Pullback bundle

A total space $E$ ,
A base space $B$ ,
A projection map $π : E \to B$ ,
A typical fiber $F$ (all fibers are homeomorphic to $F$ ),
Suppose you have a continuous map $f : B^{'} \to B$ (from a new base space into the original one),
Definition
The pullback bundle $f^{*} E$ is the set:

f^{*} E = {(b^{'}, e) \in B^{'} \times E ∣ f (b^{'}) = π (e)}

So it’s a subset of $B^{'} \times E$ , consisting of pairs where the point $e \in E$ lies over $f (b^{'}) \in B$ .
This comes with:

A projection $π^{'} : f^{*} E \to B^{'}$ given by $π^{'} (b^{'}, e) = b^{'}$ ,
A map to the original total space: $f^{*} E \to E$ , $(b^{'}, e) \mapsto e$ .

\begin{matrix} f^{*} E & \overset{π^{*} f}{\to} & E \\ π^{'} ↓ & ↓ π \\ B^{'} & \overset{f}{\to} & B \end{matrix}

$◼$
Remark
The fiber of $f^{*} E$ over a point $b^{'} \in B^{'}$ is:

(f^{*} E)_{b^{'}} = {(b^{'}, e) \in f^{*} E ∣ f (b^{'}) = π (e)} ≅ π^{- 1} (f (b^{'})) = E_{f (b^{'})}

So each fiber in the pullback is just a copy of the fiber over $f (b^{'})$ in the original bundle.

Remark
The map $π^{*} f$ is a fiberwise diffeomorphism. See @Michor2008Topics 17.5.

Example
Let’s say:

$B = S^{1}$ , and you have a Möbius bundle over it,
$f : S^{1} \to S^{1}$ is a map like $f (θ) = 2 θ$ (a double covering),

Then the pullback of the Möbius bundle via $f$ becomes a non-twisted bundle — a trivial bundle — because the twist happens every full loop, and you're going twice as fast.

Related: pullback connection.