Intuitive Introduction to Category Theory

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MetaMaths

MetaMaths

Күн бұрын

Пікірлер: 67
@thestemgamer3346
@thestemgamer3346 2 жыл бұрын
This is a very tricky concept that is critically important to functional programming. You explained it beautifully.
@mastershooter64
@mastershooter64 2 жыл бұрын
but what is functional programming important for D:
@thestemgamer3346
@thestemgamer3346 2 жыл бұрын
@@mastershooter64 a lot of things actually. First of all, Functional programming has a lot of important concepts like immutable data structures and pure functions. These are really useful because it allows you to compose code together in a way that is guaranteed to be safe and cause the least amount of side effects (a side effect is when a function changes something outside of its own scope). This is very handy for doing parsing, data transfer, and writing programs that can run on multiple threads at once. For some examples: Functional languages like elixir are pretty common for writing APIs and server code, Haskell is really handy when writing parsers, and languages like Rust try to take advantage of the Functional paradigm to write fast and safe concurrent code. Functional languages are also really common for mathematical and scientific purposes.
@girirajrdx7277
@girirajrdx7277 2 жыл бұрын
Hi there i am not able to follow this concept…can you explain or direct me to somewhere where i can learn better.
@zyansheep
@zyansheep 2 жыл бұрын
Didn't expect you here
@thestemgamer3346
@thestemgamer3346 2 жыл бұрын
@@zyansheep I am everywhere
@angelmendez-rivera351
@angelmendez-rivera351 2 жыл бұрын
Also, it should be noted that the statement that f is surjective if and only if it has a right-inverse requires the axiom of choice.
@MetaMaths
@MetaMaths 2 жыл бұрын
That is right ! But I hoped people would discover it themselves in the book.
@Achrononmaster
@Achrononmaster 2 жыл бұрын
An axiom of Set Theory for a framework that seeks to make sets secondary. Interesting. Seems one cannot truly escape the notion of a set, no?
@nomukun1138
@nomukun1138 2 жыл бұрын
Don't be shocked By the tone of my voice Check out my new axiom Axiom of Choice
@angelmendez-rivera351
@angelmendez-rivera351 2 жыл бұрын
@@Achrononmaster It depends on how elementary you want to make your primitive notions. This is why the idea of having a single foundations for mathematics is fundamentally strange.
@zapazap
@zapazap 2 жыл бұрын
I believe one can define the lambda calculus formally without recourse to sets IIRC.
@PowerhouseCell
@PowerhouseCell 2 жыл бұрын
Woah, this is such an underrated channel! As a fellow educational KZbinr, I understand how much work must have gone into this- amazing job!! Liked and subscribed :)
@Champs3443
@Champs3443 4 ай бұрын
omg! very elegant and simple explanation, amazing!
@acortis
@acortis 8 ай бұрын
Wonderful explanation! Please keep them coming!
@mathematicalmachinery7934
@mathematicalmachinery7934 2 жыл бұрын
Thank you for reading the notation out aloud at 4:32, it was really helpful :)
@gno7553
@gno7553 Жыл бұрын
Claire et succinct ❤
@metanick1837
@metanick1837 Жыл бұрын
Great analysis
@skit555
@skit555 2 жыл бұрын
This vid will only grow in value and views through time. This is great, thanks!
@joshuatilley1887
@joshuatilley1887 2 жыл бұрын
Perhaps worth stating that the equivalence of left(right) invertible, in(sur)jective, and mono(epi)morphism is false in most categories other than Set.
@MetaMaths
@MetaMaths 2 жыл бұрын
This is right, thanks. We are only talking about Set category here.
@luffis1985
@luffis1985 2 жыл бұрын
Great Video. Would've liked if you also covered surjection's three definitions aswell.
@rog7036
@rog7036 2 жыл бұрын
Im so glad I found this video, could you maybe also take a look at type theory? :)
@aditya234567
@aditya234567 2 жыл бұрын
Thank uuu
@tradingishard2038
@tradingishard2038 2 жыл бұрын
What is the analogue of a monomorphism for surjections ?
