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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.
Aug
22
comment In an abelian category,every morphism can be written as composition of epi and mono.
In the meantime, you might to look at proposition 6.3.17 here.
Aug
22
comment In an abelian category,every morphism can be written as composition of epi and mono.
For completeness, you should state the definition of abelian category Weibel uses. Of course, all you have to do is prove that the canonical $\operatorname{Ker} \operatorname{coker} f \to \operatorname{Coker} \operatorname{ker} f$ is an isomorphism. I am sure this has been asked before on this site somewhere...
Aug
21
comment What is $X^{\omega}$ where $X$ is a set?
In this context, a finite sequence is not a sequence "indexed by the nonnegative integers".
Aug
20
comment Exercise on localization as a colimit
Yes, it works. If anything what you need is that $A$ is commutative.
Aug
19
comment Ring of rational power series
If $A$ is a field then $R_A$ should be the subring of $A (t)$ consisting of those rational functions whose denominator does not vanish at 0. Or perhaps you mean to ask about Laurent series?
Aug
19
comment homology commutes with direct sum and product?
Any additive functor preserves finite direct sums/products. In particular, both cokernels and kernels (hence also images and quotients) commute with finite direct sums/products.
Aug
18
comment Cohomology of wedge equals direct sum of cohomologies
Well, now that you've restricted to path-connected spaces, you can establish the claim in degree 0 by hand and in positive degrees by reducing (hah) to reduced cohomology. No?
Aug
18
comment Cohomology of wedge equals direct sum of cohomologies
Strictly speaking direct sums are only defined for modules. Note that the formula is false if $X$ and $Y$ are not (path-)connected.
Aug
18
comment Quotient of locally free sheaf is locally free?
It also works for noetherian affine schemes. At any rate, you can verify by hand (if you like) that $\mathbb{Z} / 2 \mathbb{Z}$ is not locally free.
Aug
16
comment Examples of advancement in mathematics due to war
Do you mean the Cold War?
Aug
16
comment Examples of advancement in mathematics due to war
@AsafKaragila You are thinking of Jean Leray.
Aug
16
comment Looking for a smooth curve that is not rational
The twisted cubic is in fact rational: your parametrisation has a rational inverse, defined away from the singularities $[0 : 0 : 0 : 1]$ and $[1 : 0 : 0 : 0]$ by $[x : y : z : w] \mapsto [y : z]$.
Aug
15
comment Scheme theoretic dual of $\mathbb P^n_k$
A point of the dual projective space is an equivalence class of linear forms on the original vector space, which defines a hyperplane in the usual way.