really the codomain you want for Hom? When you say "in the same spirit as category theory", could you be more specific about what you mean? People certainly study various categories of digraphs, for example, and there are free/forgetful adjunctions between categories of digraphs and categories of categories which allow one to compare these in detail.

user 170039 (Apr 17 2020 at 14:08):

I made a typo. Thanks for pointing it out.

user 170039 (Apr 17 2020 at 14:10):

Also by in the same spirit of category theory, I mean suitable generalization of the notion of epi, mono, iso, functor, natural transformation etc. and study of their properties.

Morgan Rogers (he/him) (Apr 17 2020 at 14:25):

epi, mono, iso and related properties of morphisms no longer make sense for edges because there is no composition operation. The closest you can get to limits and colimits are the existence of cones over/under collections of objects (demanding that triangles commute is no longer meaningful) which could be made weakly universal in the sense of the existence of an edge to the "limit" from the apex of any other cone. Functors are fine, they get replaces with digraph homomorphisms (ie. mappings that preserve the structure), again no longer constrained by the need for composition compatibility. Naturality loses its power, but one can still look for the existence of edges comparing two digraph homomorphisms as the relevant transformations. The absence of the compositional restriction means there will be many more transformations of this kind between digraph homomorphisms in general.

Jules Hedges (Apr 17 2020 at 14:28):

You might be interested in "quivers", which are exactly the same thing as digraphs but the name signals that you think about them as categories minus composition https://en.wikipedia.org/wiki/Quiver_(mathematics)

Morgan Rogers (he/him) (Apr 17 2020 at 14:29):

If you deplete your digraphs, making them simple in the sense that there is at most one edge in each direction between two nodes, it is easier to recover some of the uniqueness that makes universality meaningful, and @sarahzrf might have some meaningful suggestions about that situation :information_desk_person:

sarahzrf (Apr 17 2020 at 14:45):

haha yes my terminology spreads :smiling_devil:

sarahzrf (Apr 17 2020 at 14:49):

well, anyway, in that case you just have a relation on the set of vertices :thinking:

sarahzrf (Apr 17 2020 at 14:49):

an endomorphism in Rel

user 170039 (Apr 17 2020 at 14:56):

Jules Hedges said:

You might be interested in "quivers", which are exactly the same thing as digraphs but the name signals that you think about them as categories minus composition https://en.wikipedia.org/wiki/Quiver_(mathematics)

A quiver in a category (according to Wikipedia) is just a functor, right? How can I think them as categories minus composition?

Reid Barton (Apr 17 2020 at 15:03):

You can't in general. A quiver in a category C would be a "category internal to C minus composition".

Reid Barton (Apr 17 2020 at 15:03):

I would just stick to the case C = Set.

Morgan Rogers (he/him) (Apr 17 2020 at 15:03):

Where are you reading that? The intro of that wikipedia page reads:

In category theory, a quiver can be understood to be the underlying structure of a category, but without composition or a designation of identity morphisms. That is, there is a forgetful functor from Cat to Quiv. Its left adjoint is a free functor which, from a quiver, makes the corresponding free category.

user 170039 (Apr 17 2020 at 15:05):

I just jumped to the section where they give category theoretic definiton. :stuck_out_tongue_wink:

Reid Barton (Apr 17 2020 at 15:06):

I think that section is a bit confusing. At a minimum I wouldn't use their terminology. (Though I see the nlab has the same terminology!)

Morgan Rogers (he/him) (Apr 17 2020 at 15:08):

Ah I see. The functor sends E to the set of edges, V to the set of vertices, and the morphisms to the functions sending each edge to its end-nodes. The reason for this formulation is that it allows us to replace "Set" in the codomain with another category (another topos, or a category of topological spaces, say)

Morgan Rogers (he/him) (Apr 17 2020 at 15:12):

user 170039 said:

I just jumped to the section where they give category theoretic definiton. :stuck_out_tongue_wink:

You need to be pretty comfy with category theory before "structures as functors" becomes a second nature thing :joy:

Jules Hedges (Apr 17 2020 at 15:13):

The other neat thing about the equivalent definition of digraphs as just functors $C \to \mathbf{Set}$ (where $C$ is the category with 2 objects and 2 parallel non-identity morphisms), is that by general abstract nonsense you get that the category of digraphs and digraph homomorphisms is a topos. Which is pretty neat

John Baez (Apr 18 2020 at 17:01):

Category theorists call digraphs = quivers just "graphs".

John Baez (Apr 18 2020 at 17:02):

Just as you can study enriched and internal categories, you can and should study enriched and internal graphs.

John Baez (Apr 18 2020 at 17:02):

For example, there's the "free category on a graph" functor, very important:

$F: \mathsf{Gph} \to \mathsf{Cat}$

It's part of a monadic adjunction: that is, categories are algebras of a monad on $\mathsf{Gph}.$

John Baez (Apr 18 2020 at 17:06):

But we can also look at $\mathsf{C}$-graphs for some category $\mathsf{C}$. Those are things where we have a set $V$ of vertices and an object $E(x,y) \in \mathsf{C}$ any pair of vertices $x,y \in V$.

John Baez (Apr 18 2020 at 17:07):

And if $\mathsf{C}$ is symmetric monoidal cocomplete (with tensor product distributing over colimits), we can form the free $\mathsf{C}$-enriched category on a $\mathsf{C}$-graph, just by copying that construction that works when $\mathsf{C} = \mathsf{Set}$!

John Baez (Apr 18 2020 at 17:08):

So we get

$F : \mathsf{CGph} \to \mathsf{CCat}$

John Baez (Apr 18 2020 at 17:09):

where $\mathsf{CCat}$ is the category of $\mathsf{C}$-enriched categories.

user 170039 (Apr 19 2020 at 13:40):

John Baez said:

Just as you can study enriched and internal categories, you can and should study enriched and internal graphs.``

user 170039 (Apr 19 2020 at 13:42):

Where should I start?

সায়ন্তন রায় (May 16 2020 at 13:29):

I was going through Lambek & Scott's Higher Order Categorical Logic. There they talk about the proposition that for any three categories $\mathbf{A},\mathbf{B}$ and $\mathbf{C}$, $\mathbf{C}^{\mathbf{A}\times \mathbf{B}}$ is isomorphic to $(\mathbf{C}^\mathbf{B})^{\mathbf{A}}$. Does anyone know any quick way of showing this isomorphism?

Nathanael Arkor (May 16 2020 at 14:04):

This follows from the universal property of exponential objects. $\mathscr C(A \times B, C) \cong \mathscr C(A, C^B)$ in any cartesian-closed category $\mathscr C$ because $(-) \times B \dashv (-)^B$, and $\mathbf{Cat}$ is cartesian-closed.

Jules Hedges (May 16 2020 at 14:20):

Arguably (?) it's the other way round, saying "$\mathbf{Cat}$ is cartesian closed" is a slick way to say that these isomorphisms exist (and are natural), but if you want to prove that $\mathbf{Cat}$ is cartesian closed you might still need to do it "by hand"

সায়ন্তন রায় (May 16 2020 at 14:24):

@Jules Hedges, by Currying, I suppose. I was actually looking for some sorter way to do that (i.e., prove that $\mathbf{Cat}$ is cartesian closed) because the proof is pretty long and tedious.

Nathanael Arkor (May 16 2020 at 14:33):

Ah, that's true. Awodey's Category Theory has a proof in section 7.6 that you could use as a starting point. The proof there is very short.

Jules Hedges (May 16 2020 at 15:23):

Incidentally I used to think that this was false, and I was a bit surprised when I found out it was true. (I'm not sure why I thought that, I think I thought that $\mathbf{Cat}$ usually behaved like $\mathbf{Top}$)

John Baez (May 16 2020 at 20:12):

The idea of the proof is pretty simple: if you've got a function from $\mathsf{A} \times \mathsf{B}$ to $\mathsf{C}$ you can feed it guys in $\mathsf{A}$ and be left with functors from $\mathsf{B}$ to $\mathsf{A}$. This whole process is functorial so you get a functor from $\mathsf{C}^{\mathsf{A} \times \mathsf{B}}$ to $(\mathsf{C}^\mathsf{B})^\mathsf{A}$. But you can reverse the argument and get a functor from $(\mathsf{C}^\mathsf{B})^\mathsf{A}$ to $\mathsf{C}^{\mathsf{A} \times \mathsf{B}}$... and you're back where you started from, so these categories are isomorphic.

