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Results tagged with at.algebraic-topology
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user 8032
Homotopy theory, homological algebra, algebraic treatments of manifolds.
61
votes
2
answers
3k
views
Thomason's "open letter" to the mathematical community
In 1989, Bob Thomason left his CNRS position in Orsay and moved to Paris VII. It was during this period that he composed his "Open Letter" to the mathematical community. The letter explained Thomason' …
- 22.3k
3
votes
Accepted
Explaining some detail in Wall's paper of CW-complexes
As to (1):
If we choose a basepoint in $K$, then $\phi$ can be viewed as a map of based spaces. Let $F$ be the homotopy fiber of $\phi$. Then there is
a well-defined action $\Omega K \times F \to F$ w …
- 18.5k
4
votes
Naive Z/2-spectrum structure on E smash E?
Here's a very simple way to obtain $(E \wedge E)^{\Bbb Z_2}$ without having to resort to representations (at least if $E$ is connective). Consider the functor from spectra to spectra given by $$E \map …
- 18.5k
19
votes
Is any CW complex with only finitely many nonzero homology groups homotopic to a finite dime...
In the simply connected case, the answer is yes.
In the general case, the theory was worked out in complete detail by Wall in the paper:
Wall, C. T. C. Finiteness conditions for CW-complexes. Ann. of …
- 18.5k
6
votes
Accepted
Is $\Sigma^\infty_+ O(n)^\vee$, the Spanier-Whitehead dual of the orthogonal group, an $A_\i...
The question as stated probably requires clarification. If
$X$ is a space, then the S-dual $D_+(X)$ (i.e., functions from $X_+$ to the sphere) is always an $E_\infty$-ring spectrum. In particular, it …
- 18.5k
8
votes
1
answer
202
views
Finite domination and Poincaré duality spaces
Here are some definitions:
A space is homotopy finite if it is homotopy equivalent to a finite CW complex.
A space finitely dominated if it is a retract of a homotopy finite space.
A space $X$ is a Po …
- 18.5k
8
votes
1
answer
753
views
On the Euler characteristic of a Poincaré duality space
Background. Suppose that $M$ is an oriented, connected, closed manifold of dimension $d$ with fundamental class $\mu \in H_d(M;\Bbb Z)$. Let $\Delta : M \to M \times M$ be the diagonal map. Then the p …
- 10.6k
13
votes
2
answers
489
views
"Burnside ring" of the natural numbers and algebraic K-theory
The construction of the Burnside ring $A(G)$ of a group $G$ (usually, but not always, finite) is given by taking the Grothendieck group of the commutative semi-ring of isomorphism classes of finite $G …
- 10.6k
8
votes
1
answer
310
views
Finite domination and compact ENRs
Edit: In the comments, Tyrone points out that West's positive answer to Borsuk's conjecture implies that every compact ENR is homotopy equivalent to a finite CW complex. It follows that the only finit …
- 4,699
16
votes
4
answers
1k
views
Multiplicativity of the Euler characteristic for fibrations
For a Serre fibration
$$
F\to E \to B ,
$$
with $F,E,B$ having the homotopy type of finite complexes, it is known that the Euler characteristic is multiplicative:
$$
\chi(E) = \chi(F)\chi(B) .
$$
Howe …
- 12.6k
2
votes
Multiplicativity of the Euler characteristic for fibrations
Here is an argument that the
Euler characteristic is multiplicative for fibrations
$$
F\to E \to B
$$
where $F$ and $B$ are finitely dominated and $B$ is connected.
Without loss in generality, we may …
- 18.5k
7
votes
Multiplicativity of the Euler characteristic for fibrations
Note Added March 1, 2022:
I now think there is a gap in deducing multiplicativity of the Euler characteristic from the Pedersen-Taylor result on the finiteness obstruction. I think the argument I giv …
- 18.5k
8
votes
How can I construct a closed manifold from a finite CW complex?
More generally, suppose $n \le m$ are non-negative integers,
$X$ is a CW complex of dimension $\le n$, $M$ is a non-empty, closed $m$-manifold,
and $X$ and $M$ have the same homotopy type.
It is well …
- 18.5k
2
votes
Accepted
Retractions, homology and multiplication
The role of $C$ in the problem is irrelevant.
Let $r : C \to B$ be a retraction, and set
$$m := r\circ \mu : B \times B \to B .
$$ Set $\ast := b$ and think of it as the basepoint of $B$. Then the r …
- 18.5k
9
votes
Accepted
Atiyah duality without reference to an embedding
Here is another short construction which is much simpler and just takes a few lines.
Let $M$ be a closed $n$-manifold. Consider the diagonal $M \to M \times M$. It is an embedding. Take its Pontryagi …
- 18.5k
7
votes
Atiyah duality without reference to an embedding
Assume $M$ is a closed, smooth manifold of dimension $n$. Let $\tau^+$ be the fiberwise one point compactification of its tangent bundle. This is a fiberwise $n$-spherical fibration equipped with a p …
- 18.5k
7
votes
1
answer
564
views
Filtered homotopy colimits and singular homology
Suppose I have a functor
$$
X_\bullet: I \to \text{Spaces}
$$
where $I$ is a small filtered category.