@angelmendez-rivera351
@angelmendez-rivera351 2 жыл бұрын
An epimorphism. Injective functions are the monomorphisms of the category of sets. Surjective functions are the epimorphisms of the category of sets. Bijective functions are the isomorphisms of the catgeory of sets. As such, two sets are isomorphic if and only if they are equinumerous, i.e, they have the same cardinality. Two objects A and B in a category are isomorphic if there exists a monomorphism from A to B and an epimorphism from A to B.
@igorhaladjian5718
@igorhaladjian5718 2 жыл бұрын
The last part is not completely accurate : Two objets in a category are isomorphic if there exists an isomorphism between them (which by definition is a morphism with a right-left inverse). This is not equivalent to having a monomorphism and an epimorphism from A to B (the injection Q->R in the category of topological spaces is both a monomorphism and an epimorphism but is not an isomorphism).
@angelmendez-rivera351
@angelmendez-rivera351 2 жыл бұрын
@@igorhaladjian5718 No, an isomorphism is necessarily invertible. That is what the definition of an isomorphism is. I do not know of any category-theoretic works that use a different definition. What you called an isomorphism is what I called an epimorphism.
@igorhaladjian5718
@igorhaladjian5718 2 жыл бұрын
Note that my definition of an isomorphism is a morphism with a left-right inverse ;-) This is the definition of being invertible, that is, for f:A->B an iso there exists g:B->A such that fg=id_B and gf=id_A. In this case, g is a left-right inverse to f.
@angelmendez-rivera351
@angelmendez-rivera351 2 жыл бұрын
@@igorhaladjian5718 Okay, I admit I was wrong, and you are right. I misread your comment, and you are right that for some categories, there existing a monomorphism and an epimorphism between two objects is not sufficient to guarantee the existence of an isomorphism. In fact, even in the category of sets, if the category is not defined by using the axiom of choice, then it is not true in general. This is because, at least in that category, a surjective function need not be right-invertible. So a function could be injective and surjective, but not right invertible. So it would not have a two-sided inverse.
@tombouie
@tombouie 2 жыл бұрын
Hmmmm .... most excellent; However I thought a function was a group of indep/dep pairs such that any pairs with =indep values all have =dep values Of course you can use a math expression to produce the above group of indep/dep pairs Oh I must apolozies but I missed what exactly category theory was, sorry.
@blinded6502
@blinded6502 2 жыл бұрын
Why do mathematicians love to multiply from right to left, when we read everything from left to right, most programming languages are executed from left to right, and when division works from left to right ((a*b)/b = a)?
@skit555
@skit555 2 жыл бұрын
Also, I'm seeing that math only treats lossless compressions. Using more natural domains, that perfectly bijective identity is the rare exception, not the norm.
@KevinFlowersJr
@KevinFlowersJr 2 жыл бұрын
Who needs university when you have KZbin & Discord
@williamtoner8674
@williamtoner8674 2 жыл бұрын
👍
@nyuh
@nyuh 2 жыл бұрын
hi i am here from oliver lugg
@MetaMaths
@MetaMaths 2 жыл бұрын
Hello
@bobblah78
@bobblah78 2 жыл бұрын
Me two
@angelmendez-rivera351
@angelmendez-rivera351 2 жыл бұрын
5:50 - 5:56 That is a misleading statement, if not an inaccurate one. You are still dealing with sets here. The set of all functions from A to B is denoted B^A, and the set of all functions from Z to A is denoted A^Z. The functions f°α and f°α' are functions from Z to B, and so they are elements of the set B^Z. In this context, composition ° is then a function from the Cartesian product of B^A with A^Z to B^Z. This is a necessary caveat, as otherwise, composition of functions is not even well-defined. There is some extended sense here in which one uses abstract algebraic terminology, in which we say f is cancellable if and only if, all α, α' in A^Z, f°α = f°α' implies α = α'. Thus, f is a monomorphism if and only if it is cancellable.
@MetaMaths
@MetaMaths 2 жыл бұрын
Correct ! But since it is "for any set Z", we are becoming a bit less element- centric. Monomorphism is about functions interacting and not about functions themselves.
