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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.
5
votes
Accepted
Equivalence of cosimplicial models for homotopy pullbacks
The first one has codegeneracies as well as cofaces, yes? And when you say that it computes the htpy pullback basically by definition I suppose you mean that its Tot is homeomorphic to the space whose …
26
votes
What is the algebraic geometry version of the spheres?
To expand on Will Sawin's comment in a vague sort of way (which is the best I can do):
The cofiber of the pair $(\mathbb P^n,\mathbb P^{n-1})$ (which doesn't exist as a scheme but does exist if you w …
8
votes
Accepted
Bousfield-Kan spectral sequence with local coefficients
Let LOC be the category in which an object is a space plus a local system on it, and a morphism is a map of spaces covered by a map of coefficient systems in the obvious sense. There's an obvious func …
6
votes
the hopf invariant of the hopf construction
This can be seen as follows in terms of the older geometric definitions of degree and Hopf invariant.
The degree of a smooth map $S^{n-1}\to S^{n-1}$ is the number of points mapping to a given point. …
8
votes
Accepted
The homotopy cofiber of the smash product of two maps of spectra
By factoring $g\wedge f$ as $X\wedge f$ followed by $g\wedge Y$ you see that it's in the middle of a cofiber sequence $Z\wedge C_f\to \ ?\to C_g\wedge Y$. Similarly it's in the middle of a cofiber seq …
4
votes
What is the "universal problem" that motivates the definition of homotopy limits/colimits (a...
This could have been a comment to Peter's answer, except that it's going to be long. (Maybe I should be making it a new question.)
Suppose that $Ho(\mathcal C)$ is obtained from $\mathcal C$ by (univ …
11
votes
Does Thom's J-equivalence imply Whitehead's simple homotopy?
This "J-equivalence" is usually called h-cobordism. The results on it are not scattered! They are quite complete, except in low dimensions.
Given any $M_1$, and given an element $\tau$ of the Whiteh …
16
votes
Computing homotopies
Harry, the expression "an explicit representative of the natural homotopy between the identity map and the constant map on a contractible based space" doesn't mean anything to me. Homotopies don't hav …
7
votes
Accepted
Eilenberg-Mac Lane spaces for groups that can't see $p$-groups
Yes. Your hypothesis on $G$ means that $G$ is torsion and prime to $p$: every element of $G$ has finite order prime to $p$. This in turn implies that the integral homology group $H_mK(G,n)$ is torsion …
7
votes
Singular complex = cohomology ring + Steenrod operations?
This is answering a slightly different question, but here goes:
If the question is "does knowing the singular (co)chains up to quasi-isomorphism determine the space up to weak equivalence?" then of c …
6
votes
Accepted
Lifting relative homotopies between maps $X\to Y/B$ when $B$ is contractible
Sure. The fact that $A$ is a subcomplex of $X$ implies that the restriction maps $$Map(X,Y)\to Map(A,Y)$$ and $$Map(X,Y/B)\to Map(A,Y/B)$$ are Serre fibrations. The fact that $B$ is a subcomplex of $Y …
13
votes
Classifying space for S1-bundle?
$Homeo(S^1)$ is homotopy equivalent to $O(2)$. This follows from the fact that the space of all homeomorphisms $S^1\to S^1$ having degree one and fixing a given point is contractible. In fact, the lat …
4
votes
Accepted
Suspension of an excisive pair
No. Suppose that $X$ is an $n$-sphere, $A$ is a closed hemisphere, and $U$ is a point in the interior of $A$. Then $SX$ is an $(n+1)$-sphere, $SA$ is a closed hemisphere, and $SU$ is a closed arc in $ …
5
votes
mapping spaces of diagrams
(I deleted my first attempt at an answer, as I had right and left reversed and anyway I wanted to try to say it better.)
I would advocate the following broad and comparatively low-tech view of the ma …
4
votes
The most general context of Mather's Cube Theorems
Here's a sketch proof of 2, sort of in the same spirit as Jeff Strom's answer:
These statements have equivalent formulations involving strictly commutative squares.
Denote a typical square by $\mat …