Category theory foundations 1.0 - Steve Awodey

[@dezenterrier]

This is great. Beside the fact that english is not my primary language I've understood much more from this than my native language lecture.

[@thatkindcoder7510]

When you speak better in English than in your Native language, but still suck at both (My first language is Arabic) (Wait a sec that's not what you meant)

[@samueldeandrade8535]

14:36 That's ... a terrible observation. I mean, defining compositions are not that easy. There is a definition that it is easy, concatenation of arrows of a graph. Besides that, it is not an easy problem to to "define compositions" like his statement suggests.

[@Crasshopperrr]

Great speaking!

[@Crasshopperrr]

14:00 Aren't there two candidates for *3*? ●→●←● and ●→●→●

[@gurmeet0108]

+Lao Tzu yes you are right, both of them (among others having 3 objects) are valid categories (of course, after adding identity arrows and arrows for composition which is needed in your second example), it's just category '3' is defined as second one.

[@firebrain2991]

if I recall correctly, there must be arrows from each element to another (and the composites) with no loops (or it's not finite), which means you must draw a function from A->C or C->A on the upper category, or A->C (composed of A->B and B->C) on the lower one (if you draw an arrow from C->A it creates a loop and is thus not finite) once these are applied, the two categories described become isomorphic, otherwise only the bottom one fits the definition

[@blinkybool]

@@firebrain2991 "there must be arrows from each element to another" - this is not true. Also there are infinitely many categories on 3 objects (even on one object). Picking a source and target for an arrow does not uniquely determine the arrow in an arbitrary category. You can have many non-equal arrows from object A to object B (or even back to object A), and there can be no arrows between two distinct objects. Any phrases you've heard about "you need to have this arrow", is referring either to the axiom of composition or the axiom of identity. You might have also seen someone constructing a finite categories, drawing a bunch of arrows and their compositions, and then saying "and now we're done". This is not because no more arrows can be added. It's just that they have started with some graph partially representing a category, and are aiming to construct the "simplest" categeory containing those objects and arrows they started with (this is called a free construction). I'm not quite what the advantage is to defining the category 3 as in this video, but I would imagine there is a nice embedding of the category "N" into a larger number category "M"

[@firebrain2991]

@@blinkybool I apologize that I did not clarify this (I tend to assume context that I shouldn't, sorry), but I was talking about the category that is named after the numbers, not every category which I was remembering that the category defined as the integer "N" had to be like the free construction you mentioned. I'd suggest that the main reason for this (other than the nice embeddings) is that it reflects the construction of the natural within sets (where each one contains all the previous ones, and there is exactly one incrementor) But I've also seen in some other books that you can redefine each category "N" at a whim just to give it a nice name when you're not using the default definition, so maybe it's just a nice placeholder that anybody can pick up on and mess with

[@user-wk7iq1yf4w]

I didn't really understand the monoid example. He has put the arrow around every element m of M, and then how did he define composite of arrow around m1 and m2? Wait, are our objects monoids or the elements of a certain monoid M?

[@alex1985bond]

I also got confused for a sec. There is just one object M. The elements of M define the arrows. He explains it very well in his notes www.andrew.cmu.edu/course/80-413-713/notes/chap01.pdf

[@user-wk7iq1yf4w]

Alexander Bondarenko thanks :)

[@thatkindcoder7510]

@@alex1985bond So the object, M, being the monoid, is being put through a function that multiplies all it's elements by an element in the monoid? That's my interpretation, I'll read into his notes

[@hhaavvvvii]

What did they keep asking?

[@instance3954]

"write larger," I think :))

[@יחזקאלהירשהורן]

It's POSETS, not POSTSETS. Stands for Partially Ordered SETS. He actually says that.