Surjectivity of bundle projection In some textbooks, a fiber bundle is defined as follows: for topological spaces $E$, $B$ and $F$, we say that a continuous map $\pi \colon E \to B$ is a fiber bundle with fiber $F$ if for any point $b \in B$, there exists an open neighborhood $U$ of $b$ such that $\pi^{-1}(U)$ is homeomorphic to $U \times F$ via some map $\varphi$ which satisfies $p_1 \circ \varphi=\pi|_{\pi^{-1}(U)}$. Here $p_1$ is the projection to the first factor.
I noticed that in some textbooks, a fiber bundle requires $\pi$ moreover to be surjective. I wonder if this surjectivity is necessary. I presume that if $B$ is path-connected, then the surjectivity follows from parallel transport. If not, I guess that there is an example of a fiber bundle in the sense of the above definition missing surjectivity. Is there such an example? Or, does the sutjectivity follow from the definition unless $F$ is empty?
I would also like to know the situation we really need the surjectivity.
Thank you.

When you assume the existence of the neighbourhood $U$ of $b$ homeomorphic to $U\times F$ an such that $p_1\circ \varphi=\pi$, you are assuming, also, that you can take any $f\in F$ and consider
$$
(b,f)\in U\times F.
$$
Then, $\varphi^{-1}(b,f)\in E$ is such that
$$
\pi(\varphi^{-1}(b,f))=p_1(b,f)=b,
$$
and therefore $\pi$ is surjective.