It can be tiring to check all the details but that barely matters if the idea is so simple and clear: the idea works just like it does for sets.

John Baez (May 16 2020 at 20:20):

Jules Hedges said:

Incidentally I used to think that this was false, and I was a bit surprised when I found out it was true. (I'm not sure why I thought that, I think I thought that $\mathbf{Cat}$ usually behaved like $\mathbf{Top}$)

You could say $\mathsf{Top}$ wants to be cartesian closed, there are just some cases where it fails to be. Thus, most algebraic topologists redefine $\mathsf{Top}$, replacing it with another category that's cartesian closed and calling that category $\mathsf{Top}$.

This is quite a cheeky move: shouldn't topologists study $\mathsf{Top}$? It turns out the for algebraic topology the answer is no. It's a great example of how your first guess about what you're studying may be wrong: your object of study can change as you learn more.

Basically the problem with $\mathsf{Top}$ is that there are too many kinds of topological spaces. Practically nobody seriously studies all topological spaces. Analysts do great things with compact Hausdorff spaces, and other slightly less great things with locally compact Hausdorff spaces (which thankfully include $\mathbb{R}^n$. But show them a Zariski topology and they'll hold their nose - that's for algebraic geometry.

সায়ন্তন রায় (May 25 2020 at 14:49):

I am reading @John Baez's notes on category theory (which, so far is fantastic). There is one thing which I need to clarify. In Definition 5.1. it is written that, "A cone over $F$ is a natural transformation $\alpha : G \implies F$ where $G$ sends every object of $\mathbf{D}$ to some object of $\mathbf{C}$, and $G$ sends every morphism of $\mathbf{D}$ to the identity morphism of that object." (Here both $F,G:\mathbf{D}\to\mathbf{C}$ are functors). My question is this: Does $G$ send every object of $\mathbf{D}$ to some fixed object of $\mathbf{C}$ or not?

Paolo Capriotti (May 25 2020 at 14:53):

Yes, a cone over $F$ is a natural transformation from a constant functor to $F$.

John Baez (May 25 2020 at 20:09):

Some fixed object of $\mathbf{C}$. An example of the millions of improvements that need to be made in these notes.

সায়ন্তন রায় (Jun 21 2020 at 13:38):

What is the categorical understanding of reduced product? (Apparently there is no article on reduced product in nlab.)

Jens Hemelaer (Jun 22 2020 at 08:54):

Not sure if it counts as categorical, but the reduced product has a nice interpretation in terms of sheaf theory.

A collection of sets indexed by a set $I$ is the same thing as a sheaf on $I$ (where $I$ has the discrete topology).
The product of these sets is then the set of global sections of this sheaf.

For the reduced product, you have to look at the space of filters on $I$. This space doesn't have a name as far as I know (if you are only interested in ultrafilters, then this would be the Stone–Čech compactification of $I$). In any case, the topology on the set of filters is given by taking as basis of open sets the sets of the form $\{ F \text{ filter} : F \ni S \}$ where $S$ is a subset of $I$.

Each element $i \in I$ determines a filter: the principal filter consisting of all subsets of $I$ that contain $i$. In this way, the set $I$ (with the discrete topology) is a subspace of the space of filters.

Again a collection of sets indexed by $I$ is a sheaf on $I$ (with the discrete topology). By taking the pushforward along the inclusion map, you get a sheaf on the space of filters. Then to get the reduced product with respect to the filter $F$, take the stalk at (the point corresponding to) $F$. It takes some calculation, but this agrees with the definition for reduced product on Wikipedia.

সায়ন্তন রায় (Jul 17 2020 at 05:11):

Is there any well-known converse of Proposition 10.43 of Joy of Cats?

Morgan Rogers (he/him) (Jul 17 2020 at 10:05):

Context: the referenced result follows the definitions,

10.1 DEFINITION A source is a pair $(A,(f_i)_{i \in I})$ consisting of an object $A$ and a family of morphisms $(f_i: A \to A_i)$ with domain $A$, indexed by some class $I$. $A$ is called the domain of the source and the family $(A_i)_{i \in I}$ is called the codomain of the source.

I am more accustomed to calling this a cosieve or a multispan, depending on context, but moving on,

10.41 DEFINITION Let $\mathbf{A}$ be a concrete category over $\mathbf{X}$. A source $(f_i: A \to A_i)_{i \in I}$ in $\mathbf{A}$ is called initial provided that an $\mathbf{X}$-morphism $f : |B| \to |A|$ is an $\mathbf{A}$-morphism whenever each composite $f_i \circ f : |B| → |A_i|$ is an $\mathbf{A}$-morphism.

I'll hope that readers here can deduce or identify the notation being employed for concrete categories here. Finally,

10.43 PROPOSITION If $(f_i: A \to A_i)_{i \in I}$ is an initial source in $\mathbf{A}$, then
$A = \max \{ B \in \mathrm{ob}(\mathbf{A}) \mid |B| = |A| \text{ and all } f_i: |B| \to |A_i| \text{ are } \mathbf{A} \text{-morphisms} \}$,

where the order being referred to is an order on the fibre of $|A|$ (that is, the collections of objects with $|A| = |B|$) where $B \leq A$ iff $\mathrm{id_{|A|}}: |B| \to |A|$ is an $\mathbf{A}$-morphism.

Morgan Rogers (he/him) (Jul 17 2020 at 10:11):

This formulation of concrete categories seems uncomfortably strict to me, but in answer to your question, unless there are some axioms of concrete categories that I don't know about because I haven't bothered to check, there cannot be a converse: let $\mathbf{X}$ be the commutative triangle category (three objects, $A,B,C$, and three non-identity morphisms $f: A \to B, \, g: B \to C, \, g\circ f: A \to C$). If we consider the inclusion of the subcategory $\mathbf{A}$ containing $g$ and $f \circ g$ as a concrete category over $\mathbf{X}$, it should be clear that $B$ is trivially maximal in its fibre but the source consisting of just the morphism $g$ is not an initial source.

সায়ন্তন রায় (Jul 18 2020 at 14:31):

[Mod] Morgan Rogers said:

This formulation of concrete categories seems uncomfortably strict to me, but in answer to your question, unless there are some axioms of concrete categories that I don't know about because I haven't bothered to check, there cannot be a converse: let $\mathbf{X}$ be the commutative triangle category (three objects, $A,B,C$, and three non-identity morphisms $f: A \to B, \, g: B \to C, \, g\circ f: A \to C$). If we consider the inclusion of the subcategory $\mathbf{A}$ containing $g$ and $f \circ g$ as a concrete category over $\mathbf{X}$, it should be clear that $B$ is trivially maximal in its fibre but the source consisting of just the morphism $g$ is not an initial source.

@[Mod] Morgan Rogers Is there any other way to define concrete categories? Why does it seem so strict to you (see also this nlab page and Remark 2.3)?

Morgan Rogers (he/him) (Jul 18 2020 at 17:31):

The definition implicitly treats objects up to equality rather than up to isomorphism, which is unfortunate when the main intended examples have Set as the base category, because if applied in practice it would force one to be very explicit about what construction of Set one is using. Really you want to be able to talk informally about (for example) groups with six elements without worrying about how those elements are named.

Reid Barton (Jul 18 2020 at 17:40):

In practice the forgetful functor is an isofibration (because you can transfer whatever extra structure is present across an isomorphism of sets) and it doesn't matter whether you use equality or isomorphism.

Morgan Rogers (he/him) (Jul 18 2020 at 20:36):

The definition of concrete category in Joy of Cats is just a faithful functor over the base category; strengthening that to an isofibration would certainly be one solution to the strictness problem, but it makes the categories involved less concrete in the ordinary sense of the word: I might be forced to consider essentially finite categories rather than finite categories if my category of sets satisfies the axiom of replacement, which is technically okay but unsatisfying to me.

Morgan Rogers (he/him) (Jul 18 2020 at 20:39):

And I would like to consider constructible skeleta of categories as concrete categories (eg. the category of finite dimensional vector spaces over $\mathbb{Q}$, whose underlying sets all have the same size) and having results which depend too heavily on my indexing of the elements in the objects is the source of the discomfort I was describing.

