On the universal pullback of fiber bundles - MathOverflow most recent 30 from http://mathoverflow.net2013-05-23T09:11:53Zhttp://mathoverflow.net/feeds/question/79194http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/79194/on-the-universal-pullback-of-fiber-bundlesOn the universal pullback of fiber bundlesMirco2011-10-26T21:10:58Z2011-10-26T21:50:45Z
<p>First suppose we have three smooth manifolds $M_1$, $M_2$ and $N$ with
smooth transversal maps $p_1: M_1 \rightarrow N$ and $p_2: M_2 \rightarrow N$
then its a well known fact that the categoric universal pullback
(in the category of smooth manifolds)
$$M_1 \times_{p_1 N p_2} M_2$$ is a smooth manifold, too. Hence we can say
that transversal pullbacks exist in this category.</p>
<p>Next we have a little more structure on $M_1$ and $M_2$ in that we assume
each $p_1 : M_1 \rightarrow N$ and $p_2 : M_2 \rightarrow N$ to be a locally
trivial smooth fiber bundle. In this case the categoric universal pullback
$M_1 \times_{p_1 N p_2} M_2$ is again a fiber bundle over $N$ called the fiber product
$M_1 \oplus M_2$ or $M_1 \times_N M_2$. </p>
<p>(But observe here that since $N$ is
just a manifold and not a fiber bundle this is not the universal pullback in the
category of smooth fiber bundles !)</p>
<p>This again is well known.</p>
<p>But now we assume in addition that there is a fiber bundle structure $p_N: N \rightarrow L$.
Then $M_1$, $M_2$ and $N$ are fibered manifolds
and my first question is, if the categoric pullback
$M_1 \times_{p_1, p_N, p_2} M_2$ (Used $p_N$ instead of $N$ here to show that is is a pullback in the category of fiber bundles) is defined in the category of local trivial fiber bundles?</p>
<p>In that case it should be a fiber bundle over $L$.</p>
<p>Suppose we restrict the pullback to vector bundles, then is it again a vector bundle
or is it a second order vector bundle or is the linear structure lost?</p>