Categories 5 Limits and colimits

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  • Опубликовано: 3 ноя 2024

Комментарии • 47

  • @MasterHigure
    @MasterHigure 3 года назад +19

    My favourite coequalizer is the natural numbers. Take the two categories 1 (a category with a single object and only the identity morphism), and the category 2 (a category with two objects and a single morphism from one to the other, and identity morphisms and nothing else). Then there are exactly two functors between them.
    In the category of small categories, take the coequalizer of these two, and what you get is a single object with one morphism for each natural number where composition corresponds to addition. I.e. the natural numbers realized in the standard monoid / group construction mentioned in one of the earlier videos in this series.

    • @fbkintanar
      @fbkintanar 3 года назад +1

      Does this settle the issue of whether natural numbers include zero, like engineers often prefer (I think there is even an IEEE (?) standard about this), or whether they should be confused with the counting numbers starting with one, as elementery school teachers might prefer. I figure monoids are more "natural" than semigroups!

    • @MasterHigure
      @MasterHigure 3 года назад +7

      @@fbkintanar Nah. I have a master's in algebra, and I still flip flop like crazy. It all depends on whether the 0 is needed or in the way.

    • @schweinmachtbree1013
      @schweinmachtbree1013 3 года назад +2

      @@fbkintanar category theorists tend to like 0 in the natural numbers because the natural numbers with 0 form a category with one object (and a self-morphism of that object for each natural number, with composition corresponding to addition and the morphism for 0 being the identity morphism), whereas the naturals without 0 form a *semicategory* with one object (in the same way) - a semicategory is like a category but without insisting on the existence of identity morphisms.

    • @samueldeandrade8535
      @samueldeandrade8535 3 месяца назад

      ​@@fbkintanar "the issue"??? Hahahaha. There is no such "issue".

  • @JP.Q
    @JP.Q 3 года назад +15

    My way of remembering that left adjoints commute with colimits is that both of these are meant to be mapped OUT OF. Dually, limits and right adjoints are meant to be mapped INTO.

    • @davidtriplett8105
      @davidtriplett8105 2 года назад

      I'm not very good at anything but pardon my querying.....is the left adjoint assumed to be U as in forgetful functor,???? Again I'm sorry I'm just trying to feed the mind of my hungry 7 year old... We both were abstract algebra first then programming so we are bound by my knowledge where we live. Again excuse my line of questioning good Sir

    • @henrikljungstrand2036
      @henrikljungstrand2036 2 года назад +1

      @@davidtriplett8105 The forgetful functor is a right adjoint if and only if it has a corresponding free functor. And the forgetful functor is a left adjoint if and only if it has a cofree functor.

    • @davidtriplett8105
      @davidtriplett8105 2 года назад +1

      @@henrikljungstrand2036 you explained this very simply you should have seen how you tried to make this logical in class lol.

  • @fbkintanar
    @fbkintanar 3 года назад +3

    The discussion of Freyd's adjoint functor theorem, and the "notorious counterexample" 25:00 helped clarify some things for me, I am trying to understand how Boolean algebras fit in with constructive logic and algorithms. I figure we need to explicitly model context, Boolean algebras tend to impose a global perspective.

  • @xiaohuwang4173
    @xiaohuwang4173 3 года назад +3

    How I remember that right adjoints preserve limits and left adjoints preserve colimits: The Hom-functor is covariant in the right argument and contravariant in the left argument, hence it preserves limit in the right argument and colimit in the left, hence right adjoints preserve limits and left adjoints preserve colimits.