সায়ন্তন রায় (Jul 21 2020 at 12:41):

Can anyone give an hint as to how I can show Corollary 10.50 of Joy of Cats? Since it is a corollary of Proposition 10.49 I think we need to show that the pair $(R,E)$ form a Galois correspondence where $R:(\mathbf{B},V)\to(\mathbf{A},U)$ is the concrete reflector and $E:(\mathbf{A},U)\to(\mathbf{B},V)$ is the embedding. But I have not been able to prove that yet.

সায়ন্তন রায় (Jul 21 2020 at 16:02):

Incidentally, if instead $E$ is the inclusion functor then the statement of Corollary 10.50 holds (but the argument that I have seen bypasses showing that $(R,E)$ is a Galois correspondence).

সায়ন্তন রায় (Jul 24 2020 at 12:25):

Recently I stumbled upon this nlab page on topological concrete categories. There in "Further Properties" it is written that a reflective or coreflective subcategory of $\mathbf{C}$ is topological, as long as the reflectors or coreflectors become identity morphisms in $\mathbf{D}$. Can anyone give an outline of a proof of this result?

সায়ন্তন রায় (Aug 02 2020 at 04:27):

In Joy of Cats, a diagram in a category $\mathbf{A}$ is said to be a functor $D:\mathbf{I}\to\mathbf{A}$. I am wondering as why we didn't define a diagram in a category $\mathbf{A}$ to be an embedding instead of an arbitrary functor? In the examples that I have seen so far, the notion of a diagram as an embedding seems to match more closely to intuition.

Ralph Sarkis (Aug 02 2020 at 10:51):

This would not let us define the product $X \times X$.

সায়ন্তন রায় (Aug 02 2020 at 14:09):

I don't understand. Would you mind to elaborate @Ralph Sarkis ?

সায়ন্তন রায় (Aug 02 2020 at 14:09):

For me, a diagram is essentially a graph (ignoring size issues for the time being). Now if you give me a category $\mathbf{A}$, the diagrams "inside" $\mathbf{A}$ to my mind are really the subgraphs of the underlying graph of $\mathbf{A}$. If I am not wrong then these subgraphs are essentially the underlying graphs of the subcategories of $\mathbf{B}$ (ignoring the identity arrows for the time being). Hence the question.

Ralph Sarkis (Aug 02 2020 at 15:02):

Let $\mathbf{C}$ be a category and $X \in \text{Ob}\mathbf{C}$. The product $X \times X$ is the limit of the diagram $\mathbf{X}: \mathbf{1} + \mathbf{1} \rightarrow \mathbf{C}$, where the source category only has two objects with identities only and the functor sends both objects and identities to $X$ and $\text{id}_X$ respectively. The functor $\mathbf{X}$ is not an embedding because it is not injective on objects.

I think your intuition with the graphs is great to have (and I have the same one) but it is not entirely correct as witnessed by the example above.

Morgan Rogers (he/him) (Aug 03 2020 at 18:10):

@ADITTYA CHAUDHURI I guess you meant the two distinct objects in $\mathbf{1}+\mathbf{1}$? Either way, the functor $\mathbf{1} + \mathbf{1} \to \mathbf{C}$ described by Ralph is faithful, but is a counterexample to what Sayantan was saying since it isn't injective on objects. Interestingly for @Sayantan Roy, we can actually assume without loss of generality* that the functor defining a diagram is faithful, since the conditions on cones over or under a diagram (which realistically is all we're interested in) only apply to individual morphisms, so if there are two parallel, equal morphisms in a diagram, the commutativity condition on cones holds for one if and only if it holds for the other.

*(at least in a setting with decidable equality!!)

ADITTYA CHAUDHURI (Aug 03 2020 at 18:58):

@[Mod] Morgan Rogers Sorry! It was a typo. Actually I wanted to say Full instead of Faithful. I mistakenly remember that both fullness and faithfulness are required in the definition of Embedding but now I remember that fullness is not a criteria in the definition of embedding whereas being injective on objects and being faithful are the only necessary conditions for it to be an embedding. What I said above, it is actually the definition of Full embedding. I am deleting my comment above and again I am deeply apologising for such a bad mistake and for the inconvenience caused due to it.

Morgan Rogers (he/him) (Aug 04 2020 at 09:47):

@ADITTYA CHAUDHURI no need to worry! It was only a tiny error, and I wouldn't have noticed the point I made about faithfulness without it. There's nothing wrong with making mistakes, this is a very friendly place :grinning_face_with_smiling_eyes:

ADITTYA CHAUDHURI (Aug 04 2020 at 11:02):

@[Mod] Morgan Rogers Thank you Sir.

সায়ন্তন রায় (Aug 24 2020 at 14:32):

Currently I am reading about complete categories. However, I am wondering if anyone can give me any insight/motivation/big-picture view about the reason for using the term "complete" here.

Reid Barton (Aug 24 2020 at 14:46):

A metric space is complete if it has limits, in some sense. I wouldn't look for a deeper meaning than that--although you can say a bit more about complete partial orders, this doesn't necessarily generalize in a useful way to categories that aren't partial orders.

Reid Barton (Aug 24 2020 at 14:50):

Sort of related MO question: https://mathoverflow.net/questions/9951/limits-in-category-theory-and-analysis

সায়ন্তন রায় (Aug 24 2020 at 15:27):

@Reid Barton This is interesting. But the construction looks pretty unnatural to me. If $(X,\tau)$ is a Hausdorff topological space then, I guess I was looking for a category whose objects are the elements of $X$ and whose limits are precisely the limit points of different nets. But I couldn't determine what would be a reasonable morphism in this category. I think this is what was also suggested in this answer.

সায়ন্তন রায় (Aug 24 2020 at 15:30):

The reason I included Hausdorffness is due primarily to ensure that "uniqueness of limits" (in the category) coincides with "uniqueness of limits" (in the topological space)

Jens Hemelaer (Aug 24 2020 at 18:49):

I would also say that there is no deeper meaning here, aside from the analogy "having limits = complete".

I don't think the category that you are looking for (with the elements of $X$ as objects) exists. Take for example the real line. The translations are continuous homeomorphisms of the real line, and they act transitively on the points. So all points "look the same". In the corresponding category the points should all be isomorphic, and (up to equivalence) you get a category with just one object. So limits in this category do not exist or are equal to the unique object in the category.

John Baez (Aug 24 2020 at 20:48):

Yeah it's basically a pun: a metric space is complete if all Cauchy sequence has limits, so we call a category "complete" if all diagrams have limits.

John Baez (Aug 24 2020 at 20:50):

But a more useful intuition is that we want categories to be complete because to do so math in a category we often want products, equalizers, and a terminal object - and if all of these exist, our category is complete. (This is a theorem.)

John Baez (Aug 24 2020 at 20:53):

I guess I was looking for a category whose objects are the elements of X and whose limits are precisely the limit points of different nets.

I don't think you'll succeed. I don't think limits in category theory are like limits in topology in this way.

John Baez (Aug 24 2020 at 21:01):

I think everyone tries doing this...

Morgan Rogers (he/him) (Sep 02 2020 at 11:20):

Jens Hemelaer said:

Take for example the real line.

This is actually one of the few examples where the two senses of limit can be contrived into being analogous, although not to the point of making the two different types of completeness match up nicely. Consider $\mathbb{R}$ as a topological space on one hand and as a totally ordered set (in particular, a special case of a category) on the other. Consider a monotone decreasing sequence, which corresponds to a diagram of the form:
$\cdots \rightarrow \bullet \rightarrow \bullet \rightarrow \bullet$
in the totally ordered set. The limit in the analytic sense exists if and only if the limit in the categorical sense exists, and the two coincide. We can extend this observation to limits of "cofiltered diagrams", but no further, since the limit in the categorical sense amounts to taking the minimum of all of the elements appearing as objects in the diagram.