It seems to be a "folk theorem" that the homomorphism
$$
\underset{\alpha\in I}{\text{colim }} H_ …
- 18.5k
6
votes
How can you tell if a space is homotopy equivalent to a manifold?
Sean: this gives a Poincare space which is not homotopy equivalent to a closed manifold. the idea is that the Spivak fibration of the $5$ dimensional Poincare space doesn't lift to a stable vector bun …
- 18.5k
10
votes
Accepted
Whitehead product and a homotopy group of a wedge sum
Here are some details which are related to Tyler's comment.
I recommend looking at the paper "Induced Fibrations and Cofibrations" by Tudor Ganea (1967). For connected based spaces $X$ and $Y$, there …
- 18.5k
7
votes
0
answers
132
views
Weak homotopy type of the Cech Nerve
Let ${\cal U} = \{U_i\}_{i\in J}$ be an open cover of a topological space $X$, where the indexing set $J$ is assumed to be well-ordered. Then the Cech nerve is the "simplicial space without degeneraci …
- 18.5k
1
vote
$G$-equivariant intersection theory using differential topology?
You may want to take a look at
Klein, J.R., Williams, B.
Homotopical intersection theory, II: equivariance.
Math. Z. 264(2010),849–880.
An arXiv version appears here:
https://arxiv.org/abs/0803.0017
I …
- 18.5k
2
votes
K-theory of free $G$-sets and the classifying space, and generalization
I believe you meant to write $Q(BG_+)$ in the first paragraph of your post, where $Q = \Omega^\infty\Sigma^\infty$. The this result is really a folk theorem and is sometimes called the "Barratt–Priddy …
- 10.6k
9
votes
Compelling evidence that two basepoints are better than one
I just wanted to add something to the discussion about the utility of adding additional basepoints. It turns out this is crucial for understanding
certain aspects of embedding theory. See the bottom o …
- 244
8
votes
What characteristic class information comes from the 2-torsion of $H^*(BSO(n);Z)$?
See
Brown, Edward, The cohomology of $B\text{SO}_n$ and $B\text{O}_n$ with integer coefficients. Proc. Amer. Math. Soc. 85 (1982), no. 2, 283–288
for an answer to your questions. (As for applications: …
- 18.5k
14
votes
How to prove the connected sum of two closed aspherical n-manfolds (n >2) is not asperical?
Here's a different way to see it. Let $M$ and $N$ be aspherical of dimension at least 3.
Then the wedge $M \vee N$ is aspherical (but not a manifold). Let $M\sharp N$ be the connected sum. Then we get …
- 18.5k
12
votes
1
answer
335
views
Rational homotopy invariance of algebraic $K$-theory
Suppose that $R\to S$ is a 1-connected morphism of connective structured ring spectra that induces an isomorphism on rational homotopy groups. Is the induced map of (Waldhausen) K-theory spectra
$$
K( …
- 11.9k
7
votes
Why does this construction give a (homotopy-invariant) suspension (resp. homotopy cofiber) i...
An argument showing that the two models of suspension are equivalent will probably be based on something like the following:
Assertion: Suppose we are given a commutative diagram of the form
$\requir …
- 18.5k
3
votes
Accepted
A generalization of integral Poincaré duality
Prior to Dwyer-Greenlees-Iyengar, Dwyer and myself (independently) defined Gorenstein conditions for group rings over the sphere $S[G]$, i.e., the suspension spectrum of a topological group.
The def …
- 18.5k
3
votes
Is it true, the space of embeddings segments is homotopy equivalent to the subspace of all l...
Here are some remarks which may be relevant.
First of all it seems to me that the correct topology to use is the Whitney $C^\infty$-topology on the embedding space.
Let $M$ be an closed manifold. T …
- 18.5k
10
votes
When is a homotopy pushout contractible?
Let me add some additional remarks on the enumeration question:
How many such spaces $A$ are there sitting over $B\times C$ such that the homotopy pushout
$$
B \leftarrow A \to C
$$
is contractible …
- 18.5k
5
votes
Accepted
pullback and fiber sequence
Yes. Here are some details.
The space $P$ sits in homotopy pullback diagram
$\require{AMScd}$
$$
\begin{CD}
P @>>> D \\
@VVV@VVV \\
A\times C @>>> D\times D
\end{CD}
$$
where the the right vertica …
- 18.5k
3
votes
Accepted
CW complexes obtained by attaching cells not with increasing dimension
There is a name for the kind of space you are describing: a cell complex.
A CW complex is a cell complex which has cell attachments in the increasing order of dimension.
The main advantage of having …
- 18.5k
4
votes
moving from sphere spectrum to finite spectrum
For convenience I will set $Z = Z^k$, where $Z^k$ is as in your notation.
Case 1: $Y$ is the suspension spectrum of a based finite complex $U$ having dimension $n$. Then
$\pi^Y_\ast(Z)$ is given by …
- 18.5k
10
votes
Accepted
Contraction of a family of loops simultaneously
Surely not.
