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Homotopy theory, homological algebra, algebraic treatments of manifolds.
4
votes
Accepted
Weak homotopy equivalence and Cech cohomology
$$T = \left\{ \left( x, \sin \frac{1}{x} \right ) : x \in (0,1] \right\} \cup \{(0,y)\mid y\in[-1,1]\}$$
This has trivial homotopy groups in degrees $\ge1$ but according to Wikipedia nontrivial Če …
1
vote
Accepted
Action of $\mathbb{Z}/3\mathbb{Z}$ on $P^{1}$
Not sure whether this is what you are after, but one way to give a geometric meaning is the following.
You can think of $P^1{\mathbb C}$ as the "boundary at infinity" of the 3-dimensional hyperbolic …
15
votes
Accepted
Homology sphere with $\mathbb{R}^3$ as the universal cover
In a sense, most $3$-manifolds have universal cover $R^3$. In particular, this is the case for hyperbolic $3$-manifolds. And there do exist integer homology spheres which are hyperbolic. Two explicit …
4
votes
Accepted
Bockstein homomorphism from $H^d(BG,Z_2)$ to $H^{d+1}(BG,Z)$, and Steenrod Square $Sq^1$
The natural maps from $$0\to Z\to Z\to Z_2\to 0$$ to $$0\to Z_2\to Z_4\to Z_2\to 0$$ commute with the homomorphisms in the s.e.s. and hence also with the connecting homomorphisms of the long exact seq …
10
votes
Accepted
Berger sphere theorem
By Hurewicz, (n-1)-connected implies vanishing of the first n-1 homology groups. Since the manifold is closed and (by simple connectedness) also orientable, we have $H_n={\mathbb Z}$. Of course the hi …
3
votes
simple explanation of simplicial volume=4g-4 when genus $\ge 1$
Well, the simplicial volume is not really the minimal number of simplices in a homotopy triangulation but the minimal (more precisely: the infimum) sum of (modulus of) coefficients in a fundamental cl …
-2
votes
Concrete examples of covering from the 3-torus to the 3-sphere
First I think, the covering $T^2\to S^2$ has deck group $Z/2Z\oplus Z/2Z$ generated by $(x,y)\to (-x,y)$ and $(x,y)\to (x,-y)$.
Then for the 3-dimensional case, you consider the action of $$Z/2Z\oplu …
8
votes
CW 4 manifolds with single 4 cell
For the smooth case:
Via Morse theory the claim is equivalent to having a Morse function with only one maximum, or only one minimum, or to have a handle decomposition with only one 0-handle.
Assume …
1
vote
Universal bundles: construction of the map associated to a group homomorphism
The Milnor construction realised EG as the infinite join of copies of G, see the references in http://ncatlab.org/nlab/show/Milnor+construction
It is then obvious how to construct a $\rho$-equivarian …
4
votes
What 3-manifolds can be obtained by gluing $ S^1 \times P $ and two copies of $S^1 \times D^2$
$P\times S^1$ has an obvious fibration by circles and as long as your Dehn filling does not send a meridian to the fiber of that fibration, the Dehn filled manifold will again be a Seifert fibration. …
5
votes
Four manifold without point homotopy equivalent to wedge of two-spheres?
A closed, simply connected manifold has a Morse function with one critical point of index 0 and no critical point of index 1. Thus it is homotopy equivalent to a CW-complex with one 0-cell, no 1-cell, …
6
votes
Accepted
Four manifold without point homotopy equivalent to wedge of two-spheres?
Here is a hopefully better answer which is copied from page 104 of Milnor-Husemoller's book on symmetric bilinear forms.
$X^\prime$ is also simply connected (Seifert-van Kampen reversed) and thus has …
1
vote
What are Fake Weighted Projective Spaces?
Everything you would like to know about Fake lens spaces is on http://www.map.mpim-bonn.mpg.de/Fake_lens_spaces by Tibor Macko.
A Fake lens space is the Quotient of a sphere by a cyclic group acting …
0
votes
what is the meaning of a curve $C$ representing Identity in fundamental group?
Dehn's Lemma (= Papakyriakopoulos' Theorem) asserts: if C represents 0 in $\pi_1M$ and if C is a simple closed curve, then C bounds an embedded disk.
The assumption on C being a simple closed curve i …
7
votes
Accepted
Does the Hopf construction work for $S^0$?
This is completely elementary. From the H-space structure $X\times X\to X$ you construct the Hopf fibration $X*X\to SX$ via $(x,t,y)\to (xy,t)$, where you think of $SX$ as the quotient of $X\times I$ …