Bits like 7:00 - 7:30 (or thereabouts) are some of the most useful and important parts of these videos. There are tons of free resources out there for learning mathematics, but one of the most important parts of learning any subject is figuring out what one's priorities are. This is easy when you have an established expert to tell you (like here), but hard to do by oneself; and it takes a lot of courage and time-investment for a non-expert to routinely decide such things for themselves and not be plagued by imposter syndrome. Thank you!
It is always good to watch your videos even on subjects I (supposedly) already know as you collect examples from very different subjects and show the universality of adjoints. As a side question, do you intend to treat Yoneda's lemma at some point?
Raymond Chan gave a counterexample which excludes non-existent sets, which is one third of the cases to consider! Look into cases of the empty set also, which becomes the most general under extendaboobles (retracts in category theory). He notes the withering theorem applies.
So the fixed point functor sends invariant elements of the set back to set. What does it send if every element of the set is not invariant? Presuming an empty set?
I'm not really sure how morphisms work in G-sets. Do morphisms between objects in a G-set change the group action as well as the set? Because in the exercise provided we start off with a 1-set from the trivial G-set functor. Does the morphism turn this into a G-set?
A morphism of G-sets is a morphism of the underlying sets that commutes with the group actions. I.e., if s is an action of G on X, and t is an action of G on Y, then a morphism is a set function f : X -> Y such that f(s(g, x)) = t(g, f(x)) for all g In G and x in X. This looks nicer if we write it as f(g · x) = g · f(x), where the dot on the left is s and the dot on the right is t.
I might be mistaken, but I think that the forgetful functor should be from the category of unital associative algebras over k (rather than the whole of Rings) into Lie algebras over k
For the exercises, I think I get that trivial ⊢fixed points, orbits ⊢ trivial, G x - ⊢ forget, and forget ⊢ Hom(G, -). Do people agree with me? (and how did people figure out the last one? I'm not too comfortable with actions yet, and I just got it from process of elimination)
For the last one, given f:X->S, define F:X->Fun(G,S) by F(x)(g)=f(g.x). It's kind of 'the only sensible thing to do', and you can check it's a morphism in GSet etc. (To go the other way, just take f(x)=F(x)(1))
Just looking this to begin with, is a twice-forgetful functor, by Cayley's theorem as a universal, a group in associative category theory? Extend Lie-type algebras to the nonassociative case! What is the hypergeneral nonassociative structure? There are covariant (Lie algebra) and the usual contravariant parenthesis cases. Extend to Zargon games!
Bits like 7:00 - 7:30 (or thereabouts) are some of the most useful and important parts of these videos. There are tons of free resources out there for learning mathematics, but one of the most important parts of learning any subject is figuring out what one's priorities are. This is easy when you have an established expert to tell you (like here), but hard to do by oneself; and it takes a lot of courage and time-investment for a non-expert to routinely decide such things for themselves and not be plagued by imposter syndrome. Thank you!
Adjoint functors were something I struggled to understand on my own for a while, thanks again :)
Remark: Here is a simple counterexample to the free-forgetful adjunction: the forgetful functor from Field to Set in fact has NO left adjoint.
It is always good to watch your videos even on subjects I (supposedly) already know as you collect examples from very different subjects and show the universality of adjoints. As a side question, do you intend to treat Yoneda's lemma at some point?
Absolute life saver!!
Raymond Chan gave a counterexample which excludes non-existent sets, which is one third of the cases to consider! Look into cases of the empty set also, which becomes the most general under extendaboobles (retracts in category theory). He notes the withering theorem applies.
So the fixed point functor sends invariant elements of the set back to set. What does it send if every element of the set is not invariant? Presuming an empty set?
It sends a G-set with no fixed points to the empty set, yes.
I'm not really sure how morphisms work in G-sets. Do morphisms between objects in a G-set change the group action as well as the set? Because in the exercise provided we start off with a 1-set from the trivial G-set functor. Does the morphism turn this into a G-set?
A morphism of G-sets is a morphism of the underlying sets that commutes with the group actions. I.e., if s is an action of G on X, and t is an action of G on Y, then a morphism is a set function f : X -> Y such that f(s(g, x)) = t(g, f(x)) for all g In G and x in X. This looks nicer if we write it as f(g · x) = g · f(x), where the dot on the left is s and the dot on the right is t.
I might be mistaken, but I think that the forgetful functor should be from the category of unital associative algebras over k (rather than the whole of Rings) into Lie algebras over k
For the exercises, I think I get that trivial ⊢fixed points, orbits ⊢ trivial, G x - ⊢ forget, and forget ⊢ Hom(G, -). Do people agree with me? (and how did people figure out the last one? I'm not too comfortable with actions yet, and I just got it from process of elimination)
For the last one, given f:X->S, define F:X->Fun(G,S) by F(x)(g)=f(g.x). It's kind of 'the only sensible thing to do', and you can check it's a morphism in GSet etc.
(To go the other way, just take f(x)=F(x)(1))
thank you so much for making these lectures available!
thank you professor, great lecture!
Just looking this to begin with, is a twice-forgetful functor, by Cayley's theorem as a universal, a group in associative category theory? Extend Lie-type algebras to the nonassociative case! What is the hypergeneral nonassociative structure? There are covariant (Lie algebra) and the usual contravariant parenthesis cases. Extend to Zargon games!
amazing...
Sind sie sehr gut?
Ja ist er
Das ist sehr gut!
LYSDEXIC'S NIGHTMARE!
yeeeeeeeeeeeeeeeeeeeeee