Let $S^1 \to LS^2$ be adjoint to the map
$c: S^1 \times S^1 \to S^2$ which collapses $S^1\vee S^1$ to a point.
The latter has degree one.
Let $p\in S^1$ be any point but the basepoint.
…
- 18.5k
10
votes
Why is $\mathbb{S}^1$ a cogroup object in $\mathbf{Top.}$?
The circle is not a cogroup object in the category of based spaces. If $X$ is a cogroup in based spaces, then the composition
$$
X \to X \vee X \to X\times X
$$
would necessarily coincide with the di …
- 18.5k
14
votes
Unstable manifolds of a Morse function give a CW complex
(1). Some experts tell me that Laudenbach's paper is incomplete and contains gaps.
I will retract this for now. I do recall being told this, but I am not
aware at this point in time where the gaps …
- 18.5k
5
votes
Does anyone know a basepoint-free construction of universal covers?
Here is another attempt at pinning down the meaning of "canonical" in reference to Tom's answer.
Let $X$ be a nice space (connected, locally path-connected and semi-locally simply connected).
Let …
- 18.5k
6
votes
0
answers
123
views
On the weak homotopy type of a differentiable (Chen) space
Suppose that $M$ is a differentiable space in the sense of Chen (cf. https://ncatlab.org/nlab/show/Chen+space ).
Assume that $M$ also has the structure of a topological space and that the two struct …
- 18.5k
5
votes
homotopy pushout of spaces homotopic to finite CW complexes
In Corollary 5.12 of Whitehead's book it is shown that that the cobase change of a homotopy equivalence along a cofibration (NDR pair) is again a homotopy equivalence. This ought to imply homotopy in …
- 33.2k
3
votes
Accepted
Wall self-intersection invariant for odd-dimensional manifolds?
I doubt that what you are proposing as the receptacle for the obstruction is the correct abelian group. For one thing, you are not taking into account the involution on the canonical double cover of t …
- 18.5k
7
votes
Accepted
Turning injection of homotopy groups to an isomorphism
Your question is equivalent to the following:
Given a cellular inclusion $i : X\to Y$, when is there a retraction $r:Y \to X$?
(Being a retraction means that $r\circ i: X\to X$ is the identity.)
T …
- 18.5k
7
votes
Homology of the universal cover
If we replace the field $k$ with the ring of integers $\Bbb Z$, then no.
There are non-trivial high dimensional knots $K: S^n \to S^{n+2}$, whose complements $X = S^{n+2}-K(S^n)$ have $\pi_1(X) \con …
- 18.5k
7
votes
Accepted
What is the homotopy fiber of $X \to X_{hG}$, where this is a pointed homotopy orbit?
Here is a special case which gives a partial answer:
(i). Suppose $G$ acts in a homotopically trivial way on $X$. This means that there is a trivial $G$-space $Y$ and a pair of $G$-equivariant maps …
- 18.5k
18
votes
Accepted
Realizing cohomology classes by submanifolds
Your question is just a reformulation of what Thom did, so the answer is always yes.
Since the Stokes map from de~Rham cohomology to singular cohomology (with real coefficients) is an isomorphism, yo …
- 18.5k
6
votes
Notion of linking between two general $p$ and $q$ manifolds embedded in a higher dimensional...
There are various approaches to this. One approach, developed by Bruce Williams and me, uses homotopy theory. See:
Homotopical intersection theory, I.
Geometry & Topology 11, (2007) 939–977
arXiv: m …
- 18.5k
2
votes
Accepted
Relation between transport functor of a fibration and a Hurewicz connection on it
Let $p: E\to B$ be a map. Define $\Lambda(p) = E \times_B B^I$; this is the space of pairs $(x,\gamma)$ consisting of a point $x\in E$ and a path $\gamma: [0,1] \to B$ such that $\gamma(0) = p(x)$. Th …
- 18.5k
6
votes
Accepted
$X \rtimes Y \simeq X \vee (X \wedge Y)$ for $X$ a co-H-Space
The proof is not hard, but more tedious than I would have thought.
There is a canonical identification
$$
X\rtimes Y = X\wedge(Y_+)
$$
where $Y_+$ is the effect of adding a disjoint base point to $ …
- 18.5k
6
votes
1
answer
256
views
An operad-like structure, is there a name for it?
Here is an example which I'd like to have a name for.
Let $P$ be a compact smooth manifold of dimension $p$, possibly with non-empty boundary.
Define $E(k,P)$ to be the space of smooth (codimension …
- 18.5k
3
votes
Homotopy type of smooth manifolds with boundary
I know of two proofs in the compact case. Let $M$ be a compact smooth $m$-manifold with boundary $\partial M$.
1) Morse theory (Sketch). For this I think we need $m \ge 4$.
There is a Morse functio …
- 18.5k
6
votes
Can one disjoin any submanifold in $\mathbb R^n$ from itself by a $C^{\infty}$-small isotopy?
There are lots of counter-examples.
Here's one:
The fibration $\text{SO}(3) \to \text{SO}(4) \to S^3$ splits, since it is the principal bundle of the tangent bundle of $S^3$, and the latter is paral …
- 18.5k