Category Theory III 6.2, Ends
34:26
6 жыл бұрын
Category Theory III 6.1, Profunctors
29:14
Category Theory II 9.1: Lenses
41:59
7 жыл бұрын
Category Theory II 7.1: Comonads
37:39
Category Theory II 5.2: Adjunctions
40:36
Category Theory II 3.2: Free Monoids
36:54
Пікірлер
@stevengrieco3182
@stevengrieco3182 2 күн бұрын
43:56 Bartosz leaves eraser on board.
@thomastsao7507
@thomastsao7507 12 күн бұрын
The best mathematics class I ever took. The best teacher taught a very difficult subject in a very effective way. This is amazing.
@ChelseaHays-o6u
@ChelseaHays-o6u 15 күн бұрын
"more productive" product drawing hahahah
@DthingOrg
@DthingOrg 18 күн бұрын
Good introduction 🎉
@microspacer
@microspacer 25 күн бұрын
The title became a meme
@jensl5956
@jensl5956 28 күн бұрын
I love this guy
@5pp000
@5pp000 Ай бұрын
10:06 Using the name 'eval' for the function that takes (z, a) to b is very odd. If you're trying to use traditional Lisp terminology, you've got it wrong. The correct name for that function is 'apply'. 'eval' is a function from _s-expressions and environments_ to values (I'm eliding mention of the store since we're in the functional paradigm here).
@ucManhPham-mq8jn
@ucManhPham-mq8jn Ай бұрын
Hope some one can recommend me books about this topic
@g999ar
@g999ar Ай бұрын
11.12.2024
@g999ar
@g999ar Ай бұрын
11.12.2024
@hansschenker
@hansschenker Ай бұрын
Bartosz Milewski The Tiger of Category Theory! (is he a Tiger because he is walking left and right during his talk, or is he a Tiger because he is chasing big Abstractions?)
@neonblood4658
@neonblood4658 Ай бұрын
didnt know pedro pascal retired from hollywood
@AR-ym4zh
@AR-ym4zh 2 ай бұрын
"The elementary thing is not divisible but is not a point" 😉
@josecastromora
@josecastromora 2 ай бұрын
Using the subset relation as an example of a category makes me crave to be more formal. He avoids it by stating that "it's so obvious", but it isn't. Subset is a logical relation, not a monorphism, it's only binary, it either is or is not between two sets A and B, and you can't compose it and state (here C is subset of) ... A C (B C D) = (A C B) C D. In fact, stating it like this is a type error: being a subset of is true or false, not a monomorphismt. I know this definitely has a solution, but would've wished he would have stated as much.
@bartbroek9695
@bartbroek9695 2 ай бұрын
i wonder, what does associativity even mean for the operator "is a subset of". if we say 3+4+5 is associative, this works bc + maps two numbers to another number, whereas "is a subset of" maps two sets to a boolean. and yet i understand we can compose with it
@vh9161
@vh9161 3 ай бұрын
Please, someone give him a Nobel Prize already!. what a treasure. I am just staring Part 2.
3 ай бұрын
Ok so javascript is a monoid
@Santi._.403
@Santi._.403 3 ай бұрын
I’m really glad the class clapped at the end. Thank you for the lesson.
@memegazer
@memegazer 3 ай бұрын
imo it would not be a problem to store an int in a bool...in fact that is what I am here for My hope is that this course will show why that is not a problem for some arbutrary string bc math is rational, and hyperwebsters make sense, if it just a data labling of a look up table
@tarikozkanli788
@tarikozkanli788 4 ай бұрын
There is indeed a discipline of engineering whose domain is mathematics itself.
@tarikozkanli788
@tarikozkanli788 4 ай бұрын
what would be the advantage of a multi category programming language. Haskell has one.
4 ай бұрын
Good stuff. Thank you so much.
@dddddddbable
@dddddddbable 4 ай бұрын
Dude looks like Ben Kingsley.
@lingdocs
@lingdocs 4 ай бұрын
It's interesting that in programming the bifunctor was described as CxC -> C and NOT CxC -> CxC. Because at first glance the result also looks like a cartesian product. And in haskell bimap is (a -> b) -> (c -> d) -> p a c -> p b d. p a c -> p b d seems to correspond to CxC -> CxC at first glance. But I guess it is not necessarily because the two functions might not produce all possible values, therefore we are not guarnteed to have a cartesian product of all possible types in the category. There could be pairs of functions (both completely surjective over C) that map CxC-> CxC, but a bifunctor is open to all kinds of functions that don't output the whole of C in their codomain. Therefore we have to describe the bifunctor in programming as CxC -> C. Is this correct?
4 ай бұрын
I regret not having this lecture eight years ago. Better late than never
@muttdev
@muttdev 5 ай бұрын
kzbin.info/www/bejne/hJvbY5iabbd6n9U
@huylongpham8748
@huylongpham8748 5 ай бұрын
Hello, I'm from Vietnam, I've finished watching playlist 1 of category theory, in playlist 2 there are some videos without subtitles, please add them. Thanks you so much !!!
@utof
@utof 5 ай бұрын
i feel like i actually understand epics and monic now, wow. after all these years. finally!!!!!!! thank you! the most understandable category theory introduction EVER!
@stavanjain7894
@stavanjain7894 5 ай бұрын
This was beautiful! Thank you so much for posting these lectures
@tarikozkanli788
@tarikozkanli788 5 ай бұрын
what do you call a monoid in the category of endofunctors on the category of programming languages?
@kurisumasu3329
@kurisumasu3329 6 ай бұрын
OH my! The List part is absolutely mind blowing!