  • @adityaekbote8498
    @adityaekbote8498 3 года назад +6

    So category theory is abstract algebra topology linear algebra (somewhat calculus and linear algebra) this is so cool thanks for these amazing lectures

    • @schweinmachtbree1013
      @schweinmachtbree1013 3 года назад +5

      indeed! although it is perhaps better to say "category theory is what is common to abstract algebra topology linear algebra etc.", as one cannot expect to be able to do any of these fields purely with category theory and ignoring the specific features of the structures they study

    • @kodfkdleepd2876
      @kodfkdleepd2876 2 года назад +7

      No, category theory is a codification of structure. It is not any specific mathematical subject(although many subjects elucidated it).
      Category theory is literally just objects and morphisms... but just like sets are just elements and there are more to sets than one might thing before one learns about them(cardinality, permutations, functions, continuity, etc), similar to category theory... there is far more to just "objects and morphisms" because what one is studying is how structures can relate.
      If you know anything about programming, it's more like learning UML or design patterns. Category theory is codifying certain abstract patterns that show up time and again in concrete ways. By doing this one can study the patterns without all the extra "richness" and hence get better at seeing the underlying structure in things. It's about the big picture rather than all the details. E.g., it's not about solving specific problems like proving a limit of a sequence exists or the ratio test or whatever... it's about showing what these these things really are and how they relate to other things.
      It's actually relatively simple but the language is terse and the concepts are dense... but basically everyone already knows "category theory"... we use functors("analogies") and natural transformations(translations) all the time(although in real life it's not as precise). But of course most people don't realize they are doing this and don't know the technical names(since the technical names never existed until someone defined them) or the advanced nature of them(which comes when people abstract them and can work with them in their raw state.

  • @publiced6500
    @publiced6500 2 года назад +2

    Great notes! Does anyone know which eminent mathematician is referred to towards the end of the video (the one who got confused about the direct limit and subcategory (which I certainly did!)?)

    • @boboryan1012
      @boboryan1012 Год назад +2

      I think Prof. Borcherds was referring to Shinichi Mochizuki. c.f. ruclips.net/video/Vxo51jM9khs/видео.html starting from 26:04

  • @natepolidoro4565
    @natepolidoro4565 3 года назад +2

    No idea what's going on but I'm entertained. I hope to learn it someday.

  • @kj4242
    @kj4242 Год назад

    We live in a peculiar world that allows us to learn esoteric spells that make others believe we are brilliant. Thanks.

  • @newwaveinfantry8362
    @newwaveinfantry8362 2 месяца назад

    16:01 - Don't you need R1, R2 to be vector fields with a canonical base field to take the tensor product? Or do you mean the tensor product of R1 and R2 as Z-modules as rings are Abelian groups, hence Z-modules? Or do you just mean Cartesian product?

  • @LarryBordo
    @LarryBordo 3 года назад +6

    How did limits get their name? I'm used to the limits from analysis, but I'm not sure how this is related

    • @BenSpitz
      @BenSpitz 3 года назад +12

      I'll answer in the form of a story. Let p be a prime. The cyclic p-groups ℤ/pⁿℤ (for natural numbers n) are connected by canonical surjections ℤ/pⁿ⁺¹ℤ → ℤ/pⁿℤ, this map being "reduce mod pⁿ". So, we can form an infinite sequence of surjections
      ... → ℤ/p³ℤ → ℤ/p²ℤ → ℤ/pℤ → 0 (= ℤ/p⁰ℤ).
      The kernel of each of these maps is isomorphic to ℤ/pℤ, so we can think of each stage of this infinite sequence as saying something like "ℤ/pⁿℤ can be unfurled by a factor of p to give the larger cyclic group ℤ/pⁿ⁺¹ℤ". Or alternatively, "ℤ/pⁿ⁺¹ℤ can be folded up by a factor of p to give the smaller cyclic group ℤ/pⁿℤ".
      Then we ask - what happens if we unfurl forever? In other words, what should live at the far left end of this infinite sequence, past all of our cyclic p-groups? Such a group could be thought of as the "limit" of repeatedly iterating this "unfurling" process.
      Well, this infinite sequence can be thought of as a functor ℕ^op → Grp, and we're looking for a group equipped with morphisms to all of the objects in our sequence. More precisely, the group we're looking for is the limit of the functor ℕ^op → Grp!
      In this case, the limit happens to be the p-adic integers. In general, taking a limit of a functor ℕ^op → C (a "sequential limit") gives you "an object that should live just beyond the end of your sequence". Analogously to limits in analysis!

