0
$\begingroup$

Suppose $F \to E \to M$ is a vector bundle whose base space $M$ and fiber space $F$ are both CW complexes. Let us call a CW structure on $E$ a “compatible with the bundle structure” if, for any cell $U \subset M$, the local trivialization $E \vert_U \to F \times U$ is a cellular isomorphism.

  1. Does $E$ always admit a CW structure compatible with the bundle one?

  2. Assuming $E$ has a CW structure compatible with the bundle one, is every section $M \to E$ homotopic to a cellular section, i.e., a section that is a cellular map?

$\endgroup$

1 Answer 1

1
$\begingroup$

Your definition of "compatible with the bundle structure" does not work: using that definition, $E$ never has a cellular structure compatible with the bundle structure, except in the case that the base space $M$ is zero dimensional.

To see why, consider a typical fiber inclusion $F \mapsto E$ where $F$ is the fiber over a typical point $x \in M$. Suppose that $x$ is not a vertex of the given CW structure on $M$; this is where we use that $M$ is not $0$-dimensional, in which case the vertices form a closed, proper subset of $M$, so it is certainly true that $M$ has a "typical point" that is not a vertex. It follows, from the definition of cellular map, that no point of $F$ is a vertex of $E$: the image of a vertex of $F$ under a cellular map to $E$ must be be a vertex of $E$, and the image of a vertex of $E$ under the cellular map $E \mapsto M$ must be a vertex of $M$, and so the image of the composition $F \mapsto E \mapsto M$ must be a vertex of $M$, but that image is $x$.

$\endgroup$
1
  • $\begingroup$ Oh, my bad, I see I wasn't sufficiently precise in my definition of “compatible with the bundle structure”. I meant the fiber over a vertex in the CW structure of M, of course. I updated the question. $\endgroup$
    – isekaijin
    Oct 25, 2022 at 14:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .