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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.
2
votes
LS category of the quotient of a manifold by its involution
The equivariant LS category $\operatorname{cat}_G(X)$ of a space $X$ with $G$-action is the minimal $k$ such that $X$ admits a cover by $G$-invariant open sets $U_0,\ldots , U_k$ such that each inclus …
3
votes
Loop-space functor on cohomology
For Q2, when searching it might be useful to know two things:
The map $\omega$ is often called the cohomology suspension (not to be confused with the suspension isomorphism in cohomology!), and
Eleme …
3
votes
A question on the manifold $ \{n\otimes n-m\otimes m:n,m\in S^2,(n,m)=0\} $
I'm not too comfortable with the description of $N$ in terms of matrix tensor products, but the manifold $M/(\mathbb{Z}_2\times\mathbb{Z}_2)$ you describe in your comment is the manifold of pairs of o …
5
votes
Upper bound on the Lusternick-Schnirelman category by the dimension of the space
If $X$ is path-connected, locally contractible and paracompact then $\operatorname{cat}(X)\leq\dim(X)$, the covering dimension. This is Theorem 1.7 in the book Lusternik-Schnirelmann Category by Corne …
1
vote
Accepted
fibre-preserving homotopy equivalence
The answer to the question as asked is no: a fibre-preserving map of fibrations in which the maps of total and base spaces are homotopy equivalences is neccessarily a fibre-preserving homotopy equival …
10
votes
Accepted
Why does $\iota_4^2 \in H^8(K(\mathbb Z/2,4);\mathbb Z/2)$ not come from $H^8(K(\mathbb Z/2,...
It is true, and follows from results of Browder on the mod 2 Bockstein spectral sequence for $K(\mathbb{Z}/2,4)$. (We can replace $4$ by any even integer $k$ and conclude that $\iota_k^2$ ia not the r …
4
votes
Accepted
Fundamental group of twisted loop space
Edit: The following is incorrect, see below. It might be more useful to think of $\Omega_f(M)$ as sitting in a homotopy pullback
$\require{AMScd}$
\begin{CD}
\Omega_f(M) @>>> M\\
@V V V @VVfV …
8
votes
Accepted
Can every element of a homotopy group of a smooth manifold be represented by an immersion?
There exists a simply-connected closed $6$-manifold $M$ with a homotopy class $\alpha\in \pi_4(M)$ which does not contain an immersion. The following argument is due to Diarmuid Crowley, after we real …
16
votes
Accepted
Homotopic classification of maps $M \to \mathbb{RP}^n$ where $M$ is a compact orientable $n$...
This seems to have been worked out in the 1960s by Paul Olum, see Section 1 of
Olum, P., Cocycle formulas for homotopy classification; maps into projective and lens spaces, Trans. Am. Math. Soc. 103 …
4
votes
Accepted
Homotopy type of G-CW-structure
There is a paper by Stefan Waner from 1980, I think it's called "Equivariant Classifying Spaces", in which he proves an equivariant version of Milnor's theorem. It might do what you want.
5
votes
Topology of functional spaces
Here is a nice survey article by Sam Smith (also available as https://arxiv.org/abs/1009.0804):
Smith, Samuel Bruce, The homotopy theory of function spaces: A survey, Félix, Yves (ed.) et al., Homot …
23
votes
What is classified by generalised Eilenberg MacLane spaces?
To answer your first question, take a look at the reference
Gitler, Samuel, Cohomology operations with local coefficients, Am. J. Math. 85, 156-188 (1963). ZBL0131.38006.
In particular, Theorem 7.18 …
31
votes
Accepted
An equivariant map from sphere to a Lie group of lower dimension which is not null homotopic?
The Blakers-Massey element in $\pi_6(S^3)\cong\mathbb{Z}_{12}$ can be represented by such a map. This is done explicitly on page 3 of the paper https://arxiv.org/abs/math/0501091, published as
Abres …
1
vote
A question about homotopy dimension
Spheres have this property: If $Y=S^n$ and $X\le Y$ then since the identity homomorphism on $\pi_i(X)$ factors through $\pi_i(S^n)$ for all $i$, we have that $\pi_i(X)=0$ for $i<n$ and $\pi_n(X)=0$ or …
17
votes
Homotopy pullback of a homotopy pushout is a homotopy pushout
This is Mather's second cube theorem, see Theorem 25 in
Mather, Michael, Pull-backs in homotopy theory, Can. J. Math. 28, 225-263 (1976). ZBL0351.55005.