On the universal pullback of fiber bundles - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T09:11:53Z http://mathoverflow.net/feeds/question/79194 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79194/on-the-universal-pullback-of-fiber-bundles On the universal pullback of fiber bundles Mirco 2011-10-26T21:10:58Z 2011-10-26T21:50:45Z <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>