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Homotopy theory is an important sub-field of algebraic topology. It is mainly concerned with the properties and structures of spaces which are invariant under homotopy. Chief among these are the homotopy groups of spaces, specifically those of spheres. Homotopy theory includes a broad set of ideas and techniques, such as cohomology theories, spectra and stable homotopy theory, model categories, spectral sequences, and classifying spaces.
1
vote
Multiplicative Structures on Moore Spectra
This is a very classical question, and concepts like having an $A_n$ structure were conceived with work on this problem in mind. In recent years there has been new interest and spectacular new result …
2
votes
A stable splitting of linear surjections
Greg Arone has a very general stable splitting result using Weiss calculus in PAMS, vol 129 (2000), pp 1207-1211. Take a look and see if your example fits.
12
votes
Accepted
Product structure in Milnor exact sequence
Let $P$ be the wedge of all the $X_i$s. Up to homotopy equivalence, $X$ is the homotopy coequalizer of the identity and the shift maps from $P$ to itself. The Milnor exact sequence arises by analyzing …
4
votes
What is the homotopy type of the smash power of Moore spectra $(S/2)^{\otimes n}$?
Firstly, as shown in my answer to your `old question', arbitrary smash products of copies of $S/2$ will decompose into wedges of indecomposable spectra of an infinite number of homotopy types (even af …
9
votes
Accepted
Koszul duals of n-fold loop spaces
I am not sure if this gives what you want, but maybe it is:
I went in the other direction in a paper The McCord model for the tensor product of a space and a commutative ring spectrum, in Progress in …
5
votes
Associativity of consecutive fibrations
As Tom Goodwillie has pointed out, your second statement does not imply the first. The best one can generally do is find a (homotopy) fibration sequences of the form
$$ Y_3 \rightarrow \Omega C \right …
14
votes
Accepted
Applications of equivariant homotopy theory to representation theory
There are decades and decades of algebraic results that use techniques from equivariant homotopy theory. Some examples ...
(1) Quillen's work on ring theoretic aspects of the cohomology of finite gro …
8
votes
"Phantom" non-equivalences of spectra?
I am pretty sure that it will be rather delicate to find an example like you want. Here are some thoughts:
Let $X_n$ denote $\Omega^{\infty} \Sigma^n X$. One is assuming an equivalence $X_n \simeq Y_ …
4
votes
What is an unstable dual-Steenrod comodule?
Your question seems to be equivalent to asking for a description of the unstable condition in terms of the Milnor basis. This is easy to do: Given $r = (r_1,\dots, r_s)$, let $P(r)$ be dual to $\xi_1 …
3
votes
Injectivity of rationalization on spectrum morphisms
As user171227 is pointing out, you are asking about the map
$$ F^0(E) \rightarrow F\mathbb Q^0(E).$$
This always factors as
$$ F^0(E) \rightarrow F^0(E) \otimes \mathbb Q \rightarrow F\mathbb Q^0(E).$ …
11
votes
Why do Chern classes and Stiefel-Whitney classes satisfy the "same" Whitney sum formula?
There are many ways to prove these formulae uniformly, and, of course, you need to start from an appropriate definition of these classes. But as has already been suggested in the comments, pretty muc …
12
votes
Accepted
Pointed versus unpointed maps into a topological monoid
As is implicitly pointed out in the comments, you really want to assume that $X$ ($=M$) is path connected. And then your analysis is fine. Note that $M$ will then wish to be equivalent to $\Omega BM …
8
votes
Accepted
Is there a filtered splitting of product labelling spaces?
The answer to your first question is no. And this can be seen by homology considerations. Note that this equivalence induces an isomorphism of Hopf algebras
$$ H_*(C(\mathbb R; X \vee Y \vee (X\wed …
7
votes
What are cospectra, and why have they received so little attention?
Brayton Gray had a nice preprint about cospectra and unstable $v_n$-periodic homotopy. This was maybe 20 years ago, and I can't find any online version. To confirm my memory that this existed, by s …
17
votes
Accepted
Solving polynomial equations in spectra?
Here is a simple argument that would show many finite complexes can not be `integral' in your sense.
If $Sq^{2^k}$ acts nontrivially on $H^*(X;\mathbb Z/2)$ then $Sq^{2^{k+1}}$ will act nontrivially o …