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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
awarded  Popular Question
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.
Jan
23
answered Sums of topological spaces
Jan
23
comment The problem of bound variables in mathematical definitions
This is called $\alpha$-equivalence in computer science.
Jan
21
comment About stalk of sheaves in an unbalanced category
Yes, indeed. It works as long as you have a terminal object, and you would have to if you wanted to make sense of sheaves. (What's the object of sections over the empty set?)
Jan
21
awarded  Nice Question
Jan
21
comment Is there a quicker argument from the HBL derivability conditions to the equivalence of fixed points of $\neg\Box$ to $\mathsf{Con}$?
I should adopt this quasiquotation notation!
Jan
21
answered The image of a morphism between affine algebraic varieties.
Jan
20
comment Categories with no products/coproducts
Note however that the Yoneda embedding $\mathcal{C} \to [\mathcal{C}^\textrm{op}, \textbf{Set}]$ is not guaranteed to preserve colimits.
Jan
20
comment Categories with no products/coproducts
The question is built on the misconception that all categories are concrete categories of mathematical structures of one kind or another. The fact of the matter is that, by moving to a sufficiently large universe if necessary, any fixed category can be embedded in one that is complete and cocomplete. But this is completely abstract nonsense and is not guaranteed to preserve any of the colimits that previously existed.
Jan
20
comment About stalk of sheaves in an unbalanced category
Taking stalks is a left adjoint functor $\textbf{Sh}(X; \mathcal{C}) \to \mathcal{C}$, so $\textbf{Psh}(X; \mathcal{C})$ doesn't come into play.