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 ...

17
votes

Accepted

### 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 ...

16
votes

### 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 ...

15
votes

### 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 $\...

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 ...

14
votes

### 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 $Y$, ...

14
votes

### 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 ...

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]/...

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 ...

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^...

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 ...

10
votes

### 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}\...

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.

9
votes

Accepted

### 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 ...

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 ...

9
votes

Accepted

### 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 ...

8
votes

Accepted

### 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 ...

8
votes

Accepted

### 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. ...

8
votes

Accepted

### 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 ...

8
votes

Accepted

### 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 $C$ ...

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,...

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 ...

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 $(x,0)$...

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-...

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 ...

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-...

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 ...

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 ...

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 ...

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 ...

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