Category Theory II 2.2: Limits, Naturality

[@SergeyRomanovsky]

That 4-month gap of publishing washed out all the certainty and easiness of understanding. Falling back to review it again at 2x speed. Don't hesitate to enjoy these beautiful lectures twice. Thank you Bartosz!

[@quant-trader-010]

I'm not quite following 12:35, when you said "imposing naturality condition translates to commutation condition for these triangles". What I'm confused about is the commutation of triangles only depends on that [I,C](\Delta_c, D) is a functor with contramap defined, and indeed in the commutation condition, we didn't use the \alpha_c at all and only invokes the contramap of (contravariant) functor [I,C](\Delta_c, D). What am I missing here?

[@BartoszMilewski]

Let's start with u :: c -> Lim D. alpha_c maps it to a cone mu (a cone is a natural transformation). Its component mu_i is then mapped using contramap to mu_i . f . On the other hand, u is mapped under contramap to v = u . f, which is mapped under alpha_c' to another cone nu. Naturality demands that nu_i = mu_i . f .

[@quant-trader-010]

@@BartoszMilewski Got it! I am reading Part II of your book now and re-watching these videos. Thanks a lot, Dr. Milewski!

[@abhishekaggarwal2712]

@@quant-trader-010 Also see proof in my comment, I had the exact same confusion as yourself.

[@mikesolomon7075]

Really nice! This is the first time I've seen a proof of the commutation condition for limits using a natural isomorphism between functors C->Set, and it helped my understanding of limits a lot. One question: is it possible to replace the condition of a natural isomorphism between C(-,LimD) and Nat(Δ_, D) with a weaker condition? As I understand it, the necessary conditions are: (i) A bijection between morphisms to LimD and cones (satisfied by the fact that the natural transformation is an isomorphism). (ii) A commuting triangle that has (a) on one side, the morphism from the apex of a commuting cone to D_i; and (b) on the other side, the morphism from the apex to LimD composed with the morphism from LimD to D_i. The natural isomorphism accomplishes this, but it seems to impose the additional constraint that a morphism between the apices of _any_ two cones also factorize those two cones (ie in your video you talk about apices c' and c rather than c and LimD). Intuitively, this makes sense: for example, if there is a chain of morphisms c'' -> c' -> c -> LimD, I can grok how c'' -> c' can factorize c'' a little bit, then c' -> c a bit more, and finally c -> LimD for the win, creating the equivalency (c -> LimD) ∘ (c' -> c) ∘ (c'' -> c') = c'' -> LimD. However, this seems to be a stronger statement than "there is a unique factorizing morphism from any cone to LimD" due to the additional commuting relationships between certain apices that are not LimD. My question in all this is: is this stronger condition necessary? Is it possible to weaken the statement to just talk about commuting triangles with LimD without talking about these intermediary commuting triangles? Or is the commutivity of these intermediary triangles also part of the definition of a limit? Thanks!

[@DrBartosz]

Looking at it three years later, this is really a rather trivial proof. Naturality means alpha_c'(u.f)=alpha_c(u).f and it just follows from looking at triangles in the definition of the three cones with apices c, c', and Lim D. What I should have done instead is to go the other direction: right to left, or alpha inverse. It's more interesting. To answer your question: this is not a condition I'm imposing-- naturality follows from the definition of limit.

[@mikesolomon7075]

​@@DrBartosz thanks for the quick response!

[@burnytech]

21:04 how can we use the delta c functor to pick out c, when index category is empty so we cant map any object to it?

[@gonzajuarez4918]

Ahhh the example about the terminal object is nice to see in the empty meet of a lattice. Defining a meet as the greatest lower bound of a set of objects, the empty meet is the biggest element and so it is the greatest lower bound of the empty set (nothing lies above it and every object is smaller than it - has an arrow to it in a poset-).

[@Bratjuuc]

What are morphisms between cones over different bases? Just like we had "f × g : a × b -> c × d", we need "phi : Lim D1 -> Lim D2", which I assume to be uniquely determined by the natural transformation "gamma : [D1, D2]" we chose. Is the universal construction for phi just the diagram where gamma's morphisms, phi and both of the cones commute?

[@luisdieulefait670]

At the end of the argument for the construction of limits (the goal was to show that the required commutativity conditions on the triangles "linking" C and Lim D can be rephrased in terms of the naturality of the set of maps \alpha_i) I felt that some explanations were missing (certainly due to lack of time). I mean: what was missing was to use the definition of the \alpha_i to start checking whether or not the diagram is commutative, to see that such commutativity is equivalent to that of the aforementioned triangles. I guess it is just an easy checking, and naively I can see what it means from the diagram: you do have this sort of commutativity at the level of triangles for those formed by the C, C' and each D_i, by construction: this is the formula relating the \mu_i and the u_i. The commutativity of the diagram somehow is telling us that when we "pass to the limit" this should be preserved.

