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Oct
8 |
awarded | Talkative |
Sep
24 |
awarded | Autobiographer |
Feb
21 |
asked | Failure of Noether normalization |
Oct
11 |
awarded | Yearling |
Jun
25 |
awarded | Custodian |
Jun
25 |
reviewed | Reviewed Homological Algebra texts |
Jun
10 |
comment |
Explanations for mathematicians, about the falsifiability (or not) of string theory
@JHI en.wikipedia.org/wiki/Falsifiability |
May
19 |
comment |
Composition in the category quotient
$g$ is a map $N'' \to P/P'$; taking a further quotient induces $N'' \to P/P''$; and then by definition, $N' \cap N''$ is a subobject of the kernel of this map, so it passes to $N''/(N'\cap N'')\to P/P''$. There's a good reason why arguments about abelian categories look like arguments about modules: the Freyd-Mitchell embedding theorem, which says that any small abelian category embeds into some category of modules over a ring. So any statement you can phrase in terms of an arbitrary small set of objects in an abelian category, and prove when those objects are modules, is true in general! |
May
18 |
answered | Composition in the category quotient |
May
16 |
comment |
Computation of Ext(Z^N,Z)
This should answer your question: mathoverflow.net/questions/40499/… |
May
6 |
answered | Definition of subobject classifier in presheaves |
May
3 |
comment |
Split and pure exact sequence of sheaves
Tensor products commute with direct sums, and if the sequence is split, $B \cong A \oplus C$. |
Apr
28 |
comment |
Is there a local projective model structure on simplicial sheaves? What are its fibrant objects?
This paper (math.uiuc.edu/K-theory/0462/combination2.pdf) by Benjamin Blander seems to imply that the answer to your first question is 'yes,' and you again get a Quillen equivalent model category, with sheafifying and forgetting being a Quillen equivalence. |
Mar
22 |
comment |
Intuition as to why the K-theory of a ring should be the homotopy theory of an H-space
That isn't really an answer to your questions, so for now let me just make an insinuation: what could you mean by a space 'naturally' having a multiplication on its homotopy groups, other than this being induced by a multiplication on the space itself? |
Mar
22 |
comment |
Intuition as to why the K-theory of a ring should be the homotopy theory of an H-space
In fact the $K$-groups of a ring are naturally modelled as the homotopy groups of a spectrum. (If you want to stick with spaces, you can in turn model this as an infinite loop space, i.e. a space which can be expressed as the form $\Omega^nX_n$ for some space $X_n$, for every $n$.) When you're starting with a ring, the spectrum you get is always a ring spectrum, which is the spectrum version of an $H$-space. I don't actually know why this is true, but I'd guess it pops out of one of the various constructions of this spectrum. |
Mar
4 |
awarded | Teacher |
Mar
4 |
answered | injectivity is a local property over noetherian rings |
Mar
4 |
comment |
$\pi$-cohomology class — a variant of cohomology class
In order for this to be well-defined, you'd need to say precisely what it means for two cycles to '"deform" into each other continuously.' If both cycles are subcomplexes of $X$, you could ask for them to be simplicially homotopic, but then you'd want your simplicial set to be Kan, i. e. probably the singular complex of the space $X$ rather than a triangulation. Then you'd need to make sure that your definition extends well to the abelian group of cycles. |
Feb
27 |
awarded | Critic |
Nov
19 |
awarded | Supporter |