37,993 reputation
243114
bio website
location
age
visits member for 3 years, 8 months
seen 53 mins ago

Sep
3
comment Abstract nonsense proof that stalks of $\mathcal{O}_X$ modules are modules over $\mathcal{O}_X$-stalks
The category of sheaves of sets, of course.
Sep
3
comment Abstract nonsense proof that stalks of $\mathcal{O}_X$ modules are modules over $\mathcal{O}_X$-stalks
If you think in terms of action maps, then the relevant fact is that taking stalks preserves finite limits (in particular finite products).
Sep
3
comment Do schemes have a characterisation as an etale space?
No. Read carefully: "Formally, a ringed space $(X, O_X)$ is a topological space $X$ together with a sheaf of rings $O_X$ on $X$."
Sep
3
comment Do schemes have a characterisation as an etale space?
A ring is in particular a set, so you can construct its espace étalé the same way. The rest of your question still indicates some confusion about what a scheme is. A sheaf of rings can't be locally affine – that's a property of ringed spaces; a scheme is not merely a space; etc.
Sep
3
comment Tensor product of injective ring homomorphisms
Yes, indeed. I used Gröbner bases in Mathematica for my calculations.
Sep
3
comment Recommendation on Category theory textbook
You can work purely within ZFC if you like, you just have to pay a lot of attention to size issues. It only confuses beginners.
Sep
2
comment Tensor product of injective ring homomorphisms
@MartinBrandenburg As it turns out, there is an example!
Sep
2
comment Étalé space for sheaf of sections of a fiber bundle
There is no nice description, because the espace étalé itself is not nice. For instance, the one corresponding to the sheaf of smooth functions on a smooth manifold is not even Hausdorff.
Sep
2
comment Tensor product of injective ring homomorphisms
Related question on MO.
Sep
2
comment Colimits in the category of “sets with partial mappings”
Well, having a right adjoint, the forgetful functor is a left adjoint and therefore preserves colimits. It's as simple as that.
Sep
2
comment Tensor product of injective ring homomorphisms
Hmmm, yes. So the only thing that can go wrong is that some nilpotent goes to zero in $A \otimes_R B$ but not in $A$ or $B$. Maybe there is an example after all.
Sep
1
comment Is there a minimal axiomatization of ZFC?
Take the theory with countably many constants $c_i$ and axioms $c_i \ne c_j$ for $i < j$. Then that axiomatisation is minimal, no? The same argument works for both the finite and infinite case.
Sep
1
comment Definition of “contradiction” and use of the term for “⊥”
It is convenient to have $\bot$ as a propositional constant. If you don't like it, you can replace it with any of your favourite contradictions – e.g. $0 = 1 \land 0 \ne 1$ – they are all logically equivalent. In other words, $\bot$ is the abstract contradiction.
Aug
31
comment What is the right category in which to think of adjoints?
Actually, I think the convention is that a functor $\mathcal{C}^\mathrm{op} \times \mathcal{D} \to \mathbf{Set}$ is a profunctor from $\mathcal{D}$ to $\mathcal{C}$. This is certainly the case on the cited page. A mnemonic for remembering this: the category of profunctors from $\mathcal{D}$ to $\mathcal{C}$ is equivalent to the category of left adjoints $[\mathcal{D}^\mathrm{op}, \mathbf{Set}] \to [\mathcal{C}^\mathrm{op}, \mathbf{Set}]$, at least when $\mathcal{C}$ and $\mathcal{D}$ are small.
Aug
31
comment Closed model categories in the sense of Quillen [1969] vs the modern sense
No. I'm not looking for a reference.
Aug
31
comment Sheaf associated to sheaf on basis
Verifying the stalks are the same is somewhat non-trivial, but the bigger problem is that a sheaf is not determined by its stalks alone. See here.
Aug
31
comment Two questions on kernels
Sorry, you need to assume $k'$ is a monomorphism.
Aug
31
comment Is intuitionistic naive set theory consistent?
The argument also goes through in minimal logic, because we don't use ex falso quodlibet. The only way out seems to be paraconsistent logic.
Aug
30
comment Sheaf associated to sheaf on basis
You can't define $\mathscr{G}$ arbitrarily – after all, it has to be a presheaf, so the restriction maps have to compose correctly etc. – so it's not obvious that you can construct such a $\mathscr{G}$ in the first place.
Aug
30
comment Hom functor and left exactness
That is in fact true. Take $M$ to be the free module on one generator. But my proof is more general and applies to all additive categories.