# 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 $\xi'$?

• No. The dimensions don't even have to match. – Zhen Lin Oct 10 '14 at 10:23
• 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). – mdp Oct 10 '14 at 10:24

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.

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.

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.