Morgan Rogers (he/him) (Sep 02 2020 at 11:27):

Thinking about this further, there might be an intrinsic categorical way to describe analytic completeness. If we consider the poset of open sets of a (sober) topological space $X$, and take presheaves on it (not sheaves!) then the resulting topos is equivalent to the topos of sheaves on a space having a point for each point of $X$, but also a point for each open set in $X$ (and some other points), since its points simply correspond to upward-closed collections of open sets of $X$. I think @Jens Hemelaer described this elsewhere on Zulip. This new space provides a nice setting for comparing the points and opens of the original space on an equal footing. The challenge is to work out whether the idea of a Cauchy sequence can be adequately re-expressed in this setting.

Jens Hemelaer (Sep 02 2020 at 12:52):

Here are some ideas:

Completeness doesn't make sense for topological spaces, only for metric spaces. For example the plane with one point removed is not complete, but it is homeomorphic to an infinite cylinder which is complete. So you need to keep some information about the metric. Instead of looking at all the open sets, you can maybe restrict to the open balls $B(x,\tfrac{1}{n}) = \{ y \in X : d(y,x)<\tfrac{1}{n} \}$ for $x \in X$ and $n \in \mathbb{N}$.

The open balls $B(x,\tfrac{1}{n})$ are a partially ordered set $P$ under inclusion, and as analogue of Cauchy sequences you can take subsets $S \subseteq P$ that are:

• upwards closed: $p \in S \wedge p \leq q \Rightarrow q \in S$
• downwards directed: $p,q \in S \Rightarrow r \in S,~ r\leq p,~r \leq q$).

Subsets like this are called filters, and filters are the same thing as formal cofiltered limits (the partially ordered set of filters is the category of pro-objects of the original partially ordered set).

For example, for a point $x \in X$, you can define $S_x = \{ p \in P : x \in p \}$. Now you could say that $X$ is complete if all filters $S \subseteq P$ are of the form $S = S_x$ for some point $x \in X$. But this is too strong... for example, as @[Mod] Morgan Rogers says, each $p \in P$ defines a filter as well, namely the filter $S_p = \{ q \in P : q \geq p \}$. And we can't find a point $x \in X$ such that $S_p = S_x$, except in the very special case where $p = \{x\}$.

One remedy is to add some extra conditions the filter should satisfy. There is a morphism of partially ordered sets $\mu: P \to \mathbb{R}$, sending each $B(x,\tfrac{1}{n})$ to the real number $\frac{1}{n}$. Then we say that a filter $S \subseteq P$ is shrinking if for all $\epsilon > 0$ there is some $p \in P$ such that $\mu(p) < \epsilon$.

Question 1. Is it true that $X$ is complete if and only if every shrinking filter $S \subseteq P$ can be written as $S = S_x$ for some $x \in X$?

Question 2. What does the space of all shrinking filters look like? As basis of open sets it has the sets $U_p$, where $U_p$ is the set of all shrinking filters that contain $p$.

সায়ন্তন রায় (Sep 04 2020 at 14:51):

Let $\mathbf{A}$ be a category. Suppose that $\mathbf{A}$ has finite products. What additional properties must $\mathbf{A}$ satisfy so that $\mathbf{A}$ has products as well? Are those properties necessary?

সায়ন্তন রায় (Sep 04 2020 at 15:00):

(The motivation for this question is the proof of Proposition 12.5 of Joy of Cats. I don't understand how the authors claim that the family $(A_j\overset{m_j}{\to}A)_J$ has an intersection $(B,m)$ because by Propostion 12.3 we can only say that if $J$ is finite then it has an intersection.)

Morgan Rogers (he/him) (Sep 04 2020 at 15:10):

In the statement of Proposition 12.5, the hypothesis is that $\mathbf{A}$ is complete (not just finitely complete), so you're not missing anything, don't worry :grinning_face_with_smiling_eyes:

সায়ন্তন রায় (Sep 04 2020 at 15:18):

[Mod] Morgan Rogers said:

In the statement of Proposition 12.5, the hypothesis is that $\mathbf{A}$ is complete (not just finitely complete), so you're not missing anything, don't worry :grinning_face_with_smiling_eyes:

But then how can I prove that family $(A_j\overset{m_j}{\to}A)_J$ has an intersection $(B,m)$?

Morgan Rogers (he/him) (Sep 04 2020 at 15:39):

Consider the limit of the sink diagram. You can check, using the fact that all of the $m_j$ are monic, that all of the legs of the limit cone must be monic, and so that the mono from the limit to $A$ (obtained by composing a limit leg with the relevant $m_j$) represents a subobject which is the intersection of all of the $m_j$. The step of reducing to the set $J \subseteq I$ is just to allow $I$ to be an arbitrary class.

Morgan Rogers (he/him) (Sep 04 2020 at 15:41):

The point of Theorem 12.3(3) is to demonstrate a reduced necessary and sufficient condition. That is, to check that a category is complete, you just need to verify it has all small products and finite intersections. Then you get arbitrary small limits (including arbitrary small intersections) for free.

সায়ন্তন রায় (Sep 09 2020 at 13:04):

Can anyone give me some an intuitive/big-picture view of a colimit-dense category? Its definition, according to Joy of Cats is,

A full subcategory $\bf{B}$ of a category $\bf{A}$ with embedding $E : \bf{B} \to \bf{A}$ is called colimit-dense in $\bf{A}$ provided that for every $\bf{A}$-object $A$ there exists a diagram $D : \bf{I}\to \bf{B}$ such that the diagram $E \circ D : \bf{I} \to \bf{A}$ has a colimit with codomain $A$."

Roughly speaking to me an informal way of understanding the definition is to say that the category "has enough colimits", or in other words, every object can be "reached" via colimits of $E\circ D$. Is it a good intuition to have regarding this notion?

Also I have some questions,

1. Why does $\mathbf{B}$ need to be a full subcategory? The definition would still make good sense if $\bf{B}$ is just a subcategory of $\mathbf{A}$ (not necessarily full).
2. Is there any generalization of this notion via functors? The following generalization almost immediately comes to my mind (although I am not sure about its usefulness),

Let $\mathbf{A}$ and $\mathbf{B}$ be two categories and $F : \bf{B} \to \bf{A}$ is a functor. Then $F$ is called colimit-dense in $\bf{A}$ provided that for every $\bf{A}$-object $A$ there exists a diagram $D : \bf{I}\to \bf{B}$ such that the diagram $F \circ D : \bf{I} \to \bf{A}$ has a colimit with codomain $A$."

Morgan Rogers (he/him) (Sep 09 2020 at 13:32):

I think Tom Leinster would agree with scepticism of the need for fullness and your generalization, based on his n-cat cafe post where he mentions his "suspicion of subcategories". Your summary is a good one; the definition is saying that every object of $\mathbf{A}$ can be expressed as a colimit of $\mathbf{B}$-objects and $\mathbf{B}$-morphisms.

Since it doesn't seem to be explicitly pointed out in that chapter of Joy of Cats, I should mention that the canonical example of colimit denseness is the subcategory of representable functors in a presheaf (or more generally sheaf) category; the fact that the representables are colimit-dense in a presheaf category is sometimes called the co-Yoneda Lemma. Since the Yoneda embedding is full and faithful, perhaps there was not much incentive for the authors to think too hard about this.

সায়ন্তন রায় (Sep 09 2020 at 14:06):

That post by @Tom Leinster is truly excellent and I find myself agreeing with a lot of points.

সায়ন্তন রায় (Sep 20 2020 at 06:46):

Let $\mathbf{A}$ be a category and $(A_i)_{i\in I}$ is a family of $\mathbf{A}$-objects. Let $\emptyset\subset J\subseteq I$ and let $\mathcal{P}:=\left(\displaystyle\prod_{i\in I}A_i\overset{\pi_k}{\to} A_k\right)_{k\in I}$ and $\mathcal{Q}:=\left(\displaystyle\prod_{i\in J}A_i\overset{p_k}{\to} A_k\right)_{k\in J}$ be the products of $(A_i)_{i\in I}$ and $(A_i)_{i\in J}$ respectively. Is it true that $\displaystyle\prod_{i\in J}A_i$ is a retract of $\displaystyle\prod_{i\in I}A_i$? What if all $A_i$'s are same?

সায়ন্তন রায় (Sep 20 2020 at 06:55):

I can only observe so far that since $\mathcal{Q}$ is a product, it follows that there exists a unique $\mathbf{A}$-morphism, say $\displaystyle\prod_{i\in I} A_i\overset{\varphi}{\to}\displaystyle\prod_{i\in J} A_i$ such that $p_k\circ\varphi=\pi_k$ for all $k\in J$. Apparently I need to show that $\varphi$ is a retraction. But I am stuck at this point.

