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Mar
22
revised isomorphism between divisible, totally ordered, abelian groups
edited tags
Mar
22
answered Is there a first order theory for equivalences classes?
Mar
22
comment Relating Ext groups of abelian groups and group cohomology
I don't see why that would work.
Mar
22
comment Relating Ext groups of abelian groups and group cohomology
Yes, you're right. $B G$ can't be obtained by a bar construction because the underlying simplicial set of a bar resolution admits an extra degeneracy hence is homotopically discrete.
Mar
22
comment Is there a first order theory for equivalences classes?
What are the morphisms under consideration? What do you mean by an "equivalence relation"? A category is not just a collection of objects.
Mar
21
comment When do finite sets embed in a category?
In a certain sense. Read the link page.
Mar
21
comment Question about accessibility of category of free abelian groups.
It's pure ZFC. All you need is a non-principal ultrafilter and Łoś's theorem.
Mar
21
answered When do finite sets embed in a category?
Mar
21
comment When do finite sets embed in a category?
Your second question is poorly phrased. It is not even true for $\mathbf{Set}$: one side is cocomplete, the other is not.
Mar
21
comment Question about accessibility of category of free abelian groups.
I just gave you a ZFC proof that being a free abelian group is not a first-order property!
Mar
21
comment Relating Ext groups of abelian groups and group cohomology
Depends on what you apply it to. If you apply it to $G$ with the permutation self-action then you get $E G$. If you apply it to $1$ you get $B G$.
Mar
21
comment Question about accessibility of category of free abelian groups.
In principle, yes. But as you say, we do not know whether it is accessible or not.
Mar
21
comment Finitely generated k-algebra is Noetherian
$k [x]$ is noetherian, but $k [x^2]$ is not an ideal. (It is a subring.)
Mar
21
comment Question about accessibility of category of free abelian groups.
The property of being a free abelian group is not first-order. Indeed, consider a non-trivial ultrapower of $\mathbb{Z}$. It contains an element $(1, 2, 6, 24, 120, \ldots)$ that is divisible by every integer, hence contains a copy of $\mathbb{Q}$.
Mar
21
awarded  category-theory
Mar
21
comment Correct Definition of Concrete Category over Set
Awodey's suggestion is not bad, but excludes examples like $\mathbf{Set} \times \mathbf{Set}$.
Mar
21
comment Prime Ideals as Ring theoretic Ultrafilters
The statement is simply incorrect. Prime ideals are generalisations of ultrafilters, but not for the given reason.
Mar
21
answered A proof using Yoneda lemma
Mar
20
answered Relating Ext groups of abelian groups and group cohomology
Mar
20
revised Definition of locally presentable category
added 247 characters in body