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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.
73
votes
Do we still need model categories?
Here are some rough analogies:
Model Category :: $(\infty, 1)$-category
Basis :: Vector space
Local coordinates :: Manifold
I especially like the last one. When you do, say, differential geometry …
68
votes
Accepted
Is there an accepted definition of $(\infty,\infty)$ category?
One thing that might interest you is my result with Clark Barwick which gives an axiomatiation + uniqueness result for the homotopy theory of higher categories:
arXiv:1112.0040
(i.e. $(\infty,n)$-cat …
49
votes
What are surprising examples of Model Categories?
The category of sets admits precisely nine model category structures, no more no less.
I learned this fact from Tom Goodwillie's comments on a different MO question. It always shocks people when I m …
45
votes
Difference between represented and singular cohomology?
This is a good question because it really hits on a subtle issue. It turns out that Johannes and Ben are both correct and incorrect at the same time unless we settle some very subtle issues. Let me ex …
40
votes
Accepted
Classifiying sphere eversions
Answer Summary
The fundamental group of the space of immersions of $S^2$ into $\mathbb{R}^3$ is
$$ \pi_1 Im(S^2, \mathbb{R}^3) \cong \mathbb{Z}/2 \times \mathbb{Z}$$
This means that there are infini …
30
votes
Accepted
"Homotopy-first" courses in algebraic topology
I was a heavily involved TA for such a graduate course in 2006 at UC Berkeley.
We started with a little bit of point-set topology introducing the category of compactly generated spaces. Then we move …
21
votes
Accepted
Is super-vector spaces a "universal central extension" of vector spaces?
This will be a little imprecise, but hopefully if you need a precise result like this you can fill in the details.
First of all Vect has not only the symmetric monoidal structure but also the direct …
20
votes
Accepted
For what topological groups $G$ can we take $EG \rightarrow BG$ to be of the form $S^{\infty...
I like to think of $EG$ and $BG$ in terms of configuration spaces.
The space $BG$ can be identified with the following configuration space. It consists of configurations of finitely many points in the …
18
votes
Accepted
What is the free symmetric monoidal $\infty$-category on one object?
Yes, it is the same as $\mathbb{F}$.
As John Baez points out, it is the same as the free symmetric monoidal $\infty$-groupoid on one object. (This can also be seen by playing around with the adjoints …
16
votes
Accepted
Super-cobordisms
There are a number of technical issues with making what you describe precise, for example: what precisely is a supermanifold with boundary? how can you glue/compose bordisms? etc. I am going to ignore …
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 …
11
votes
Accepted
Diffeomorphisms and homotopy equivalences sliced over BO(n)
I wanted to say I think this is a great question, though phrasing things in terms of stacks might scare off some of the people who can best answer this question. I think in general understanding the …
11
votes
Accepted
I think I have a category enriched in $(\infty,n-1)$-categories. Is it an $(\infty,n)$-categ...
First, as Rune pointed out in the comments, his paper with David Gepner gives a very general approach to your wish list. However to make it so general that it applies to arbitrary monoidal $(\infty,1) …
10
votes
Accepted
Classifying Space of a Group Extension
Yes. The principal bundles are the same and your guess that $BA$ is an abelian group is exactly right. A good reference for this story, and of Segal's result that David Roberts quotes, is Segal's pape …
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 …