does a commutative diagram implies pull-back? Let $\xi=(E,p,B),\xi'=(E',p',B')$ be fibre bundles. Let $f: B\to B'$, $\bar f: E\to E'$ be maps such that the diagram commutes
$\require{AMScd}$
\begin{CD}
    E @>\displaystyle\bar f>> E'\\
    @V \displaystyle p V V\# @VV\displaystyle p' V\\
    B @>>\displaystyle f> B^{\,\prime}
    \end{CD}
Does this alway imply that $\xi$ is isomorphic to the pull-back bundle ( the pull-back bundle, denoted by $\eta$, of $\xi'$ to $B$ through $f$ is: $E(\eta)=\{(b,e')\mid b\in B, e'\in E' \text{ such that } f(b)=p'(e')\}$, $p(\eta)(b,e')=b$, $B(\eta)=B$) of $\xi'$?
 A: No, that would imply that all fiber bundles are trivial:
Let $p\colon E\rightarrow B$ be a fiber bundle. The unique map from a one point set to a one point set is a fiber bundle too and one gets a commuting diagram of the shape
$$
\begin{array}{ccccccccc}
E & \xrightarrow{} & \{*\} \\
%
\downarrow & & \downarrow \\
%
B & \xrightarrow{} & \{*\} 
\end{array}.$$
Since pullback bundles of trivial bundles are trivial, your claim would imply that $E\rightarrow B$ is trivial.
A: If you add the hypothesis that $\bar f$ restricts to a homeomorphism on each fiber, then it is true that $\xi$ is isomorphic to the pullback bundle. To see this note that the total space of the pullback bundle is
$$f^*E' = \{(b,e')\in B\times E': p'(e') = f(b)\}.
$$
Define a bundle map $\phi\colon E\to f^*E'$ by
$$
\phi(e) = (p(e),\bar f(e)).
$$
Then the hypotheses guarantee that $\phi$ is a bijective bundle map covering the identity of $B$, and it is continuous because $p$ and $f$ are. You can show that $\phi^{-1}$ is continuous by expressing it locally in terms of local trivializations.
A: Propably the most enlightening answer would be the following:
among all such $E$ that you get a commutative diagram as above, the best one (i.e. the universal one) would be the pullback (intuitively speaking). This is in fact the categorical definition of a pullback. You should check the catsters' video on youtube on this topic!
What this implies, is that given your $E$, you will be a unique map to the pullback $f^*E'$ s.t. it commutes with all the relevant maps involved.
