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Homotopy theory, homological algebra, algebraic treatments of manifolds.
4
votes
Accepted
Is a left Bousfield localization of simplicial presheaves a locally cartesian closed model c...
Section 2 of this paper of Rezk addresses exactly the question of when the localization by S yields a Cartesian model category. For that the relevant property is that that if you take the product of a …
8
votes
Accepted
A step in Lurie's treatment of $L$-theory
Here is a way to see it which might constitute "just unfolding the definitions" (at least if one were sitting in on the class at Harvard when it was being taught).
Set $Z(T) = Y(T^c)$, (compliment tak …
9
votes
1
answer
316
views
Framed version of the "copants bordism"?
The "pants" bordism in dimension n is a bordism which goes from $S^n \sqcup S^n$ to $S^n$ witnessing the connected sum operation - equivalently by attaching 1-handle to the trivial bordism, equivalent …
10
votes
1
answer
280
views
Induced map on $H_4$ of Eilenberg–MacLane spaces
$\DeclareMathOperator\Hom{Hom}$It is well-known (see Breen, Mikhailov, Touzé - Derived functors of the divided power functors for example) that for $A$ a free abelian group we have
$$ H_i(K(A,1); \mat …
14
votes
1
answer
901
views
What is known about exotic spheres up to stable diffeomorphism?
In even dimensions $n=2k$ we can define two smooth manifolds $M$ and $N$ to be stably diffeomorphic if they become diffeomorphic after the connect sum with $r$ many copies of $S^k \times S^k$ for some …
10
votes
2
answers
647
views
Homotopy equivalent smooth 4-manifolds which are not stably diffeomorphic?
Recall that two 4-manifolds $M$ and $N$ are stably diffeomorphic if there exist $m,n$ such that
$$M \#_n (S^2 \times S^2) \cong N \#_n (S^2 \times S^2).$$
That is, they become diffeomorphic after taki …
17
votes
Accepted
How aggressive is the fibrant replacement of $\mathrm{Bord}_n$?
The completeness condition is not really about making things invertible which weren't already. It is about where the information about invertible morphisms is stored.
We can already see this with $(\i …
29
votes
4
answers
1k
views
Which stable homotopy groups are represented by parallelizable manifolds?
The Pontryagin-Thom construction allows one to identify the stable homotopy groups of spheres with bordism classes of stably normally framed manifolds. A stable framing of the stable normal bundle ind …
7
votes
1
answer
202
views
Groupoid completion of a topological category vs its homotopy category?
Given a category $\mathcal{C}$ enriched in spaces, we can take the nerve (a simplicial space) and then geometric realization to get a space $B\mathcal{C}$. If we view spaces as $\infty$-groupoids, the …
1
vote
Accepted
Locally trivial fibration over a suspension
It is independent of the choice of base point.
Let $Map((X,x_0), (G_F,id))$ be the based mapping space (based at $x_0$).
Let $Map(X, G_F)$ be the free mapping space. Then we have a split short exact …
6
votes
Accepted
Space of sections of a fibration under weak homotopy equivalence
This is not true in general, unless you assume the base is sufficiently nice (eg a CW-complex). Here is a counter-example.
Let $B = \mathbb{Q}$, the rationals with its topology as a subspace of the …
18
votes
Accepted
A search for a sequence of $6$-manifolds
I looked at Wall's paper Classification problems in differential topology. V
On certain 6-manifolds. In theorem 3 of that paper Wall describes some invariants of 6-mainfolds, and the relation between …
2
votes
Accepted
Relation between Morse Theory and integration against Euler Characteristic
I glanced at the paper briefly. Let me try to explain what I understand. Let us suppose that M is compact and without boundary. Let $f: M \to \mathbb{R}$ be a Morse function. Let us further suppose fo …
12
votes
Is any CW complex with only finitely many nonzero homology groups homotopic to a finite dime...
As requested I am writing this as an answer.
No there are spaces with vanished homology which are not homotopy equivalent to finite CW-complexes.
For example if $G$ is an acyclic group, then the cl …
8
votes
Accepted
Understanding model independently the equivalence of two ways of obtaining homotopy types fr...
Here is an argument, which is basically Denis Nardin's comment.
To have a model independent proof you need model independent definitions of the hocolim and of the localization. You can define them …