Does the total space of a fibre bundle have the homotopy type of a CW complex if the base and the fibers have?

Question

Let $$ F\to E\xrightarrow{\pi} B $$ be a fibre bundle over a connected and compact base $B$.

Is it true that the total space $E$ has the homotopy type of a CW complex, if the fibre $F$ and the base $B$ have the homotopy type of a CW complex?

I don't think that this is true for Serre fibrations $\pi$, but I don't know a proof either.

Answer 1

Here is a sketch of an argument. I've never been much good with CW complexes, so treat it with caution!

---

First, we may in fact assume that $B$ is a CW complex. (Pull-back by a homotopy equivalence on the base.) Now our fibre bundle will be trivial when pulled back to each cell of CW (since a cell is contracible), and so I think this should let us write the total space as being glued out of pieces of the form

$$\text{cell} \times F.$$

Now replacing $F$ by a CW complex, and using that the product of CW complexes is a CW complex, we should be in good shape.

Answer 2

I don't know the answer in general, but in the case that we (a) restrict to compactly-generated weak Hausdorff spaces (not a big deal), and (b) restrict to the case where F is compact, I think I have an answer.

Namely, by a result of May there is a space, $BG$, with the homotopy type of a CW-complex (since $F$ has this property), such that fibrations over $B$ with fiber having the homotopy type of $F$ are classified up to equivalence by homotopy-classes of maps $B \rightarrow BG$.

More specifically, there is a universal fibration $EG \rightarrow BG$ with fiber $F$, and pulling back this fibration by the map $B \rightarrow BG$ gives the fibration you started with (up to equivalence). Since $EG$ also has the homotopy type of a CW-complex, then the total space will as well.

So if a counterexample exists, $F$ will have to be non-compact.