[@amitsett8117]

Nice lecture. You gave the terminal object as an example of a limit. But doesn't the definition of a limit depend on the definition of the terminal object (the limit is the terminal object in the category of cones)? Is there a way then to define a limit without this circular definition?

[@abhishekaggarwal2712]

In the end Bartosz mentions a case where there is no morphism between object c and limD, while there could still be cones from c to diagram D. So it is unclear how we can assume bijection between these two sets?

[@DrBartosz]

There could be two cones, but there won't be a cone morphism between them (a _natural_ transformation from Delta_c to D).

[@abhishekaggarwal2712]

@@DrBartosz Thanks for the prompt response. My understanding is that when we say there is natural isomorphism between sets Hom(X, limD) and Cones(X, D), there is no restriction imposed on the cones being natural transformations on Cones(X, D). Similarly, there is no restriction imposed on morphisms in Hom(X, limD) to be only those which are morphisms between "natural" cones. So I am not sure I understand your comment.

[@DrBartosz]

@@abhishekaggarwal2712 Sorry for the confusion. If we are talking about two arbitrary cones, then there may be no morphism between them. But if the second cone is the limiting cone, then there is a unique morphism to it. Another way of looking at it is that the limit is the terminal cone in the category of cones over D.

[@TernaryM01]

The clip seems to be trimmed... ._.

[@siyuanchen659]

Thanks for your great lectures. I have a question: In 12:31, you said "imposing naturality on this transformation, it is translated to the commutation condition", but in 13:03, you said "natural isomorphism". Does it necessarily require natural isomorphism rather than natural transformation? And natural transformation is not enough? For the definition of limit existence: A diagram (functor D from I to C) in C has a limit LimD if only if there is a cone over it such that given any another cone there are an unique morphism m : c'->limD and make the triangles (cone to cone) commute. So there are two naturality conditions: (1) unique (2) triangles commute I try to translate it to higher-level language: A diagram (functor D from I to C) in C has a limit LimD if only if (1) For any c, there is a bijection between two sets C(c,LimD) and [I, C](Δc, D) (2) There is natural transformation α : C(-,LimD) -> [I, C](Δ-, D). The first condition represent "unique", the second represent "triangles commute". Is my understanding correct? It seems unnecessary to introduce natural isomorphism. Natural transformation need every component has a invertible morphism, but what these invertible morphism do? What role do these invertible morphism in satisfying these naturality conditions (eg: triangles commute")? Or I might have missed some key things? Thanks a lot.

[@DrBartosz]

But it is a natural isomorphism because, as you correctly state, every component of it is a bijection.

[@abhishekaggarwal2712]

I provided the proof in my comment, but I think isomorphism is enough, and naturality is stronger condition not required strictly to define a limit

[@ilyarezvov2043]

It is obvious for me how to define one direction of naturality isomorphism: suppose we define Lim D cone as tuple (Di, Dj, Dk) projections. So if we have morphism m(c -> Lim D) and just compose it with Lim D cone components then we will get C-cone natural transformation and it will define our Alpha direction. But it is not clear for me how to extract factorising morphism m(c -> Lim D) from a bunch of functions (c -> Di, c -> Dj, c -> Dk) and define Betta(opposite) direction for naturallity isomorphism. Any suggestions?

[@DrBartosz]

There is no general procedure for finding the morphism m given a cone at c and the limiting cone. For every category and every limit you have to define a different procedure. If you find one, the limit exists. (In category theory you reason as if you had infinite amount of time. Imagine that you try all possible cones and all possible morphisms one by one, and see if any of them works.)

[@prontopuntor]

How can constant functor produce objects (cone apexes) from an empty category?

[@DrBartosz]

This is one of these math tricks. A constant functor always produces just one object no matter how many objects there are in the source category. So it's natural to assume that it does the same with an empty category. That's the best explanation I could come up with off the top of my head. See, for instance, ncatlab.org/nlab/show/terminal+object

[@DrBartosz]

Frankly, I'm nor really happy with this answer, and you may notice that I did a little double take during the lecture. Feel free to post a question on stack overload.

[@jwj410]

Sounds a bit like the 'selection' function () -> a... but then why wouldn't that be a functor from the category of a single object to C?

[@konstantinskachkov3516]

For small categories this mapping is just a function. The existence of such a function follows from the definition of function: subset of the cartesian product of two sets A and B such that for every a from A there is one pair (a, b) in it. In this case there is only one subset which is empty. I think this can be extended to a mapping from a small category to any category? But what about an arbitrary mapping between arbitrary categories?