Ralph Sarkis (Sep 20 2020 at 07:52):

Here is a counterexample. Take $\mathbf{A} = \mathcal{P}(\{0,1,2\})$ the category corresponding to the poset $(\mathcal{P}(\{0,1,2,\}, \subseteq)$. In this category, the product is the intersection. Taking the intersection of all two element sets $\{0,1\} \cap \{1,2\} \cap \{0,2\}$ yields the empty set while taking the intersection of two of these sets yields a singleton. We have $\emptyset \subseteq \{x\}$ for any $x \in \{0,1,2\}$, but this inclusion cannot be a retratct.

A condition that would be helpful is the existence of morphisms $p_i: \prod_{j \in J} A_j \rightarrow A_i$ for any $i \in I\setminus J$. If you have these, then you can consider the family $\{p_j\}_{j \in J} \cup \{p_i\}_{i \in I \setminus J}$ and use the universality of $\prod_{i \in I} A_i$ to obtain $\psi: \prod_{j \in J} A_j \rightarrow \prod_{i \in I} A_i$ such that $\pi_j \circ \psi = p_j$. We can conclude that $p_j \circ \varphi \circ \psi = p_j$, thus $\varphi \circ \psi$ is the identity by universality of $\prod_{j \in J} A_j$.

সায়ন্তন রায় (Sep 20 2020 at 08:02):

@Ralph Sarkis Indeed. Actually, the problem at which I am stuck is precisely in trying to extend the class of morphisms indexed by $J$ to the class of morphisms indexed by $I$. Even in the case when all $A_i$'s are same, I have no clue regarding how to do that.

Ralph Sarkis (Sep 20 2020 at 09:30):

In my derivation, I only use the existence of such $p_i$'s. In particular, it doesn't matter what they are and if $A_i = A_j$, then you can take $p_i = p_j$.

সায়ন্তন রায় (Sep 20 2020 at 13:54):

So, just to confirm whether I have got it right, in any category $\mathbf{A}$, if $J\subseteq I$ then $A^J$ is always a retract of $A^I$, right? (Provided of course both of them exist.)

Ralph Sarkis (Sep 20 2020 at 13:56):

Are $A^I$ and $A^J$ the functor categories ?

সায়ন্তন রায় (Sep 20 2020 at 13:56):

No, they are just the $I$-th and $J$-th power of $A$ in $\mathbf{A}$.

Ralph Sarkis (Sep 20 2020 at 13:57):

Then yes.

Reid Barton (Sep 20 2020 at 14:14):

Not quite always, there's a problem if $J$ is empty (and $A$ is the empty set, say).

John Baez (Sep 20 2020 at 15:32):

Indeed, $\emptyset^\emptyset$ is not a retract of $\emptyset^1$ because the 1-element set is not a retract of the empty set.

সায়ন্তন রায় (Sep 21 2020 at 14:53):

Indeed. I forgot to mention that $\emptyset \subset J$. This result is actually used in the proof of Theorem 12.13 of Joy of Cats which I am now trying to understand. I am however stuck at the following phrase, "Since $\mathbf{A}$ is co-wellpowered, (2) (see below) immediately implies that $\bf{A}$ is wellpowered." Since the authors say that the conclusion is immediate, it must be easy, but I can't figure it out. Any hint(s)?

For the sake of completeness, the proof of Theorem 12.13 uses the following two results, (1) For every $\mathbf{A}$-source $\mathcal{S}$ there exist an epimorphism $e$ and a mono-source $\mathcal{M}$ with $\mathcal{S} = \mathcal{M}\circ e$ and (2) If $\mathcal{M}:=(A\overset{m_i}{\to}A_i)$ is a small mono-source, then $A$ is a quotient object of $^{\prod_I\text{hom}(S,A_i)}S$ or a quotient object of $^\emptyset S$ where $S$ is a separator in $A$.

সায়ন্তন রায় (Oct 11 2020 at 15:31):

I am having trouble understanding a terminal object as a limit over the empty diagram. Can anyone outline any formal proof of it?

Dan Doel (Oct 11 2020 at 15:34):

I believe a 'cone' on the empty diagram is just an object. Does that help?

সায়ন্তন রায় (Oct 11 2020 at 15:35):

Dan Doel said:

I believe a 'cone' on the empty diagram is just an object. Does that help?

You beat me by a couple of seconds. I have heard that too. But it is precisely this point that I don't get. Can it be proven formally?

Dan Doel (Oct 11 2020 at 15:36):

I'll have to go look at the precise definition of a cone. :)

Dan Doel (Oct 11 2020 at 15:41):

Okay, so, given a functor $F : J → C$, a cone on $F$ is an object $A : C$ and maps $ψ_X : A → F(X)$ that commute with $F(f) : F(X) → F(Y)$. However, when $J$ is empty, there are no objects $X$ nor maps $f$, so that extra structure is just absent.

Dan Doel (Oct 11 2020 at 15:41):

And that leaves you with just the object $A$.

Dan Doel (Oct 11 2020 at 15:50):

If you use the natural transformation definition of $ψ$, then it is the empty natural transformation, which is empty because its indexing set is empty; there are no components $ψ_X$ because there are no objects $X$ of the empty category.

সায়ন্তন রায় (Oct 11 2020 at 15:51):

Dan Doel said:

Okay, so, given a functor $F : J → C$, a cone on $F$ is an object $A : C$ and maps $ψ_X : A → F(X)$ that commute with $F(f) : F(X) → F(Y)$. However, when $J$ is empty, there are no objects $X$ nor maps $f$, so that extra structure is just absent.

What is $A:C$? Also when $J$ is empty then what the definition of the functor explicitly? I am confused at this point because I always read the definition of a functor $F:\mathbf{A}\to \mathbf{B}$ as saying, e.g., the following:

(1) If $a$ is an $\mathbf{A}$-object then $F(a)$ is a $\mathbf{B}$-object.
(2) If $a$ is an $\mathbf{A}$-object then $F(id_{a})=id_{F(a)}$.
(3) If $g,f$ are $\mathbf{A}$-morphisms and if $g\circ f$ exists then $F(g\circ f)=F(g)\circ F(f)$.

Dan Doel (Oct 11 2020 at 16:01):

$A : C$ was meant to mean that $A$ is an object of $C$. If $J$ is empty, then we know that $F$ is the unique empty functor. All the conditions 1-3 are vacuous, because there are no objects $a$, and no morphisms $f,g$.

Dan Doel (Oct 11 2020 at 16:04):

It is like the empty function. There are no values to map anywhere.

সায়ন্তন রায় (Oct 11 2020 at 16:08):

I see. I think I understand now. Thanks.

Dan Doel (Oct 11 2020 at 16:11):

No problem.

সায়ন্তন রায় (Oct 18 2020 at 03:53):

Theorem. Embeddings of dense subcategories preserve limits.

Part of the Proof. Let $\bf{A}$ be a dense subcategory of $\mathbf{B}$ with embedding $E : \bf{A}\to\bf{B}$, and let $D : \bf{I}\to \bf{A}$ be a diagram with a limit $\mathcal{L} = (L\overset{l_i}{\to}D(i))_{i\in \text{Ob}(\mathbf{I})}$. Then $\mathcal{L}$ is a cone over $E\circ D$. Let $\mathcal{S} = (B\overset{f_i}{\to}D(i))_{i\in \text{Ob}(\mathbf{I})}$ be an arbitrary cone over $E\circ D$. By density there exists a diagram $G : \bf{J}\to \bf{A}$ and a colimit $(G(j)\overset{c_j}{\to} B)_{j\in \text{Ob}(\mathbf{J})}$ of $E \circ G$. For each object $j$ of $\text{Ob}(\bf{J})$, $(G(j)\overset{f_i\circ c_j}{\to} D(i))_{i\in \text{Ob}(\mathbf{I})}$ is a cone over $D$. Hence for each $j \in \text{Ob}(\bf{J})$ there exists a unique morphism $g_j : G(j)\to L$ with $f_i \circ c_j = \ell_i \circ g_j$ for each $i \in \text{Ob}(\bf{I})$....

