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Aug
17
comment Category of profinite groups
The other thread has an answer now, so voting to close this one as a duplicate.
Aug
17
awarded  Popular Question
Aug
17
comment Category of profinite groups
@user48900 Yes, as I said. My point is that there is no reason to expect it to be accessible – if it is then it is some kind of accident. (For example, the category of profinite abelian groups is definitely not accessible: because it is equivalent to the opposite of the category of torsion abelian groups, which is locally presentable.)
Aug
17
comment Category of profinite groups
Yes, in the category of topological spaces. But in the category of compact Hausdorff spaces, the coproduct of countably many copies of $1$ is the ultrafilter space $\beta \mathbb{N}$.
Aug
17
comment Category of profinite groups
@user48900 I'm inclined to believe the answer is no, but for a different reason: the category of profinite groups is the pro-completion of the category of finite groups, so its opposite is an accessible category. There is a result that says that the opposite of a locally presentable category is locally presentable if and only if it is a preorder, but unfortunately that is not applicable in this case.
Aug
17
comment Category of profinite groups
@PeteL.Clark That only shows that profinite groups are not closed under colimits in topological groups, however. For instance, the category of compact Hausdorff spaces is cocomplete but not closed under colimits in the category of all topological spaces.
Aug
17
comment Is it possible to formalize (higher) category theory as a one-sorted theory, just like we did with set theory?
The problem, really, is that there is not even a commonly-accepted first-order theory, multi-sorted or otherwise, of the category of categories yet.
Aug
17
comment Is it possible to formalize (higher) category theory as a one-sorted theory, just like we did with set theory?
Any multi-sorted first-order theory can be reformulated as a single-sorted one, but that is essentially just a trick and is conceptually meaningless.
Aug
15
comment Dependent choice does not imply “the reals are well-ordered”; citation?
So, you're not looking for the Solovay model, then?
Aug
15
awarded  Nice Answer
Aug
15
comment Trying to prove structure result for ${\rm Hom}(A,B)$
In that case, anon's comments are right: the algebra structure on $A$ is irrelevant, as is the homomorphism $\pi$.
Aug
15
answered The sum of the first $n$ natural numbers is $n(n+1)/2.$ How do we express this in Peano arithmetic?
Aug
15
comment Trying to prove structure result for ${\rm Hom}(A,B)$
What do you mean by $\mathrm{Hom}$: module homomorphisms or algebra homomorphisms?
Aug
15
comment Exactness of direct image functor
The direct image functor is a right adjoint functor, which means it is left exact.
Aug
15
answered Epistemic disjunction, axiom or rule?
Aug
14
comment categorical interpretation of quantification
I'm afraid adjunctions require some reasonable understanding of categories and functors. But then again, there's no reason to expect quantifiers to be easy! This is explained briefly in Awodey's Category theory.
Aug
14
comment categorical interpretation of quantification
The introduction/elimination rules for quantifiers exhibit them as adjunctions between appropriate categories of formulae.
Aug
14
comment Examples of $\mathcal{O}_X$-modules that are not quasi-coherent sheaves
The statement about "every presheaf" is, strictly speaking, only valid if you use Hartshorne's definition of presheaf (as opposed to the one appearing in the 1971 edition of EGA I).
Aug
14
comment Please list a few topological groups that I should learn about.
Profinite groups (and Galois groups in general) should be added to the list.
Aug
13
comment Notations that are mnemonic outside of English
I always thought $B$ stood for "boundary".