37,934 reputation
242112
bio website
location
age
visits member for 3 years, 8 months
seen 47 mins ago

Aug
26
comment Definition of quotient category
No. You may have to add new composites in the quotient.
Aug
26
comment How does a section of a stack give a sheaf?
I think you should go back to the case of a section of a constant sheaf. Yes, a section of a constant sheaf is a locally constant function – think about what its codomain is. Once you do so, it will be clear in what sense a section of a constant stack is a locally constant sheaf.
Aug
26
comment Definition of quotient category
Because it's the easy case where the functor to the quotient is surjective.
Aug
25
comment insight into the definition of intersection multiplicity for two plane curves
The number of generators is related to the dimension of the intersection of the tangent spaces. Since we are dealing with curves, that means the number of generators is at most one.
Aug
24
comment When is the map “attaching irreducible components” an effective isomorphism?
This appears to be the definition of effective epimorphism, rather than strict epimorphism. Of course there is no difference in this context.
Aug
24
comment insight into the definition of intersection multiplicity for two plane curves
On Q1: $\mathscr{O}_P / (f, g)$ is isomorphic to a localisation of $k [x, y] / (f, g)$: more precisely, it is isomorphic to the ring obtained by inverting in $k [x, y] / (f, g)$ the image of any function not vanishing at $P$. But $k [x, y] / (f, g)$ is (usually) artinian, hence is isomorphic to the direct product of some number of copies of $k [\epsilon] / (\epsilon^n)$; what this localisation does is to extract the copy corresponding to $P$.
Aug
23
comment Building a function $p : \mathcal{D} \rightarrow \mathbf{Ord}$ from a faithful functor $U : \mathcal{C} \rightarrow \mathcal{D}.$
This appears to be an impoverished version of categorical topology. See Abstract and concrete categories.
Aug
22
comment Category of Hilbert Spaces
It seems unlikely. The kernel of a bounded linear operator is a closed subspace, but I don't think the image of a bounded linear operator is necessarily closed.
Aug
22
comment When may the bar resolution be truncated for calculating $\operatorname{hocolim}$?
Yes, I expect it should work, but I have never seen it written out explicitly.
Aug
22
comment Regular function on a variety which is not globally rational
The only thing to prove is what I said in my first comment, which you seem to already know.
Aug
22
comment When may the bar resolution be truncated for calculating $\operatorname{hocolim}$?
No, you can't even do that for $\mathbb{N}$. You have to restrict to a suitable subcomplex of the 1-truncated bar complex.
Aug
22
comment Regular function on a variety which is not globally rational
Yes, that's what I said.
Aug
22
comment When may the bar resolution be truncated for calculating $\operatorname{hocolim}$?
Because $\mathbb{N}$ is a free category.
Aug
22
comment When may the bar resolution be truncated for calculating $\operatorname{hocolim}$?
The second formula. The first formula is the definition.
Aug
22
comment Regular function on a variety which is not globally rational
Look: every element of the coordinate ring of $V$ comes from some element of the coordinate ring of $\mathbb{A}^n$, and geometrically this means that every regular function $V \to \mathbb{A}^1$ comes from the restriction of some regular function $\mathbb{A}^n \to \mathbb{A}^1$, which are always polynomial functions.
Aug
22
comment Regular function on a variety which is not globally rational
But it's true. Every regular function $V \to \mathbb{A}^1$ comes from a (not necessarily unique) regular function $\mathbb{A}^n \to \mathbb{A}^1$, when $V$ is closed in $\mathbb{A}^n$.
Aug
22
comment Regular function on a variety which is not globally rational
Your question could be phrased better. Do you mean to ask whether every regular function $V \to \mathbb{A}^1$ occurs as the restriction of some rational function $\mathbb{A}^n \to \mathbb{A}^1$ whose domain contains $V$?
Aug
22
comment When may the bar resolution be truncated for calculating $\operatorname{hocolim}$?
What I am saying is that the formula works if you change the coproduct.
Aug
22
comment Regular function on a variety which is not globally rational
If $V$ is affine then there is a natural bijection between the regular functions $V \to \mathbb{A}^1$ and the elements of the coordinate ring of $V$. (If $V$ is projective then the only regular functions $V \to \mathbb{A}^1$ are constants.)
Aug
22
comment When may the bar resolution be truncated for calculating $\operatorname{hocolim}$?
You have to understand the formula correctly. The subtlety is in the indexing of the first coproduct: it's not a coproduct over all arrows in $\mathbb{N}$, just the generators.