In this proof of Proposition 13.11 of Joy of Cats, I don't understand the reason for writing, the last line. Of course this will happen if $f_i \circ c_j$ is an $\mathbf{A}$-morphism. But I am not sure how that is true unless $\mathbf{A}$ is a full subcategory of $\mathbf{B}$. Could anyone explain?

Morgan Rogers (he/him) (Oct 18 2020 at 09:49):

Indeed, the proof does rely on fullness; the definition on nLab is bad, since it doesn't rely make sense to talk about "a colimit of objects". Fortunately, the definition in use in Joy of Cats, Definition 12.10, includes fullness in the definition of colimit-dense subcategory. :+1:

সায়ন্তন রায় (Oct 18 2020 at 14:07):

I don't know how I forgot that. :sad:

John Baez (Oct 18 2020 at 15:18):

Please fix the definition on the nLab if it's bad!

Reid Barton (Oct 18 2020 at 15:25):

I think the first nLab definition intentionally does not assume that $D$ is a full subcategory, though I never really understood what happens when it isn't.

Reid Barton (Oct 18 2020 at 15:36):

For example, from the start of section 5.11 of Kelly's book:
"From the beginning of 5.4 until now we have dealt only with fully-faithful dense functors–exploiting the notion of a density presentation, which does not seem to generalize usefully to the non-fully-faithful case. Now we drop this restriction, and consider an arbitrary dense $K:A \to C$."
Kelly is using the same definition of density (see Theorem 5.1), but in the enriched context.

Reid Barton (Oct 18 2020 at 15:45):

Also, the "dense" being referred to on the nLab is different from the "colimit-dense" of Joy of Cats Definition 12.10 (not just not the same by definition, but actually different). Mike Shulman has a useful article which talks about the relationships between these and related notions.

Dan Doel (Oct 18 2020 at 16:27):

Is the nlab definition saying that every object of $C$ is a colimit of some $F : J → C$ where just $F_0$ factors through $i_0 : D → C$ or something?

Dan Doel (Oct 18 2020 at 16:33):

I guess that's related to the density functor $\int^{d\in D} C(i(d), -) \otimes i(d)$ not involving the homs of $D$?

Dan Doel (Oct 18 2020 at 16:36):

(I mean, it does, in that the coend involves talking about the homs of $D$ but...)

Reid Barton (Oct 18 2020 at 16:37):

The nLab definition of dense says that every object of $C$ is canonically a colimit of objects of $D$--it doesn't explain what that means but what it means is this: For any object $X$ of $C$ there's always a canonical cocone with vertex $X$ and one object in the diagram for every object of $D$ together with a map to $X$. The maps in the diagram are the maps in $D$ which commute with the specified map to $X$. When this diagram is a colimit, we say that $X$ is "canonically a colimit of objects of $D$".

Reid Barton (Oct 18 2020 at 16:37):

The coend you wrote down is the colimit of this diagram, when it exists

Reid Barton (Oct 18 2020 at 16:39):

and it does depend on which maps belong to $D$, but what I guess you mean is that it makes sense even when $D$ is not a full subcategory, which is true. However, I've never seen it come up except when $D$ is in fact a full subcategory.

Reid Barton (Oct 18 2020 at 16:40):

I think the notion of colimit-dense also makes sense even when $D$ is not full, and it's what you wrote in your first message.

Dan Doel (Oct 18 2020 at 16:40):

Ah, okay.

Morgan Rogers (he/him) (Oct 18 2020 at 21:53):

Reid Barton said:

and it does depend on which maps belong to $D$, but what I guess you mean is that it makes sense even when $D$ is not a full subcategory, which is true. However, I've never seen it come up except when $D$ is in fact a full subcategory.

Might that be because as soon as $D$ has coequalizers (say), every morphism between objects of $C$ must lie in a diagram of the form you describe?

সায়ন্তন রায় (Dec 03 2020 at 04:24):

I think I will need to reformulate my question in more detail. So I removed the previous version and here we go.

In Joy of Cats the definition of concrete limits is given as follows (cf. Definition 13.12):

Let $(\mathbf{A},U)$ be a concrete category. A limit $\mathcal{L}$ of a diagram $D:\mathbf{I}\to\mathbf{A}$ is called a concrete limit of $D$ in $(\mathbf{A},U)$ provided that  it is preserved by $U$.

How do we understand the definition? Is it saying that a source $\mathcal{L}$ in $\mathbf{A}$ is called a concrete limit of $D:\mathbf{I}\to\mathbf{A}$ in $(\mathbf{A},U)$ iff,

1. $\mathcal{L}$ is a limit of $D:\mathbf{I}\to\mathbf{A}$ and $U(\mathcal{L})$ is a limit of $U\circ D$.

or,

2. $U(\mathcal{L})$ is a limit of $U\circ D$ whenever $\mathcal{L}$ is a limit of $D$.

সায়ন্তন রায় (Dec 03 2020 at 04:29):

In the former case, I think one can prove the following theorem (which I believe is one way of interpreting Theorem 13.15),

If $(\mathbf{A},U)$ is a concrete category and $D : \mathbf{I} \to \mathbf{A}$ is a diagram, then an $\mathbf{A}$-source $\mathcal{L}$ is a concrete limit in $(\mathbf{A},U)$ if and only if $U(\mathcal{L})$ is a limit of $U \circ D$ and $\mathcal{L}$ is an initial source in $(\mathbf{A},U)$.

However, I don't see how I can prove this result if I assume (2).

Fawzi Hreiki (Dec 03 2020 at 09:02):

I understand the definition to take for granted that the limit already exists in the concrete category. Then such a limit is said to be concrete iff it’s preserved by the faithful functor to sets.

Fawzi Hreiki (Dec 03 2020 at 09:04):

So concreteness in this case is not a property of cones but of limits

সায়ন্তন রায় (Dec 03 2020 at 13:37):

So it's (2), right?

Reid Barton (Dec 03 2020 at 13:39):

I think it's (1). I don't even understand what (2) would mean.

Reid Barton (Dec 03 2020 at 13:40):

$D$ and $\mathcal{L}$ are fixed things. So there is no "whenever".

Reid Barton (Dec 03 2020 at 13:41):

Unless you mean "if $\mathcal{L}$ is a limit of $D$, then $U(\mathcal{L})$ is a limit of $U \circ D$". But this doesn't make sense because then anything which isn't a limit would be a concrete limit.

Reid Barton (Dec 03 2020 at 13:42):

Or as Fawzi said, the original definition tells you when a limit is a concrete limit. If you start with just a source (which I think is what the rest of us call a cone) and ask when it is a concrete limit, then it first needs to be a limit.

সায়ন্তন রায় (Dec 03 2020 at 14:20):

Reid Barton said:

Unless you mean "if $\mathcal{L}$ is a limit of $D$, then $U(\mathcal{L})$ is a limit of $U \circ D$". But this doesn't make sense because then anything which isn't a limit would be a concrete limit.

Is it because of the false antecedent? I get it. Although I wasn't aiming that. Now when you point that out, I feel really bad!

Anyway, I understood the definition as (1), however, I wasn't sure that my understanding is what the original definition was saying.

John Baez (Dec 03 2020 at 21:55):

Sayantan Roy said:

In Joy of Cats the definition of concrete limits is given as follows (cf. Definition 13.12):

Let $(\mathbf{A},U)$ be a concrete category. A limit $\mathcal{L}$ of a diagram $D:\mathbf{I}\to\mathbf{A}$ is called a concrete limit of $D$ in $(\mathbf{A},U)$ provided that  it is preserved by $U$.

How do we understand the definition? Is it saying that a source $\mathcal{L}$ in $\mathbf{A}$ is called a concrete limit of $D:\mathbf{I}\to\mathbf{A}$ in $(\mathbf{A},U)$ iff,

1. $\mathcal{L}$ is a limit of $D:\mathbf{I}\to\mathbf{A}$ and $U(\mathcal{L})$ is a limit of $U\circ D$.

or,

2. $U(\mathcal{L})$ is a limit of $U\circ D$ whenever $\mathcal{L}$ is a limit of $D$.

I think they mean 1.

The second alternative doesn't even make sense, because it involves an (implicit) universal quantifier over $\mathcal{L}$, while the phrase being defined does not. I read your "whenever" as meaning "for all $\mathcal{L}$".

