It's well known that there are non-trivial bundles such as the Mobius strip. In these cases it is known the bundle $E$ with bundle map $\pi : E \to B$ over base space $B$ with fiber $F$ cannot be expressed as $E = B \times F$. This is because the topology of $E$ and $B$ might be such that the map $\pi$ is not continuous as it needs to be for a bundle map if $E$ is $B\times F$ endowed with the usual product topology.
However, can it be shown that bundle $E$ over base space $B$ with bundle $F$ always has a bijection with $B\times F$, ignoring topology/continuity? I'm wondering if a bundle can, in general be written as $E=B\times F$, but with the understanding that the topology on $E$ is NOT the product topology that would typically come with $B\times F$.
As noncontinuous bijections we have $$E=\bigcup_{b\in B}\pi^{-1}(b)\cong \bigcup_{b\in B}F\cong B\times F.$$
The following turns out to be useful in the homotopical classification of locally trivial bundles with fixed fibre.
Application: The collection of isomorphism classes of bundles over $B$ with fibre $F$ forms a set (and not a proper class).
Proof: The total space $E$ of any $F$-bundle over $B$ is homeomorphic to a space whose underlying set is $B\times F$. But the family of all topologies on any given set is a set. For any one of these topologies one may need to consider different projections $E\rightarrow B$, but there is again only a set of these. Finally, a set-indexed union of sets is a set. $\quad\blacksquare$