    • @shiina_mahiru_9067
      @shiina_mahiru_9067 3 года назад +4

      It has absolutely nothing to do with the limit in analysis (so you can try to make joke with your analysis professor). It is just one of those unfortunate misnaming that survives to modern mathematics. If I remember correctly, before category theory was developed, the first (co)limit appearing in literature are the inverse / direct limits, and the word "limit" just stay. What's even worse, in my opinion, is that an inverse LIMIT is a LIMIT but a direct LIMIT is a COLIMIT. (I often got this stupid convention wrong half of the time.) If you learn algebraic geometry, you might have experienced this confusion: a stalk of a sheaf is constructed using "direct limit" in most texts. (Vakil's FOAG is probably the only text avoilding this non-modern confusing convention.)

    • @fbkintanar
      @fbkintanar 3 года назад

      The answers of Raymond Chan and Ben Spitz seem to contradict each other, but they both make sense. I don't know whether the preplexities are compounding, or slowly starting to unravel in my still-uncategorical brain.

    • @BenSpitz
      @BenSpitz 3 года назад +5

      @@fbkintanar I think we're both right :) Formally, limits in analysis/topology are not just a special case of limits in category theory. On the other hand, the term "limit" arose in category theory precisely because the first kinds of (non-discrete) limits people studied were sequential limits, and in the sequential case there *is* at least an analogy between these two meanings of "limit".
      In other words, the historical origin of the term "limit" in category theory does come via comparison with limits in analysis/topology, but the precise notion of "limit" in category theory doesn't have a strong formal connection to the notion of limit in topology.
      I should also make one other fun note: if you consider ℝ as a poset, and take a bounded sequence F : ℕ → ℝ, then lim(F) is the infimum of the sequence and colim(F) is the supremum of the sequence. If the sequence is monotonically decreasing (resp. increasing), then the categorical limit (resp. colimit) object is equal to the analysis-limit of the sequence!

  • @officialEricBG
    @officialEricBG 3 года назад +2

    Typo in the description: "We also describe how **adoint** functors ... "
    Thanks for the incredible content!

  • @YEOMSIJIN
    @YEOMSIJIN Год назад

    2:28 the composition is an identity because there is only one arrow from P1 to P2 by the universality and every object has at least one arrow from it to itself

  • @tim-701cca
    @tim-701cca 11 месяцев назад

    At 5:04, why category theory can explain product topology is used instead of box topology for the product space?

    • @samueldeandrade8535
      @samueldeandrade8535 3 месяца назад +2

      Because it explains what the product of topological spaces is supposed to be.
      Let (X_i) be a family of topological spaces. The product of (X_i), if it exists, must be a topological space X with projections
      p_i : X→X_i
      supposed to be continuous maps satisfying the property of product. In particular, this implies X must be the cartesian product, X=Prod_i X_i. What topology X must have to be the product? Considering the projections p_i must be continuous, for each open set U in X_i, the inverse image
      (p_i)^{-1}(U) = prod U_j,
      U_j = X_j, for j≠i,
      U_i = U,
      must be an open set in X. So the topology has to contain the topology PT_X generated by such sets, which is the so called Product Topology.
      Now, for any other topology T in X, if T strictly contains PT_X, then (X,T) can NOT be the product of (X_i). Because, considering (X,T) as candidate for product, if we try to apply the property of product for (X,PT_X) and the morphisms
      f_i = p_i : (X,PT_X)→X_i
      the unique function f satisfying p_i f = f_i must be the identity
      f = id_X : (X,PT_X)→(X,T)
      but THIS identity is NOT continuous, so it is NOT a morphism in Top, because T strictly containing PT_X means there is some open set V in T such that V is NOT in PT_X, so
      (id_X)^{-1}(V) = V
      is not open in (X,PT_X), proving this identity id_X is NOT continuous. So, (X,T) is not a product of (X_i).
      Finally, the box product Bo_X strictly contais PT_X when the family (X_i) is infinite. By our reasoning, (X,Bo_X) can NOT be the product of (X_i) in this case.
      Thaaaaat's it.

