Zhen Lin
Reputation
392/400 score
 Apr16 comment Looking for info on power set functor Not especially. You can start here. Apr16 comment Is the product of all objects of a finite category an initial object? @armchairprogrammer Perhaps the result you are thinking of is that if the limit of the identity diagram exists, then it is an initial object. Apr16 comment Reference request for the independence of $\text{Con}(\mathsf{ZFC})$. If ZFC is $\omega$-consistent, then $\lnot \mathrm{Con}(\mathrm{ZFC})$ is not provable. (If it were, then by $\omega$-consistency, ZFC is inconsistent – a contradiction!) Apr16 comment Correct definition of subframe That is correct. Apr16 comment Looking for info on power set functor For better or worse, that's what it's called here as well. Apr15 comment Looking for info on power set functor The observation you speak of is called Frobenius reciprocity: $\exists_f (X' \cap f^{-1} Y') = \exists_f X' \cap Y'$. It is equivalent to the fact that $f^{-1}$ preserves the Heyting implication. Apr15 comment Looking for info on power set functor These are not functors between categories but rather monotone maps between posets. Apr15 comment Looking for info on power set functor $\exists_f$ is the left adjoint of pullback $f^{-1}$, $\forall_f$ is the right adjoint of $f^{-1}$. Apr15 comment Is there a more general obstruction to the existence of moduli spaces than the existence of automorphisms? The $X$-th power. Or the category of functors $X \to \mathbb{B} \mathrm{Aut} (A)$. Apr15 comment In Ring Theory, does a 'power' of a morphism represent composition? You can't make the set of continuous functions $\mathbb{R} \to \mathbb{R}$ into a ring with pointwise addition and composition. You would have to restrict to linear functions, in which case you get something isomorphic to $\mathbb{R}$ again. Apr15 comment Questions about the function fields of complex algebraic surfaces Well, for a start, you should take $f : X \to Y$ to be dominant. And then you should probably also impose a finiteness condition on $f$. Apr15 comment Why do counits go that way? I think the easiest answer is that it would break self-duality. Apr15 comment Why is the internal hom of a Kan complex also a Kan complex? There is an argument of that type in [Goerss and Jardine, Ch. I, §4]. Apr15 comment Why is the internal hom of a Kan complex also a Kan complex? No. It only works for sufficiently nice model categories, e.g. cartesian model categories in which all objects are cofibrant and cofibrations = monomorphisms. Apr14 comment When is $\mathbb{Q}(x)$ a finite extension of $\mathbb{Q}$? @RobertLewis Use Zorn's lemma to construct transcendence bases of $\mathbb{C}$ and $\overline{\mathbb{Q}}$, and then use Zorn's lemma again to construct the appropriate automorphisms by extending automorphisms of $\overline{\mathbb{Q}}$. Apr14 comment Functorizing a choice of sections I think $\mathcal{T}'$ is supposed to be the category obtained from $\mathcal{T}$ by freely adjoining a section for every morphism in $\mathcal{T}$ subject to the constraint that the sections compose. Apr14 comment Subobjects(A) $\cong \operatorname{Hom}(A,\Omega)$ in a topos is a natural transformation? Well, it is a correct observation, but I don't think it is especially interesting. You can get rid of $\Omega$ and turn it into a statement about pullbacks: the pullback of $n \circ m$ along $n$ is $m$ again, which is what you expect. Apr14 comment Subobjects(A) $\cong \operatorname{Hom}(A,\Omega)$ in a topos is a natural transformation? The directions of your morphisms are mixed up. Think through everything carefully and re-ask your question. Apr14 comment First axiom of sheaves: in noetherian topological spaces the direct limit presheaf is a sheaf. The point is indeed that every cover of every open subset of a noetherian topological space has a finite subcover. So everything that needs to be amalgamated can be amalgamated by going "far enough" in the direct system. Apr14 comment Representability criterion with universal element You are correct: a universal element for $X$ is precisely an initial object in $(1 \downarrow X)$.