18
votes
Accepted
Classifying space for fibrations with Eilenberg-MacLane space fibers and nontrivial fundamental group actions
Mark Grant's excellent answer already resolves the question. However, let me sketch how this arises as a special case of the more general problem of classifying fibrations with a given fiber.
For any ...
- 4,544
17
votes
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Foliation of $\mathbb R^n$ by connected compact manifolds
There does not, even if you don’t require the fiber and base to be manifolds (or even connected, just that $F$ is not a single point). See
Borel, Armand; Serre, Jean-Pierre,
Impossibilité de fibrer ...
- 41.2k
16
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Can Homotopy Type Theory or algebraic geometry deal with homotopy fibers in terms of families?
I am answering your "later addon" only, although it seems actually to be a very different question than your original one.
This is perhaps one of the most misunderstood aspects of HoTT and ...
- 62.7k
15
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Classifying space for fibrations with Eilenberg-MacLane space fibers and nontrivial fundamental group actions
Denote $\pi=\pi_1(X)$ and fix the monodromy action $\rho:\pi\to \operatorname{Aut}(A)$. There is a generalized Eilenberg-Mac Lane space $L_\rho(A,n+1)$, whose only non-trivial homotopy groups are $\...
- 34k
14
votes
cup product and Steenrod operations in Serre spectral sequence
The behavior of the Steenrod squaring operations in the Serre spectral sequence was determined by Araki and independently by Vázquez (whose article I cannot locate online). However, it's a little work ...
- 49.6k
14
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Can Homotopy Type Theory or algebraic geometry deal with homotopy fibers in terms of families?
Here is an answer to your original question in the context of HoTT. An arbitrary map $f:X\to Y$ that isn't a fibration can't be viewed literally as a family of spaces varying continuously over $Y$, ...
- 62.7k
14
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Foliation of $\mathbb R^n$ by connected compact manifolds
On the other hand, if you only mean "foliation" as in your title, and not "fibration", then there is Vogt's foliation of R^3 by circles! (But it is not C^1, only differentiable).
Vogt, Elmar, "A ...
- 2,059
13
votes
Accepted
cup product and Steenrod operations in Serre spectral sequence
1) No in general. A counterexample is the projective space bundle associated to a vector bundle. For a rank $n$ vector bundle, the fiber, $\mathbb C \mathbb P^{n-1}$, has cohomology ring $\mathbb Z[x]/...
- 126k
11
votes
Accepted
Shafarevich conjecture for abelian varieties
Let $B$ be a smooth projective curve over an algebraically closed field of characteristic zero. Let $K$ be the function field of $B$. Let $S$ be a finite set of closed points of $B$.
You might find ...
- 9,127
11
votes
Accepted
Obstructions for the lifting problem after a pull-back
First, note that since you are assuming $F$ is $d-1$-connected, the primary obstruction lies in $H^{d+1}$, not $H^d$.
Now, consider the diagram $\require{AMScd}$
\begin{CD}
& & & & S^...
- 1,321
11
votes
Accepted
Smooth morphism (algebraic geometry) vs. Submersion (differential geo) & Ehresman's Lemma
One of the many equivalent definitions of smoothness of a morphism $f\colon X\to Y$ of varieties over a field $k$ is that $f$ is smooth if and only if it is formally smooth. The latter means the ...
- 1,680
10
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Where does the primary obstruction of a fibration show up in its spectral sequence?
In the general case of integral coefficients and possibly non-trivial local coefficient system, let $\pi=\pi_1(B)$.
A cocycle for the obstruction class is an element $o\in Hom_{\mathbb Z\pi}(C_{k+1}\...
- 451
10
votes
Accepted
A fibration equivalent to having a terminal object
A category $C$ has a terminal object if and only if the canonical functor $C \star \mathbf{1} \to \mathbf{1} \star \mathbf{1} = \mathbf{2}$ is a fibration.
- 4,955
9
votes
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Circle Action on Quaternionic Projective Space
$\mathbb{HP}^n\cong \mathrm{Sp}(n+1)/(\mathrm{Sp}(n)\times \mathrm{Sp}(1))$ is a symmetric space, so every one-parameter subgroup of $\mathrm{Sp}(n+1)$ acts on it. As was noted in the comments, such ...
- 4,544
9
votes
Accepted
Half-dimensional torus fibration vs Lagrangian torus fibration
This doesn't need to hold. For example, if one takes a $(T^4,\omega)$ with a constant symplectic structure $\omega$, in order for it to have a fibration by Lagrangian tori one should be able to find a ...