    • @tim-701cca
      @tim-701cca 3 месяца назад +1

      Thanks you for your explanation❤. I got the idea.

  • @MichaelFJ1969
    @MichaelFJ1969 3 года назад +1

    Thank you for making these video lectures. Will you be making a video series on Homotopy Type Theory ?

    • @schweinmachtbree1013
      @schweinmachtbree1013 3 года назад

      I expect not; professor borcherds is not super into mathematical logic/foundations of mathematics as far as I know - he might mention it as an aside though (or equally he may not mention type theory at all, who knows)

  • @gammaknife167
    @gammaknife167 3 года назад

    You say the result about tensors with Y being right exact is now trivial, doesn't this just mean the theorem is particularly hard to prove?

    • @schweinmachtbree1013
      @schweinmachtbree1013 3 года назад +3

      Not necessarily; when people say that something is "made trivial", that just means it is made much easier/more immediate than it was before. Richard got the saying at 29:07 slightly wrong: it is "category theory makes trivial results trivially trivial" - so going off of the saying I expect that the result about tensors with Y is not too hard to prove without category theory*.
      (*although I have not attempted such a proof so feel free to take that with a grain of salt)

  • @jimadams8385
    @jimadams8385 2 года назад

    Just to start, there exist projective limits taken twice which are contravariant! We know 0 has Topologic structure. There exists 0-variance as a 'withered boundary' between co-variance and contra-variance! This is a real (-1,0,1) Zargon game.

    • @ethanjahan780
      @ethanjahan780 2 года назад

      This doesn't even begin to make sense

    • @jimadams8385
      @jimadams8385 2 года назад

      @@ethanjahan780
      No, it probably doesn't. I was ill when I wrote it. But consider that contravariant functors map multiplicatively to - 1, as in the diagram which inverts the order of functors. Then if we extend the algebra as we do the complex numbers then there should be an i-variant functor, two of which map in conjunction to a contravariant functor.
      So the definitions we are using is that an opposite category inverts all arrows, but the swap between covariance and contravariance inverts just one arrow.
      Structures are generally non associative categories.
      There is a covariant functor in an exponential structure C to B^(C - 1) mapping (A^B) ^C to A^(B^C).
      The entailment arrow functor is not the implication operator in propositional calculus.
      A tree structure has direction. A tree root is an inverted tree. Attaching a tree root single terminating node to the single node of a tree forms the diagram for a multifunctor arrow.
      The dual of a function is generally a type of multifunction.
      Functors may be glued together to form Topologic manifolds with handles and crosscaps.
      The {uncooperative, competitive, cooperative} game triple maps to games which are {-, 0,+} sum games respectively.
      When the payoff array is generally non associative, this is a Zargon game.

    • @ethanjahan780
      @ethanjahan780 2 года назад

      @@jimadams8385 hum, I don't want to be arrogant or mean, but it seems that you just glued together random words hahaha
      What the hell is an i-variant functor, is the i supposed to be the imaginary unit?
      Also, "structures are generally non associative categories" I'm pretty sure a non associative category isn't a very real concept.
      Nothing does any mathematical sense haha, maybe you're still ill, watch out bro