[@ShimshonDI]

Someone recently asked this question on stackexchange: math.stackexchange.com/questions/2716347/how-do-we-reconcile-these-two-definitions-of-categorical-cones. Apparently there is a different way to define cones which avoids this issue, but which uses terms that Milewski hasn't introduced as of this lecture. idk whether he has introduced them by now in future lectures.

[@clarejang9088]

I thought the point was "For given cones in [I,C](Δ_c, D) and [I,C](Δ_c', D), and morphism m in C(c', c), find a way to express "m is valid morphism between given cones"". However, as far as I understand, the conclusion is more like "For given cone in [I,C](Δ_c, D), and morphism m in C(c', c), there is a corn in [I,C](Δ_c', D) which accept m as a valid morphism to it." It looks quite different to me. Could you please explain about it? Sorry for my poor english.

[@clarejang9088]

And I can't understand what α(alpha) does in 11:47.... The triangle on the bottom of the screen commutes because of the definition of ν_i itself, and looks like having nothing to do with α(alpha)...

[@clarejang9088]

And always thank you for great category lectures and awesome c++/haskell/... videos!

[@DrBartosz]

We start with some u :: c -> Lim d. We map it with alpha_c to get mu_i. Then we use contramap that lifts f to get mu_i . f. That's one side of the naturality condition for alpha. The other side: we use contramap f to act on u and get u . f. That's our v. Then we act on it with alpha_c' to get nu_i. Thus naturality of alpha demands that nu_i = mu_i . f. If you want, you can draw a bigger diagram that includes Lim D, v and u. Alpha would then connect v to nu_i and u to mu_i.

[@clarejang9088]

Thank you for your kind answer! However, because of my ignorance, I still cannot understand some points. Don't we know what is ν_i until naturality condition applied? We already know that ν_i = μ_i . f when we define a contramap, don't we? And as far as I know, contramap is only related with [I,C](Δ_c, D) and [I,C](Δ_c', D)... are there any important things I missed?

[@clarejang9088]

If you let me summarize my question, in short, what I cannot find is, what is the actual translation for "m is morphism between cones if corresponding morphism between apexes make some morphisms commute" into terms of naturality condition? Could you please give a solid translation? Thank you for reading my messy english question.

[@abhishekaggarwal2712]

I think I understand the connection between Limits and Natural Isomorphism. My proof is as following Let `D` be a diagram in a category `C`, and let `(L, φ)` be a limit of `D`. We want to show that for any object `X` in `C`, there is a natural isomorphism between the hom-sets `Hom(X, L)` and the set of cones from `X` to `D`. **Step 1: Construct the natural isomorphism** Define two functors `F, G: C^op -> Set` as follows: 1. `F(X) = Hom(X, L)` for each object `X` in `C`. 2. `F(f) = f*` for each morphism `f: X -> Y` in `C`, where `f*: Hom(Y, L) -> Hom(X, L)` is the post-composition with `f`, i.e., `f*(g) = g ∘ f` for any `g: Y -> L`. 3. `G(X)` is the set of cones from `X` to `D` for each object `X` in `C`. 4. `G(f)` is the function that maps a cone `(Y, ψ)` to the cone `(X, ψ ∘ f)` for each morphism `f: X -> Y` in `C`. Now, we define a natural transformation `α: F -> G` as follows: 1. For each object `X` in `C`, `α_X: F(X) -> G(X)` is a function that maps a morphism `u: X -> L` to the cone `(X, ψ)` where `ψ = φ ∘ u`. Here u is the unique morphism between X and Limit object L. And φ is the morphisms that form the cone of limit object L. **Step 2: Show that α is a natural isomorphism** To prove that `α` is a natural isomorphism, we need to show that `α_X` is an isomorphism for each object `X` in `C` and that the following naturality square commutes for any morphism `f: X -> Y` in `C`: ``` F(X) Y` in `C` and a morphism `g: Y -> L` in `Hom(Y, L)`, we need to show that `G(f)(α_Y(g)) = α_X(F(f)(g))`. Let's start by computing `α_X(F(f)(g))`: 1. Apply `F(f)` to `g` to get `F(f)(g) = g ∘ f: X -> L`. 2. Apply `α_X` to `g ∘ f` to get `α_X(g ∘ f)`, which is the cone `(X, φ ∘ (g ∘ f))`. Now let's compute `G(f)(α_Y(g))`: 1. Apply `α_Y` to `g` to get `α_Y(g)`, which is the cone `(Y, φ ∘ g)`. 2. Apply `G(f)` to the cone `(Y, φ ∘ g)` to get the cone `(X, (φ ∘ g) ∘ f)`. Now we need to show that these two cones are equal, i.e., `(X, φ ∘ (g ∘ f)) = (X, (φ ∘ g) ∘ f)`. Since composition is associative, `φ ∘ (g ∘ f) = (φ ∘ g) ∘ f`. Thus, the cones are equal, and the naturality square commutes.

