1
$\begingroup$

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 $\xi'$?

$\endgroup$
  • $\begingroup$ No. The dimensions don't even have to match. $\endgroup$ – Zhen Lin Oct 10 '14 at 10:23
  • $\begingroup$ I'm not fluent enough with bundles to write down a particular example, but the answer should be no; all you should know is that there is a unique map $\varphi$ from $\xi$ to the pull-back bundle such that $\overline{f}$ is $\varphi$ composed with the map from the pull-back to $E'$, and similarly for $p$. This is the universal property of the pull-back; see en.wikipedia.org/wiki/Pullback_(category_theory). $\endgroup$ – mdp Oct 10 '14 at 10:24
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.