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Apr
16
comment Looking for info on power set functor
Not especially. You can start here.
Apr
16
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.
Apr
16
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!)
Apr
16
comment Correct definition of subframe
That is correct.
Apr
16
comment Looking for info on power set functor
For better or worse, that's what it's called here as well.
Apr
15
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.
Apr
15
comment Looking for info on power set functor
These are not functors between categories but rather monotone maps between posets.
Apr
15
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 $f^{-1}$.
Apr
15
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 $X \to \mathbb{B} \mathrm{Aut} (A)$.
Apr
15
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.
Apr
15
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 $f$.
Apr
15
comment Why do counits go that way?
I think the easiest answer is that it would break self-duality.
Apr
15
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].
Apr
15
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.
Apr
14
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 $\overline{\mathbb{Q}}$.
Apr
14
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.
Apr
14
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.
Apr
14
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.
Apr
14
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.
Apr
14
comment Representability criterion with universal element
You are correct: a universal element for $X$ is precisely an initial object in $(1 \downarrow X)$.