CW structures compatible with fiber bundle structures

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?

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 $$x$$.
• 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. Oct 25, 2022 at 14:11