Zhen Lin
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 Apr15 answered Is there a more general obstruction to the existence of moduli spaces than the existence of automorphisms? 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 answered Why is the internal hom of a Kan complex also a Kan complex? 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 revised Subobjects(A) $\cong \operatorname{Hom}(A,\Omega)$ in a topos is a natural transformation? deleted 31 characters in body 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 answered Representability criterion with universal element 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 $(1 \downarrow X)$. Apr14 comment Is GRP a subcategory of SET, or not? See also this question. Apr14 comment Is GRP a subcategory of SET, or not? Well, is a group a set, or not? Think very carefully. Apr13 comment What are monomorphisms in the category of real vector bundles over a fixed base space $X$? But there are non-projective finitely presented modules! Apr13 comment What are monomorphisms in the category of real vector bundles over a fixed base space $X$? Compact objects in the category of modules over a ring (commutative or otherwise), in the sense of Definition 1 on the linked page, are precisely finitely presented modules. Compact objects in the sense of additive categories (esp. triangulated categories) are an even weaker notion. You need to impose projectivity separately. Apr13 comment What are monomorphisms in the category of real vector bundles over a fixed base space $X$? The definition of "compact object" you link to actually refers to finitely presentable objects, not dualisable objects. Apr13 comment What are monomorphisms in the category of real vector bundles over a fixed base space $X$? I would be inclined to say that the category of vector bundles is too small for the notion of monomorphism/epimorphism to behave well. It is better to look at the category of $\mathscr{O}_X$-modules, say.