[@abhishekaggarwal2712]

However, the proof means that if for any object X, there is natural isomorphism between Hom(X, L) and Cones(X, D), then L is the limit. But if I understand correct, as long as only the isomorphism between Hom(X, L) and Cones(X, D) exists, it is sufficient for the universality of Limit L, that is there is a unique morphism in Home(X, L) there factorises Cones(X, D) via L. Am I correct that Naturality is a stronger condition, but isomorphism itself is sufficient? The only reason we insist of Natural Isomorphism is because we really "like to use" natural transformation is a way to express Limit, but in the process we added more constrains (or coherence when morphisms in C are considered) ? Or is it that Naturality is a free by product for the choice of functors we defined? That Naturality is not required but a pleasant bonus that happens automatically?

[@DrBartosz]

@@abhishekaggarwal2712 Cones over D form a category. Morphisms in that category are those morphisms between apexes that make the triangles between the cones commute. This is why you need naturality.

[@abhishekaggarwal2712]

​@@DrBartosz Thanks for the prompt response and this amazing course. I don't have mathematical training but following your course I really feel like I am getting to understand this beautiful theory. Based on your comment I think what I need to show is the equivalence relationship. That is to prove that A (lim L) B (natural isomorphism), I have already proven A implies B. If I can also prove the converse B -> A and in the proof if I should need to use both isomorphism and naturality, Here is the converse proof if I understand this correctly. 1. Assume there exists an object L such that there is a natural isomorphism Φ: Hom(-, L) ≅ Cones(-, D) for any object X in C. Our goal is to show that L is the limit of D. 2. Consider the functors F: C^op → Set and G: C^op → Set, where F(X) = Hom(X, L) and G(X) = Cones(X, D). For any morphism f: X → Y in C, F(f): Hom(Y, L) → Hom(X, L) is defined as F(f)(g) = g ∘ f, and G(f): Cones(Y, D) → Cones(X, D) is defined as G(f)(ψ) = ψ ∘ f. The natural isomorphism Φ provides a family of bijections Φ_X: Hom(X, L) → Cones(X, D) for any object X in C. 3. Let f: X → L be a morphism in C. We want to show that both f and ψ_L are unique, and that ψ_X = ψ_L ∘ f, where ψ_X and ψ_L are the components of the cones at X and L, respectively. 4. Examine the naturality square for the morphism f: X → L in C: ``` Hom(X, L) --Φ_X--> Cones(X, D) ^ ^ | | F(f) G(f) | | Hom(L, L) --Φ_L--> Cones(L, D) ``` 5. Path 1: Consider the identity morphism 1_L in Hom(L, L). Apply Φ_L to get Φ_L(1_L) in Cones(L, D), which is a cone (L, {π_j}). The component ψ_L is unique since Φ_L is a bijection. 6. Path 2: Apply F(f) to 1_L to get F(f)(1_L) = 1_L ∘ f = f in Hom(X, L). Then apply Φ_X to get Φ_X(f) in Cones(X, D), which is a cone (X, {ψ_j}). 7. By naturality, the two paths from Hom(L, L) to Cones(X, D) must be equal. Therefore, Φ_X(f) = G(f)(Φ_L(1_L)), which implies ψ_X = ψ_L ∘ f for all objects j in J. 8. Since Φ_X is a bijection, f is the unique morphism that factorizes the cone at X via the cone at L. Thus, (L, {π_j}) is a limit cone for D, and L is the limit of D. In summary, the existence of a natural isomorphism Φ between Hom(-, L) and Cones(-, D) implies that L is the limit of D. The naturality condition and the bijection between Hom(X, L) and Cones(X, D) are both essential in establishing this result. The uniqueness of both f and ψ_L is guaranteed by the bijections Φ_X and Φ_L. Is that correct?

[@DrBartosz]

@@abhishekaggarwal2712 Looks correct. This use of id_L is sometimes called the Yoneda trick. This is an example of a more general pattern: the adjunction. So there is a more general proof valid for any adjunction.

[@mechadense]

Here's the correct pronunciation of apexes (audio): www.dictionary.com/browse/apexes Not like a-piece-es but more like ape-a-seize.

[@TernaryM01]

It's simply like "apex" but with the 'x' replaced with 's' and adding 'is' sound. Why does "vertices" follow a different rule? I don't know. And maybe it's the exception, not the rule. Anyway, who cares?

[@izutsm]

"apices" though