John Baez (Dec 03 2020 at 21:56):

Okay, good: Reid Barton agrees with me:

Reid Barton said:

I think it's (1). I don't even understand what (2) would mean.

John Baez (Dec 03 2020 at 21:59):

Definition (2) has the same logical structure as this:

"A cat is cute if whenever you look at a cat you smile."

John Baez (Dec 03 2020 at 22:00):

That doesn't make sense because "whenever you look at a cat you smile" doesn't refer at all to the cat being discussed: it refers to all cats.

John Baez (Dec 03 2020 at 22:00):

Here's a definition that avoids this logical problem:

"A cat is cute if whenever you look at that cat you smile."

Chad Nester (Dec 03 2020 at 22:03):

The onus is on you to find a cat that is not cute.

Dan Doel (Dec 03 2020 at 22:06):

Even if all cats are cute, it doesn't mean that #2 makes sense as a definition.

John Baez (Dec 03 2020 at 22:08):

Yes, it's possible that

"A cat is cute if whenever you look at a cat you smile."

is a way of defining "cute" that secretly means "all cats are cute", or perhaps "no cats are cute". But I think we can agree it's a bad style of definition, just like #2.

সায়ন্তন রায় (Dec 06 2020 at 06:14):

In Remark 13.26 of Joy of Cats, an idea for "limit-closed" subcategories of a category is given. However, the definition for the same is not mentioned explicitly. From the proof of Proposition 13.27 I get the following idea,

Let $\mathbf{B}$ be a category and $\mathbf{A}$ be a full(?) subcategory of $\mathbf{B}$. Then $\mathbf{A}$ is said to be limit-closed if for any diagram $D:\mathbf{I}\to\mathbf{A}$ and for any $\mathbf{B}$-source $\mathcal{L}:=(L\overset{l_i}{\to}D(i))_{i\in \text{Ob}(\bf{I})}$ if $\mathcal{L}$ is a limit of $E\circ D$ (where $E:\mathbf{A}\to\mathbf{B}$ is the inclusion functor) then $\mathcal{L}$ is an $\mathbf{A}$-source.

Is this correct?

সায়ন্তন রায় (Jun 10 2021 at 15:21):

So, after a gap of six months, I have begun my self-study of category theory. Currently I am trying to read Chapter IV of Joy of Cats. On page 237, they say that,

[C]ategorists have created an axiomatic theory of factorization structures $(E, M)$ for morphisms of a category $\mathbf{A}$. Here $E$ and $M$ are classes of A-morphisms such that each $\mathbf{A}$-morphism has an $(E, M)$-factorization $A\overset{f}{\rightarrow} B=A\overset{e\in E}{\longrightarrow} f[A]\overset{m\in M}{\longrightarrow} B$. Naturally, without further assumptions on $E$ and $M$ such factorizations might be quite useless.

I am wondering about the last remark. Of course, it is true that "without further assumptions on $E$ and $M$ such factorizations might be quite useless", but I don't seem to understand how the requirement of putting "further assumptions" lead us naturally to Definition 14.1. For example, what would be the problem(s) with the following definition,

Definition. A category $\mathbf{A}$ is said to have an $(E,M)$-factorization structure if,

• $E\cap M\ne \emptyset$ and $E\cup M\ne \sf{Mor}(\bf{A})$.

• $\text{id}_A\notin E\cup M$ for all $\bf{A}$-object $A$.
  

David Michael Roberts (Jun 10 2021 at 15:26):

I think the issue is then what are you going to use this factorisation structure for? The existing (stronger) notions satisfy properties that get used in practice (for instance existence of lifts in commuting squares).

সায়ন্তন রায় (Jun 10 2021 at 15:29):

I am more interested in the purely theoretical aspect. As an example, I would be very much interested to know that given $E$ and $M$, when we do have an $(E,M)$-factorization.

সায়ন্তন রায় (Jun 10 2021 at 15:31):

Also, for me the definition of "diagonal" condition doesn't seem intuitive enough. Is there any alternative way to phrase it using more familiar concepts (like, e.g., (co)limits, initial/terminal objects etc.) so that I get a big picture understanding of its essence?

Morgan Rogers (he/him) (Jun 10 2021 at 15:34):

Re your proposed alternative, what would you expect the factorization of an identity morphism to be, if identities are excluded from both classes?

Morgan Rogers (he/him) (Jun 10 2021 at 15:37):

সায়ন্তন রায় said:

As an example, I would be very much interested to know that given $E$ and $M$, when we do have an $(E,M)$-factorization.

Interestingly, the verification that an $(E,M)$-factorization exists is typically a non-categorical result, unless it is inherited from another category (via monadicity, say). Orthogonality (the left and right lifting properties) are used to show uniqueness once existence is established.

Morgan Rogers (he/him) (Jun 10 2021 at 15:39):

(I would be happy to have that claim refuted, btw)

সায়ন্তন রায় (Jun 10 2021 at 15:41):

Morgan Rogers (he/him) said:

Re your proposed alternative, what would you expect the factorization of an identity morphism to be, if identities are excluded from both classes?

Unless I am making a big mistake, could it not simply be any two morphisms $e$ and $m$ such that $\text{id}_A =m\circ e$ where $e\in E$ and $m\in M$?

Spencer Breiner (Jun 10 2021 at 15:49):

Hi @সায়ন্তন রায়,

This isn't specific to factorization systems, but to get some intuitions for diagonal lifts, I drew some pictures.

First we have the extension condition all by itself. An example is finding a curve between two given points:

Spencer Breiner (Jun 10 2021 at 15:51):

ext2.jpg

Spencer Breiner (Jun 10 2021 at 15:52):

ext6.jpg

Spencer Breiner (Jun 10 2021 at 15:53):

Next we have the lifting condition all by itself. An example here is setting the altitude for a flight plan:
lift2.jpg

Spencer Breiner (Jun 10 2021 at 15:53):

lift6.jpg

Spencer Breiner (Jun 10 2021 at 15:53):

The general diagonal condition allows us to put extension and lifting problems together. Here the example is that we would like our flight plan to start and end on the ground:
square0a.jpg

Spencer Breiner (Jun 10 2021 at 15:53):

square3.jpg

সায়ন্তন রায় (Jun 10 2021 at 15:56):

I am sorry, but could you please add some comments on the pictures explaining what is going on. I would like to know whether my understanding of them is essentially same as what you intended.

Morgan Rogers (he/him) (Jun 10 2021 at 16:28):

সায়ন্তন রায় said:

Morgan Rogers (he/him) said:

Re your proposed alternative, what would you expect the factorization of an identity morphism to be, if identities are excluded from both classes?

Unless I am making a big mistake, could it not simply be any two morphisms $e$ and $m$ such that $\text{id}_A =m\circ e$ where $e\in E$ and $m\in M$?

Right, so this $e$ and $m$ would have to be a split mono and split epi, respectively. So now consider their composite $g_1 := e \circ m$, which will be an idempotent on some other object, so $g_1^2 = g_1$. Factorizing this, we get $e \circ m = m' \circ e'$, and hence an endomorphism $g_2 := e' \circ m'$ of another object with $g_2^3 = g_2^2$, and so on, obtaining endomorphisms of successive objects with $g_k^{k+1} = g_k^k$. So the existence of a factorization system in your sense causes each identity morphism to generate a sequence of endomorphisms. While some of these endomorphisms could be identities (or satisfy stronger identities), as soon as that happens, one has both split monos and split epis in both $E$ and $M$, which is fraught territory if one is hoping for $E$ and $M$ to be classes which are definable categorically. Can you find an example of a category with a pair of classes of morphisms satisfying your definition?

Already, I should mention that you should also exclude isomorphisms from $E$ and $M$ if you're excluding identities from them, since otherwise the existence of your factorization system will fail to be stable under equivalences of categories.

Reid Barton (Jun 10 2021 at 16:38):

I'm not sure this particular subthread is going in a useful direction.

Morgan Rogers (he/him) (Jun 10 2021 at 16:39):

Can you find an example of a category with a pair of classes of morphisms satisfying your definition?

Important point regarding your earlier comment of being "more interested in the purely theoretical aspect": if you can't easily find any examples, or the examples you find are very convoluted to the point of not having any obvious relation to familiar concrete examples of categories, then you should stop, or be at risk of rather missing the point/spirit of category theory.

