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$S^2$ bundles over $S^2$ (smooth, PL, topological) are in bijective correspondence with homotopy-classes of maps: $$ S^2 \to BSO_3 $$ which up to homotopy is $$\pi_2 BSO_3 \simeq \pi_1 SO_3 \simeq …

Another way to describe Anton Petrunin's example is to start with the trivial $S^1$-bundle over $\mathbb RP^2$ and to take the fibrewise connect sum with the Hopf fibration $S^3 \to S^2$. By fibrewis …

The answer to (1) is no. The easiest example I've been able to think of is the case of a bundle over $S^1$ with fibre a compact surface (genus $\geq 2$). For your monodromy, choose a non-trivial el …

A small note on extending the argument I gave in the previous (linked) thread. You get a free involution on $\mathbb CP^{2n+1}$ by using the fibrewise antipodal map for your bundle $$ S^2 \to \mathb …

I believe the answer is yes (although I haven't actually checked). Here's the idea: Given a Seifert-fibred space you can think of it as being fibred over a $2$-orbifold. You can de-singularise that …

It's not clear what you mean by "various refinements and generalizations". Cerf has a huge paper published by IHES "Topologie de certains espaces de plongements" which goes into many related details. …

A variation of your question has a positive answer. If you take any compact manifold that is a smooth bundle over another compact manifold $\pi : M \to N$, there is a smooth embedding $$f : M \to N \t …

Let $M$ be a smooth $m$-dimensional manifold, $TM$ its tangent bundle and $SM$ its unit sphere bundle. Are there some simple examples where $SM$ is fibrewise homotopy-equivalent to the trivial bundl …

Let's consider the case $B = \mathbb R^n$ and $E = B \times \mathbb R$, with bundle map $\pi(x,t) = x \in B$. This isn't much of a simplification as your fibre bundle is locally diffeomorphic to suc …