@k3rnel-p4n1c
@k3rnel-p4n1c 6 ай бұрын
Circa 26:32, is it therefore also true that h•g•f can be asociated with (h•g)•f and h•(g•f) ?
@samueldeandrade8535
@samueldeandrade8535 6 ай бұрын
30:16 This is fundamentally wrong.
@jensl5956
@jensl5956 6 ай бұрын
I wosh this guy was my professor
@pablovaldes6022
@pablovaldes6022 7 ай бұрын
The more I watch the lecture the more it reminds me the machete movie.
@alexbalandi
@alexbalandi 7 ай бұрын
I'm trying to grasp 41:00 and running into problems. So for natural numbers and multiplication what would be the "morphism"? If I think correctly. I have element 5, I have morphism "multiply by 4", I get element 20. Obviously, "multiply by 4" will map set of natural numbers N into N, same for any morphism "multiply by k". Which leads me to question. In such formulation, the morphism itself in cathegory theory isn't actually tied to the element of N. so " multiply by 1" is identity morphism, but from the standpoint of cathegory theory we don't really know that this "1" is from that dot "m", because we don't really know what's inside the dot. Doesn't this mean that it's possible for cathegory to have identity morphism but for the derived set to not have identity element? If you take the previously talked about "<=" relation and construct monoid in N with it, it seems you won't get a "set theory monoid" from "cathegory theory monoid" with it.
@alexbalandi
@alexbalandi 7 ай бұрын
I guess for me the "unintuitive" part is that in cathegory theory we don't postulate that "arrows" are the members of "m" dot here. I understand why they are members of hom-set, but we have to postulate (?) their relation to elements in m when we convert from cathegory theory to set theory, right? So we have to postulate that "multiply by k" morphism has one-to-one relation to elements of m. I wonder if it is possible to do this in more minimal way? Like axiomatically say "identity morphism corresponds to 1" , and then i'm not sure what extra things we need to postulate to get to "monoid in set theory". But if you read this, I'd really like to check - am I correct in my understanding that we need to postualte the mapping between elements of m and morphism or is there a way around it?
@alexbalandi
@alexbalandi 7 ай бұрын
Oh, I think I used term "morphism" in a wrong way here. I meant "arrows". So "multiply by 4" is an arrow and the set of all of these arrows is morphism.
@tylerbakeman
@tylerbakeman 7 ай бұрын
5:30, Mentions a DAG as a Poset, then draws a Directed-Cyclic Graph lol Great video btw
@MrRobot-pv6mo
@MrRobot-pv6mo 5 ай бұрын
Where exactly have you seen the cyclic graph on the board?
@bnchi
@bnchi 2 ай бұрын
There's no cyclic dependencies he just drew them the opposite way
@gumikjuggler
@gumikjuggler 7 ай бұрын
I've watched all the Haskell videos. Although I knew Haskell before (up to parallelism) your way of presenting things is so clear and it was so cool to refresh my knowledge! Thank you for that!
@cadenzah93
@cadenzah93 7 ай бұрын
1:01 talk about sets in terms of category so that we can generalize it to an arbitrary category 2:52 'any' means if exists, so if not exists, then it's automatically true 6:25 the value of empty category
@cadenzah93
@cadenzah93 8 ай бұрын
12:47 Void <=> False 13:43 proposition as a type 21:49 replace elements by morphism to a set
@cadenzah93
@cadenzah93 8 ай бұрын
2:44 pure function 8:48 definition of relation 10:11 definition of relation, cartesian product 11:32 comparison between relation and function 34:34 if I can't tell about the internals(elements), how can I talk about morphism's properties? (in terms of category theory)
@cadenzah93
@cadenzah93 8 ай бұрын
4:30 definition of category 6:36 barber's paradox(Russell's paradox) 13:25 language constrains us 14:58 category is like a graph 15:54 "how is it possible that you can have more than one arrow? Aren't they all the same?" / "no, they are different ... you just give them different names" 18:18 definition of 'composable' 27:40 meaning of "two compositions are isomorphic" 33:19 bottom value for infinite loop 35:20 definition of types in programming 37:45 using category theory as a abstract and simplified(high-level) model for understanding programming language
@cadenzah93
@cadenzah93 8 ай бұрын
21:21 practical motivation of learning category theory in terms of programming 37:23 abstract approach used in most of academic subjects 46:04 meaning of category theory
@JayColeman-nd1xb
@JayColeman-nd1xb 8 ай бұрын
But, how to find the rule to construct the arrow
@dorle3046
@dorle3046 8 ай бұрын
The insights and perspectives shared in this lecture are truly beautiful. Once seen, they stay with you forever. My sincere thank you for the entire series, and especially for this lecture!
@jeffreyhowarth7850
@jeffreyhowarth7850 8 ай бұрын
I'm looking for help anywhere. My question is I see (T(X)) and (T(T(X))) are parentheses around Monads, is this the wrap that is mentioned in Monad literature?
@aristotales4906
@aristotales4906 8 ай бұрын
Incredible
@0Tsutsumi0
@0Tsutsumi0 8 ай бұрын
Great lecture, the only improvement I'd suggest is a clearer distinction between "Monoid" and "Monoidal Category"
@ForestQuest-f9w
@ForestQuest-f9w 9 ай бұрын
This is epic, so much thanks!
@ForestQuest-f9w
@ForestQuest-f9w 9 ай бұрын
The rare and truly motivated tutor to see. If you have such a tutor, you also get full of fuel from a feel, nothing is really hard to get through.