@deyanirasaez9540
@deyanirasaez9540 Жыл бұрын
Perhaps his explanation can be extrapolated to the study of the human sciences, between sociology (focused on the whole, ignoring the individual, such as freedom, passions... -->objects, elements, identities) and psychology, focused on the individual (elements , objects). And perhaps to look at the tree or the forest. In this way, every theory looks at either the tree or the forest as a method to draw conclusions where both "looks" are needed to reach more general "truths." The social "kills" the individual, and looking only at the individual does not explain certain social facts (the important thing is what the person does, his social function, not his motivations or whys). The family is the measure of all things, because it is the middle state between the individual and the social. Maybe we have to look for the comparable concept "family" in mathematics. Sorry for the possible nonsense, I come from the human sciences
@geekoutnerd7882
@geekoutnerd7882 2 жыл бұрын
WHERE’s the rest :’(
@MetaMaths
@MetaMaths 2 жыл бұрын
I am working hard on the next chapter !
@angelmendez-rivera351
@angelmendez-rivera351 2 жыл бұрын
1:28 - 1:40 This is technically inaccurate. What this is describing is a functional graph, not a function. A function F is a set ((X, Y), G) such that G is a subset of the Cartesian product of X with Y, and such that for all elements x in X, there exists exactly one y in Y, such that (x, y) is in G. Here, I would like to clarify that the ordered pairs (x, y), (X, Y), and ((X, Y), G), are Kuratowski pairs, in set theory. The Kuratowski pair is defined as (x, y) := {{x}, {x, y}}, and it can be proven that this is always a well-defined object, by the axiom of extensionality and the axiom of pairing. The axiom of power set and the axiom of union guarantee the existence of the Cartesian product of X with Y, and the restricted axiom of comprehension guarantees the existence of the uniqueness restriction above. Therefore, this is a well-definition of a function.
@tonaxysam
@tonaxysam 2 жыл бұрын
Wait but you don't need the triplet... Given a relation R (a set whose elements are ordered pairs) you can get it's domain and it's range by taking the union twice
@angelmendez-rivera351
@angelmendez-rivera351 2 жыл бұрын
@@tonaxysam Again, what you are describing is a functional graph, not a function. What you described is not a function.
@tonaxysam
@tonaxysam 2 жыл бұрын
@@angelmendez-rivera351 but why not?
@angelmendez-rivera351
@angelmendez-rivera351 2 жыл бұрын
@@tonaxysam Because the codomain is a defining part of the function, and the graph cannot determine the codomain. Two functions ((X, Y), G) and ((U, V), H) are equal if and only if X = U, Y = V, and G = H. With this, the idea of surjectivity actually has meaning.
@tonaxysam
@tonaxysam 2 жыл бұрын
@@angelmendez-rivera351 okay, I can buy that the codomain information is not built into the function. But why would you keep the domain?
@lhpl
@lhpl 2 жыл бұрын
Hmm. Is this merely an advertisement for a book? First you talk about injection, surjection and bijection based in sets, then you rephrase them in left and right inverses, so far so good, then finally you go full abstract nonsense, but suddenly it's just about injections. Where did surjection go? I have watched a few "introductions to category theory", and it is my impression that most of them appear to only be understandable if one knows everything in advance - but then one would not need to watch an intro... I am still looking for an intro video on category theory that is as good and thorough as sudgylacmoe's _A Swift Introduction to Geometric Algebra_ is for geometric algebra...
@zapazap
@zapazap 2 жыл бұрын
Sets are primary and functions are secondary according to traditional logic? No. Traditional logic had no concept of sets or of functions. Although TFL does have the notion of truth-functions and PC has the idea of domains if discource. But TFL precedes PC Now in axiomatic set theory (eg ZFC) sets do precede functions. But set theory is not logic. Cheers! :)
@dembeto
@dembeto 2 жыл бұрын
I do not understand why french-speaking scholars dont explain things like that. It’s always a mess!!
@kennethcarvalho3684
@kennethcarvalho3684 2 жыл бұрын
Too much jargon for a non math
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