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Results tagged with homotopy-theory
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user 6666
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.
36
votes
What determines a model structure?
Among the nine model structures on the category of sets, there are:
Two in which cofibrant=empty and every set is fibrant.
Two in which cofibrant=empty and fibrant={empty or singleton}.
One in whic …
- 55.9k
32
votes
Accepted
Integral cohomology (stable) operations
$HZ^nHZ$ is trivial for $n<0$. $HZ^0HZ$ is infinite cyclic generated by the identity operation. For $n>0$ the group is finite. So you know everything if you know what's going on locally at each prime. …
- 55.9k
31
votes
Why is BG infinite dimensional for G finite ?
For every subgroup $H\subset G$, $BH$ occurs as a covering space of $BG$. If $BG$ were finite-dimensional then every covering space would be finite-dimensional. But for $C_p$ cyclic of prime order $p$ …
- 55.9k
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 …
- 55.9k
23
votes
Accepted
Is there a good way to understand the free loop space of a sphere?
Stably the free loop space of the suspension of a connected space splits up, just as the based loop space does. Just as $\Omega\Sigma X$ is stably the wedge of the smash products $X^{\wedge n}$, $L\Si …
- 55.9k
21
votes
Total spaces of $TS^2$ and $S^2 \times R^2$ not homeomorphic
This may be overkill, but to elaborate on Ryan's answer in another way:
Without mentioning either boundaries or any other compactifications, we can define the intersection number of $x\in H_p$ and $ …
- 55.9k
21
votes
Accepted
Are Eilenberg-MacLane spaces limits of manifolds?
Any homotopy type that is represented by a countable CW complex is also represented by an increasing union of closed manifolds:
First consider a finite CW complex $X$. Let $M\sim X$ be a compact manif …
- 55.9k
19
votes
Plus construction considerations.
I don't really know if this helps, but you can in effect give the plus-construction definition of $K$-groups without explicitly mentioning homotopy groups, and without ever doing the plus construction …
- 55.9k
19
votes
Accepted
When does rationalization commute with homotopy fixed points?
The statement appears to me to be false. The difficulty, from some point of view, is that in the spectral sequence going from $H^{-i}(G;\pi_j(M))$ to $\pi_{i+j}(M^{hG})$ an infinite number of torsion …
- 55.9k
18
votes
Basic questions on the homotopy category
The inclusion $\mathbb RP^2\to \mathbb CP^2$ has no kernel in the pointed homotopy category.
Proof:
Suppose $X\to \mathbb RP^2$ is such a kernel. Then maps $S^1\to X$ (in that category) correspond …
- 55.9k
18
votes
A homotopy commutative diagram that cannot be strictified
One thing to be aware of is that, even when liftings to strict diagrams exist, their non-uniqueness is a serious matter. For example, consider the square pushout diagram in which $S^n$ maps to a point …
- 55.9k
17
votes
Accepted
Homotopy of space of immersions, Smale-Hirsch theorem
No. For example, if $M$ is a Moebius band then, at least for even $k$, $Imm(M,\mathbb R^{2+k})$ is not homotopy equivalent to $Imm(S^1\times \mathbb R,\mathbb R^{2+k})$.
The latter is equivalent to …
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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 …
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14
votes
Non-examples of model structures, that fail for subtle/surprising reasons?
The answer to Peter's answer is no, as expected. Here's why: The intersection of {homotopy equivalences} and {Serre fibrations} is not closed under pullback (base change).
It's easy to make a contin …
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14
votes
Poincare duality spaces vs. manifolds via lifting maps, the obstruction theory and the role ...
Let's talk first about the smooth and simply connected case. As you say, Poincare duality for $X$ yields a spherical fibration, or map $X\to BG$. A lifting $X\to BO$ is necessary but not sufficient fo …
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