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Jan
30
comment Proof that $B^{A \times A'}$ is isomorphic to $(B^A)^{A'}$ in a CCC
$\textrm{Hom}(X, -)$ doesn't reflect isomorphisms in general, but the collection of all $\textrm{Hom}(X, -)$ jointly reflect isomorphisms.
Jan
29
comment classical topology but with lattices
If you demand that your frame $\tau$ is embedded as a subframe of $P X$ for some set $X$, then you are asserting that $\tau$ actually is a topology on $X$. In effect, you are suggesting that we restrict the study of locales to those that actually come from topological spaces. In other words there is no added generality at all! So why not just study these things qua topological spaces?
Jan
28
comment Finding a functor satisfying a recursive equation
Perhaps Kelly's 1980 paper, A unified treatment of transfinite constructions, is relevant here.
Jan
27
answered Images of simple modules under exact endofunctors
Jan
27
revised How to define the dual sheaf
added 34 characters in body
Jan
27
comment Images of simple modules under exact endofunctors
The answer to your question is extremely simple and is contained in my previous comments. If you do not see it, then most likely this is because your understanding of category theory is incomplete. For instance, is it obvious to you that submodules are the same thing as monomorphisms?
Jan
27
comment Defining the Riemann-Roch space of a divisor
$K(U)^*$ means units. (I prefer to write $K(U)^\times$ to avoid confusion.)
Jan
27
answered How to define the dual sheaf
Jan
27
comment Images of simple modules under exact endofunctors
A simple module is a module $M$ such that every submodule is either $0$ or $M$. There's not much more to it than that! The trouble is that $\mathcal{C}$ may not contain all the submodules of $M$, so from the perspective of $\mathcal{C}$, $M$ may appear to be simple but not actually be simple.
Jan
27
comment Images of simple modules under exact endofunctors
Your functor is not merely exact but an equivalence of categories. Being a simple module is something that can be detected purely in terms of the category of modules, so assuming $\mathcal{C}$ has enough submodules, your functor will preserve simple modules.
Jan
27
comment Is the appropriate notion of homomorphism between models of set theory the “obvious” one?
@user18921: You could do that, but then homomorphisms won't preserve the truth of more complicated formulae. I think the usual notion of homomorphism here is that of elementary embedding.
Jan
27
comment Forcing and generic extensions
You could do that if $\mathbf{V}$ is really just a countable transitive model. But we could equally well take $\mathbf{V}$ to be the actual universe of all sets.
Jan
27
comment Forcing and generic extensions
The trouble is that $\mathbf{V}[G]$ is a mathematical fiction (at least from the perspective of $\mathbf{V}$) and so we must first define what it means for something to be true or false in $\mathbf{V}[G]$.
Jan
26
comment Can exponentials be distinct from hom-functors in enriched categories?
The internal logic of toposes is better understood than the internal logic of general symmetric monoidal closed categories, as far as I know.
Jan
26
comment Is the completeness theorem for first-order logic relative to one's choice of set theory?
@PeterSmith It seems to be a general phenomenon that restricting attention to things coded by natural numbers makes theorems provable without choice. I recall some posts by Simpson along those lines...
Jan
26
answered Can exponentials be distinct from hom-functors in enriched categories?
Jan
25
comment What properties are preserved under a measurable mapping?
No, it does not. The definition I use is the one implicit in the notion of a functor "reflecting $P$" in category theory. You should read a category theory textbook.
Jan
25
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Jan
25
comment What properties are preserved under a measurable mapping?
A more precise definition would say, the property $P$ is reflected by a map $F$ if, when $F x$ has property $P$, then $x$ has property $P$. Typically one also demands that $F$ preserves property $P$ as well. So, for example, a continuous map does not reflect openness of subsets, but a continuous open map does.
Jan
24
comment What properties are preserved under a measurable mapping?
@Tim No. A property being reflected is not the same thing as being preserved by the inverse image. It's a bit more sophisticated than that.