MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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 $M_1 \times_N M_2$.

(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 !)

This again is well known.

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?

In that case it should be a fiber bundle over $L$.

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?

share|cite|improve this question
Ok the answer to the first question is yes because $p_N \circ p_1$ and $p_N \circ p_2$ are transverse. Hence $M_1 \times_{(p_N \circ p_1) L (p_N \circ p_2)} M_2$ is a fiber bundle. – Mirco Oct 26 '11 at 21:48
with the universal property. – Mirco Oct 26 '11 at 21:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.