- 28.5k
9
votes
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Can we show that a functor is a fibration without choosing a cleavage?
Just as an example, given a category $\mathcal{C}$ with finite limits, showing $\mathrm{cod}\colon \mathcal{C}^\mathbf{2}\to \mathcal{C}$ is a fibration does not involve choosing a cleaving. All that ...
- 32.2k
8
votes
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Cup product of cohomology in a Serre spectral sequence
Maybe the simplest example is the following. There are two fiber bundles with base and fiber both $S^2$. One is the product $S^2\times S^2$ and the other consists of two copies of the mapping cylinder ...
- 18.9k
8
votes
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Serre and Hurewicz fibrations definition for pointed spaces?
This all works as long as the basepoint $x_0$ of $X$ is nondegenerate. The general context for this is due to Arne Strom who showed that the category of topological spaces, with the classic (i.e. ...
- 10.3k
8
votes
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Universal property of the codomain fibration
First a note about terminology: older literature sometimes uses the term "cofibration" for a functor $p:E\to B$ such that $p^{\rm op} : E^{\rm op} \to B^{\rm op}$ is a fibration, but it's ...
- 62.7k
8
votes
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Associativity of consecutive fibrations
You are right: they are not equivalent. For an example, choose a group $G$ that has a normal subgroup $H$ in which there is a subgroup $K$ that is normal in $H$ but not in $G$. Take $A$, $B$, and $C$ ...
- 50.7k
7
votes
Cup product of cohomology in a Serre spectral sequence
A good example showing the answer to question 1 is no is the case of the projectivization of a complex vector bundle $V$ over $B$. This has $F=\mathbb C P^{n-1}$, and it is a theorem (see e.g. Hatcher,...
- 556
7
votes
Multiplicativity of Euler characteristic for non-orientable fibrations
Since this question seems to be attracting some renewed interest, I may as well point out that a few years after I asked it, Kate Ponto and I proved a generalization of this formula by purely ...
- 62.7k
7
votes
Accepted
About fibrations with fibre Eilenberg-MacLane spaces
No. If this were the case then there would be a section $s: B \to E$ to $f$ induced by the $G$-equivariant map $\widetilde{s}:\widetilde{B} \to \widetilde{B} \times {\rm K}(M,n)$ sending $x$ to $(x,0)$...
- 9,026
7
votes
Accepted
Does the Eilenberg Moore Construction Preserve fibrations?
Let $C$ be a 2-category and $Mnd(C)$ the 2-category of monads in $C$. As explained by Street in the formal theory of monads, the Eilenberg-Moore construction is right 2-adjoint to the inclusion 2-...
- 55.4k
6
votes
Fibrations of projective varieties
Another proof: The assumptions imply that $X\times_YX$ is irreducible and that the two composite maps $X\times_YX\rightrightarrows X\xrightarrow{g}Z$ coincide over the generic point of $Y$. By density ...
- 19.1k
6
votes
Constructively, are all fibrations cloven?
If the axiom of choice holds then every fibration is cloven, and I would not be surprised if the converse holds. So this talk about not using the axiom of choice is best understood as closet-...
- 45.3k
6
votes
Accepted
Can Homotopy Type Theory or algebraic geometry deal with homotopy fibers in terms of families?
Yes, there is a certain sense in which your statements are true. As Mike Shulman and Qiaochu Yuan said, the strict fiber of a map cannot be defined in HoTT and doesn't make sense, but you can work ...
- 4,710
6
votes
Accepted
Space of sections of a fibration under weak homotopy equivalence
This is not true in general, unless you assume the base is sufficiently nice (eg a CW-complex). Here is a counter-example.
Let $B = \mathbb{Q}$, the rationals with its topology as a subspace of the ...
- 26.6k
6
votes
Trivialization of Pontryagin square on oriented $4$-manifolds
The original question was whether $\mathcal{P}(x)$ is non-zero when $x$ is. This answers that question, and was aimed at generic $r$, $p$ and $K$. Special cases may have special answers.
By Brown ...
- 3,446
6
votes
Accepted
elliptic fibration over $\mathbb{P}^1$ with exactly two fibres with monodromy of unipotency rank 1
If you look at the global monodromy action $\pi_1 ( \mathbb P^1 -\{0,1\}) \to SL_2(\mathbb Z)$, you see that $\pi_1 ( \mathbb P^1 -\{0,1\}) = \mathbb Z$ so the image is abelian and therefore has ...
- 126k
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