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May
11
comment Are there any nontrivial examples of contradictions arising in non-foundational or applied math due to naive set theory?
@TrevorWilson The whole enterprise of mathematical foundations only exists because there is the intuition that there is no "natural-seeming" way to prove the inconsistency of naïve set theory. I think if there were, we would have long ago given up on rigour.
May
11
comment Textbooks on higher category theory
Lurie's book is a textbook, but as you say, it not a textbook on $\infty$-categories. Leinster's book is not a textbook on $\infty$-categories in the same sense that CWM is a textbook on categories – you won't even find a higher Yoneda lemma in there. The same goes for the book of Cheng and Lauda.
May
11
revised Some questions about écarts
deleted 8 characters in body; edited title
May
10
comment What is the Eilenberg-Moore category of this diagonal-like monad?
Yes, of course.
May
10
comment Textbooks on higher category theory
The question is far from being settled. That is why there are no textbooks.
May
10
comment Pullbacks in the Ind-completion
@AdeelKhan It's not a set-theoretic obstacle. $\mathbf{Ind}(\mathcal{C})$ is reflective in $\mathbf{Psh}(\mathcal{C})$ if and only if it is cocomplete, but that doesn't always happen.
May
10
answered What is the Eilenberg-Moore category of this diagonal-like monad?
May
10
comment Pullbacks in the Ind-completion
@MeesdeVries It is true that the canonical fully faithful embeding $\mathbf{Ind}(\mathcal{C}) \to \mathbf{Psh}(\mathcal{C})$ preserves filtered colimits and all limits. But that doesn't tell you which limits exist or not.
May
10
comment Pullbacks in the Ind-completion
@AdeelKhan No. First you would need to know that $\mathbf{Ind}(\mathcal{C})$ is reflective and secondly you would need to know that the reflector preserves finite limits. The first doesn't always happen and I don't know whether the second ever happens.
May
10
answered Pullbacks in the Ind-completion
May
7
comment Definition of (left) resolution
First, build a morphism between $P$ and $A[0]$ using $\epsilon$. Then show it is a quasi-isomorphism if and only if $P_1 \to P_0 \to A \to 0$ is exact.
May
7
comment Definition of (left) resolution
Well, you haven't used the quasi-isomorphism condition...
May
7
comment An example of a coproduct of sheaves in the category of presheaves that is not a sheaf
Well, consider something silly like decomposing the empty set into two copies of itself vs three copies of itself...
May
7
comment An example of a coproduct of sheaves in the category of presheaves that is not a sheaf
Don't you have to quotient out by a suitable equivalence relation?
May
7
comment Question on the definition of a locally presentable category
No, it does not.
May
6
comment De Rham-Weil theorem
There are lots of different cohomology groups lying around. For instance, yes – an acyclic complex has vanishing cohomology groups. But taking global sections of an acyclic complex doesn't yield an acyclic complex.
May
6
answered Question on the definition of a locally presentable category
May
6
comment Continuous functors
Sure, take $\mathcal{D} = \emptyset$. Then the condition is vacuous.
May
6
revised Module of differentials in the functorial approach to schemes and quasi-coherent modules
added 78 characters in body
May
6
comment Is the diagonal map $\mathbb{C} \to \prod_{i=1}^\infty \mathbb{C}$ an etale map of rings?
No. Your confusion comes from the fact that Spec does not take infinite products to disjoint unions.