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Feb
4
comment Exactness of a right adjoint functor
Actually, that direction is not true, but fortunately we do not need it.
Feb
4
revised Exactness of a right adjoint functor
deleted 12 characters in body
Feb
4
comment “Coforgetful” functors?
If you don't send the morphisms somewhere then you don't have a functor. So you have to send the morphisms somewhere. I wouldn't say $F S$ has no morphisms – identity morphisms are still morphisms...
Feb
4
comment “Coforgetful” functors?
It is not correct. A functor from a category to a set is just a mapping of objects, but the morphisms still have to go somewhere...
Feb
4
comment Proof that the dual of a coalgebra is an algebra via commutative diagrams
@DavidHill One also needs to know that dualisation is compatible with the tensor product...
Feb
3
answered Exactness of a right adjoint functor
Feb
3
comment Exactness of a right adjoint functor
You haven't used the fact that $B$ is a cogenerator.
Feb
3
comment Name for categories with a certain property on coproducts
In some ways this is the complement of Bourn's notion of unital category, which asks for the canonical $X_0 + X_1 \to X_0 \times X_1$ to be a strong epimorphism.
Feb
3
answered Does the forgetful functor from smooth manifolds to sets preserve colimits?
Feb
1
comment Complete atomic boolean algebras as coalgebras of some endofunctor on Set
There are ways to prove such results, yes. For instance, the category of coalgebras for an endofunctor will be closed under retracts in a strong sense.
Feb
1
comment Complete atomic boolean algebras as coalgebras of some endofunctor on Set
Complete atomic boolean algebras are algebras for a monad. What makes you think they should be coalgebras for an endofunctor?
Feb
1
comment How do you know two morphisms are equal (without using elements)
Given two elements in a set, how can you know if they are equal?
Jan
31
comment Presheaves,simplicial sets, evaluation functor, Yoneda lemma,hom-functor
In fact this statement is just another way of stating the Yoneda lemma. You just have to unfold the definitions.
Jan
31
comment How do you define such map $(C^B \times B^A) \to C^A$?
Of course they are not literally equal. They don't even have the same domain and codomain.
Jan
30
comment A Grothendieck topology on $\Delta$
The Segal conditions are not really a sheaf condition – remember, covering sieves in a Grothendieck topology have to be closed under pullbacks. (In particular, neither quasicategories nor categories form a topos.)
Jan
30
answered Morphisms of varieties equal over algebraic closure
Jan
29
revised Algebraic closure with no nontrival automorphism
edited tags
Jan
29
answered How do you define such map $(C^B \times B^A) \to C^A$?
Jan
29
comment The cofree coalgebra using adjoint functor theorems
In Q2, shouldn't you be looking for a weakly terminal set of objects?
Jan
29
comment The cofree coalgebra using adjoint functor theorems
I suppose it's possible in principle. If you know that $\mathbf{Coalg} (\mathcal{C})$ is $\kappa$-accessible then the $\kappa$-presentable objects form a dense subcategory. The problem is that, in the standard textbooks, no estimate of $\kappa$ is given – you have to chase through the proofs carefully. However, you might have better luck looking at the 1977 preprint of Ulmer (check your email).