The reason the definition of factorization system is what it is, is that there are many many examples of such systems, and an abstract framework for understanding those examples clarifies what they have in common, and what features the examples of 'almost factorization systems' (weak factorization systems, or orthogonal classes which fail to be factorization systems) are missing.

Reid Barton (Jun 10 2021 at 16:50):

There is some kind of relation to initial/terminal objects. Fix one of these factorization systems $(E, M)$, and a map $f : A \to B$. Then we can consider all the factorizations $A \to C \to B$ of $f$ as an arbitrary map followed by an $M$-map. These form a category whose morphisms are the diagrams of the shape in (for instance) the diagram in 14.4 (1) (except we don't require the $A \to C$ part to belong to $E$).

Among all such factorizations, there is a terminal one which is $A \xrightarrow{f} B \xrightarrow{\mathrm{id}} B$, since $\mathrm{id}$ belongs to $M$. That's not very interesting. There's also an initial one, which is the one produced by the factorization system. The statement that it's initial is the unique lifting property for $E$ against $M$. So the factorization produced by the factorization system is also the factorization through a map of $M$ which is "as long as possible". By a dual argument, it's also the factorization through a map of $E$ which is "as long as possible".

Reid Barton (Jun 10 2021 at 16:56):

Morgan Rogers (he/him) said:

সায়ন্তন রায় said:

As an example, I would be very much interested to know that given $E$ and $M$, when we do have an $(E,M)$-factorization.

Interestingly, the verification that an $(E,M)$-factorization exists is typically a non-categorical result, unless it is inherited from another category (via monadicity, say). Orthogonality (the left and right lifting properties) are used to show uniqueness once existence is established.

The small object argument produces factorizations in rather general settings, once you know that each of $E$ and $M$ is determined by the orthogonal lifting property with respect to the other one, and that $E$ is generated by a set of maps.

Morgan Rogers (he/him) (Jun 10 2021 at 21:01):

Reid Barton said:

Morgan Rogers (he/him) said:

সায়ন্তন রায় said:

As an example, I would be very much interested to know that given $E$ and $M$, when we do have an $(E,M)$-factorization.

Interestingly, the verification that an $(E,M)$-factorization exists is typically a non-categorical result, unless it is inherited from another category (via monadicity, say). Orthogonality (the left and right lifting properties) are used to show uniqueness once existence is established.

The small object argument produces factorizations in rather general settings, once you know that each of $E$ and $M$ is determined by the orthogonal lifting property with respect to the other one, and that $E$ is generated by a set of maps.

True, but showing that the generated system isn't one of the trivial ones is then necessary, and I'd maintain that that part is typically not categorical (since there may not be any non-trivial factorization systems on a generic locally finitely presentable category).

Spencer Breiner (Jun 10 2021 at 21:46):

সায়ন্তন রায় said:

I am sorry, but could you please add some comments on the pictures explaining what is going on. I would like to know whether my understanding of them is essentially same as what you intended.

No problem.

We're basically starting from the simplest possible situation, three arrows satisfying $g\circ f=h$. Composition gives us the forward map: if we know $f$ and $g$, then we can get $h$.

The first pair of pictures is an "inverse problem" for that: given $h$ and $f$, find $g$. The first picture is just geometric intuition. The second one breaks the intuitive picture up into three spaces: a pair of points, an interval and the ambient space. The green arrow is $f$ and the blue arrow is $h$. The dashed red arrow is the $g$ that you are supposed to find.

In the second pair of pictures we're playing the same game, but this time the red arrow at the bottom is $h$ and the pink arrow on the side is $g$, which just projects a path in 3D down to 2D. We need to find the dashed yellow arrow $f$, which determines (just) the z-coordinate for each point in the path.

For the last pair, we need to fill the diagonal, and that lets us combine elements of both extensions (specifying the endpoints) and lifts (path projection).

To work this back to factorization systems, note that the inclusion of endpoints on the left (green arrow) is monic/injective and the projection from 3D to 2D (purple arrow) is epic/surjective. It's worth thinking about what can go wrong otherwise; if either of those things fails, you can cook up a square that has no diagonal filler.

সায়ন্তন রায় (Jun 11 2021 at 04:39):

Morgan Rogers (he/him) said:

সায়ন্তন রায় said:

Morgan Rogers (he/him) said:

Re your proposed alternative, what would you expect the factorization of an identity morphism to be, if identities are excluded from both classes?

Unless I am making a big mistake, could it not simply be any two morphisms $e$ and $m$ such that $\text{id}_A =m\circ e$ where $e\in E$ and $m\in M$?

Right, so this $e$ and $m$ would have to be a split mono and split epi, respectively. So now consider their composite $g_1 := e \circ m$, which will be an idempotent on some other object, so $g_1^2 = g_1$. Factorizing this, we get $e \circ m = m' \circ e'$, and hence an endomorphism $g_2 := e' \circ m'$ of another object with $g_2^3 = g_2^2$, and so on, obtaining endomorphisms of successive objects with $g_k^{k+1} = g_k^k$. So the existence of a factorization system in your sense causes each identity morphism to generate a sequence of endomorphisms. While some of these endomorphisms could be identities (or satisfy stronger identities), as soon as that happens, one has both split monos and split epis in both $E$ and $M$, which is fraught territory if one is hoping for $E$ and $M$ to be classes which are definable categorically. Can you find an example of a category with a pair of classes of morphisms satisfying your definition?

Already, I should mention that you should also exclude isomorphisms from $E$ and $M$ if you're excluding identities from them, since otherwise the existence of your factorization system will fail to be stable under equivalences of categories.

Many thanks for the detailed reply. It is very helpful (to me at least). Yeah, you are right. I should have put more thoughts in crafting the definition. My original motivation was to craft a definition which doesn't allow for trivial factorizations (like e.g. , $f=f\circ \text{id}$). But anyway my central interest was not to provide an alternative definition per se, but just to understand the essence of the usual definition.

In any case, would you mind to elaborate a bit on the remark, "....which is fraught territory if one is hoping for $E$ and $M$ to be classes which are definable categorically" @Morgan Rogers (he/him)?

সায়ন্তন রায় (Jun 11 2021 at 04:47):

Reid Barton said:

There is some kind of relation to initial/terminal objects. Fix one of these factorization systems $(E, M)$, and a map $f : A \to B$. Then we can consider all the factorizations $A \to C \to B$ of $f$ as an arbitrary map followed by an $M$-map. These form a category whose morphisms are the diagrams of the shape in (for instance) the diagram in 14.4 (1) (except we don't require the $A \to C$ part to belong to $E$).

Among all such factorizations, there is a terminal one which is $A \xrightarrow{f} B \xrightarrow{\mathrm{id}} B$, since $\mathrm{id}$ belongs to $M$. That's not very interesting. There's also an initial one, which is the one produced by the factorization system. The statement that it's initial is the unique lifting property for $E$ against $M$. So the factorization produced by the factorization system is also the factorization through a map of $M$ which is "as long as possible". By a dual argument, it's also the factorization through a map of $E$ which is "as long as possible".

Why should $\text{id}\in M$?

Reid Barton (Jun 11 2021 at 05:41):

It's one of the axioms. Or it follows from the fact that $M$ contains all the maps with the unique right lifting property with respect to $E$. Anyways it's not important, it's just meant to motivate looking at the initial factorization through a map of $M$, because the terminal one would be boring.

Reid Barton (Jun 11 2021 at 05:45):

For instance, in the central example of Set with E = epis and M = monos, suppose you have a function $f : A \to B$. The smallest monomorphism $C \to B$ you could factor $f$ through is the image of $f$, and the image factorization is the epi-mono factorization. The largest monomorphism you could factor $f$ through is all of $B$; that's not interesting.

সায়ন্তন রায় (Jun 16 2021 at 09:30):

In this note, in the definition of orthogonal factorization systems, why (III) is termed as functorial?

Morgan Rogers (he/him) (Jun 16 2021 at 10:59):

Because it mean that we have functors from the full arrow category to the subcategories consisting of the left- and right-hand parts of the factorization system.

সায়ন্তন রায় (Jun 16 2021 at 11:15):

Great. Thanks!