Homotopy theory, homological algebra, algebraic treatments of manifolds.
Questions tagged [at.algebraic-topology]
8237 questions
151
votes
13 answers
Why is the fundamental group of a compact Riemann surface not free ?
Consider a compact Riemann surface $X$ of genus $g$.
It is well-known that its fundamental group $\pi_1(X)$ is the free group on the generators $a_1,b_1,...,a_g,b_g$ divided out by the normal subgroup generated by the single relator…
Georges Elencwajg
- 46,833
78
votes
7 answers
Cubical vs. simplicial singular homology
Singular homology is usually defined via singular simplices, but Serre in his thesis uses singular cubes, which he claims are better adapted to the study of fibre spaces. This young man (25 years old at the time) seemed to know what he was talking…
Georges Elencwajg
- 46,833
54
votes
6 answers
"Why the heck are the homotopy groups of the sphere so damn complicated?"
This is a quote from a dear friend asking the rest of us on Facebook. I gave him some half-baked response, but the truth is I don't really know enough about this to give him a good response.
So why ARE they so complicated? The topologists here want…
52
votes
10 answers
Intuition behind Thom class
The Thom class and Thom isomorphism theorem for oriented vector bundles are proven ( at least to my knowledge) by induction on the open covers and some manipulation with Mayer-Vietoris sequences.
What is the "actual reason" behind the existence of…
Axel
- 1,297
50
votes
10 answers
Definition of "simplicial complex"
When I think of a "simplicial complex", I think of the geometric realization of a simplicial set (a simplicial object in the category of sets). I'll refer to this as "the first definition".
However, there is another definition of "simplicial…
Kevin H. Lin
- 20,738
43
votes
8 answers
What part of the fundamental group is captured by the second homology group?
Let $X$ be a connected CW complex. One can ask to what extent $H_\ast(X)$ determines $\pi_1(X)$. For example, it determines its abelianization, because the Hurewicz Theorem implies that $H_1(X)$ is isomorphic to the abelianization of $\pi_1(X)$.
I'm…
Daniel Moskovich
- 21,887
40
votes
1 answer
Proof that a local fibration is a fibration, in May
I was reading "A Concise Course in Algebraic Topology" by J.P.May (page 52) and found the proof of the following theorem incomprehensible:
Let $p:E\rightarrow B$ be a map and let $\mathcal{O} $ be a numerable open cover of B. Then $p$ is a fibration…
Gheehyun Nahm
- 909
38
votes
3 answers
Why do the homology groups capture holes in a space better than the homotopy groups?
This is a follow-up to another question.
A good interpretation of having an $n$-dimensional hole in a space $X$ is that some image of the sphere $\mathbb{S}^n$ in this space given by a mapping $f: \mathbb{S}^n \rightarrow X$ cannot shrink down to…
Akela
- 3,579
- 3
- 32
- 41
35
votes
3 answers
Incorrect information in an old article about the Kervaire invariant
In the Soviet times there was a famous Encyclopedia of Mathematics. I think it is still familiar to every Russian mathematician maybe except very young ones, and yours truly is in possession of all 5 volumes. Browsing it recently (with…
Alex Gavrilov
- 6,851
33
votes
5 answers
Coefficients in cohomology
(Sorry if this is too elementary for this site)
I’m having some trouble understanding sheaf cohomology. It’s supposed to provide a theory of cohomology “with local coefficient”, and allow easy comparison between different theories like singular,…
user14800
- 473
33
votes
3 answers
No matter how many algebraic invariants we attach to topological spaces, there will always be nonhomeomorphic spaces agreeing on all their invariants
A while ago a professor of mine said something along the lines of
No matter how many algebraic invariants we attach to topological spaces, there will always be nonhomeomorphic spaces agreeing on all their invariants
Where algebraic invariants are…
user2520938
- 2,768
- 1
- 14
- 21
30
votes
3 answers
Examples for non-naturality of universal coefficients theorem
Does anyone have good examples of a space $X$ and a map $f: X \to X$ so that $f_*: H_*(X) \to H_*(X)$ is the identity but (e.g.) $f_*: H_*(X; \mathbb{F}_2) \to H_*(X; \mathbb{F}_2)$ is not the identity?
Edit: As mentioned in the comments, $f_*$ is…
Dylan Thurston
- 10,033
30
votes
1 answer
What is, really, the stable homotopy category?
When you try to understand the fuss behind the new good categories of spectra that arose on the 90's, you read things such as the following paragraph written by Peter May (from "The Hare and the Tortoise"):
All consumers are now in agreement:…
Bruno Stonek
- 2,914
- 1
- 24
- 40
29
votes
1 answer
Two questions on rational homotopy theory
I'm trying to read Quillen's paper "Rational homotopy theory" and am a little confused about the construction. As I understand, he associates a dg-Lie algebra over $\mathbb{Q}$ to every 1-reduced simplicial set via a somewhat long series of Quillen…
Akhil Mathew
- 25,291
29
votes
4 answers
Which stable homotopy groups are represented by parallelizable manifolds?
The Pontryagin-Thom construction allows one to identify the stable homotopy groups of spheres with bordism classes of stably normally framed manifolds. A stable framing of the stable normal bundle induces a stable framing of the stable tangent…
Chris Schommer-Pries
- 27,144