    • @jimadams8385
      @jimadams8385 2 года назад

      @@ethanjahan780
      I need medical drugs to control my thinking process. It is difficult to self-monitor.
      Also I cannot parse functions from right to left. That is why I write fg(x) for functions f then g acting on domain (x).
      There is a niche market for innovation in mathematics, provided it does not get out of hand. Sometimes it is good to experiment with ideas.
      It is possible to introduce a new notation for functor arrows. Label an arrow +1 if it goes from left to right or down to up and - 1 otherwise. Between two parallel arrows introduce one functor arrow between them. Then a functor is an arrow between arrows, and a diagram may be labelled covariant or contravariant depending on whether there are an even or odd number of contravariant arrows in it. Hence contravariance acts multiplicatively as - 1. We can now observe that though normally two identical functors do not compose as a contravariant - 1 functor, we can introduce the idea of an i-variant functor, two of which compose to - 1 contravariance, by analogy with imaginary numbers.
      Diagrams may be labelled with twisted or untwisted strips, and glued in surgery as in Massey diagrams for a manifold with m handles and n crosscaps.
      Then a logical deduction becomes a diagram. This is Topologic.
      (2) A cograph of C to D has number of functions D^C. Thus categories involve exponentiation, which is non associative. We can keep in a standard theory by always referring to (A^B) ^C = (covariantly) A^(B×C), but there is the other arrangement A^(B^C), to which we have described a mapping.
      It is possible to extend the sequence {addition, multiplication, exponentiation} on the right, or horribly, on the left.

    • @henrikljungstrand2036
      @henrikljungstrand2036 2 года назад

      @@ethanjahan780 Mmm it seems more like some mathematically themed dream than remotely rigorous, consistent or even intelligible mathematics. But just as there are non-associative analogues of groups i.e. loops (especially Moufang loops), non-associative versions of fields and division rings i.e. division algebras, and non-associative versions of linear logic i.e. Lambek calculus etcetera, we may create non-associative versions of category theory where "composition" is not always exactly the same as "doing one morphism after another in the sense of functions", but that this obvious down-to-earth composition of two morphisms may not always even produce a new morphism in the same "category".
      Some restricted forms of associativity that might be true for "composition" in such a "category" would be e.g. for all m : B -> D, n : D -> B, f : A -> B, g : D -> E, x : D -> C, y : C -> B
      it is true that
      1) (m ° n) ° m = m ° (n ° m) = m ° n ° m [this is only to simplify notation by removing excessive parentheses, otherwise it follows from slightly more elucidated versions of the following axioms]
      2) m ° (n ° (m ° f)) = (m ° n ° m) ° f
      3) ((g ° m) ° n) ° m = g ° (m ° n ° m)
      4 (m ° y) ° (x ° m) = m ° (y ° x) ° m
      These two also hold, by deduction:
      for all e : B -> B, f : A -> B, g : B -> C
      it is true that
      5) e ° (e ° f) = (e ° e) ° f = e ° e ° f
      6 (g ° e) ° e = g ° (e ° e) = g ° e ° e
      This was inspired from the Moufang laws of Moufang loops, plus alternative and flexible laws.

  • @Anujkumar-my1wi
    @Anujkumar-my1wi 3 года назад

    Hey ,I want to ask that If a differentiable function is locally linear ,then shouldn't the differentiable function be piecewise as differentiable function is made up of pieces of locally linear regions?

    • @MasterHigure
      @MasterHigure 3 года назад +5

      Not that this is terribly on topic for the video, but... Differentiable functions aren't locally linear. They are locally APPROXIMATELY linear, with a certain rigorous definition of what "approximately" actually means in this context.

    • @Anujkumar-my1wi
      @Anujkumar-my1wi 3 года назад

      @@MasterHigure yes the linear approximation is indeed a approximation but derivative function are made up of these approximation thus shouldn't the derivative function be piecewise not the function whose derivative is taken but the derivatiev function itself as it is made up of pieces of linear approximation of local regions of a nonlinear function?

    • @MasterHigure
      @MasterHigure 3 года назад +3

      @@Anujkumar-my1wiThe derivative of a function is defined from pointwise linear approximations, not piecewise.

    • @bentoomey15
      @bentoomey15 3 года назад +1

      Do you know about "germs"? If I understand what you're asking, they're exactly the "local pieces" that make up functions you might want, but unfortunately they aren't quite linear. Informally, they need to store information about second derivatives, third, etc. to capture enough information to rebuild the whole function.

  • @migarsormrapophis2755
    @migarsormrapophis2755 3 года назад +6

    yeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

  • @madhavinomula7430
    @madhavinomula7430 2 года назад

    くさんありますありがとうございます」、