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 $\xi'$?

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.