# Fiber bundle = principal bundle + fiber?

This question is heavily related to this question.

Fix a sufficiently nice and connected topological space $B$ and let $FB$ be the category of fiber bundles over $B$. A morphism $f: (E\to B)\to (E'\to B)$ in this category is a map $E\to E'$ over $B$ and hence it maps fibers to fibers.

Edit: A fiber bundle $p:E\to B$ is a continuous map such that there exists a local trivialization. Since $B$ is nice and connected, all the fibers $p^{-1}(x)$ are isomorphic.

According to the very helpful answers here,

• a fiber bundle over $B$ with fiber $F$ determines a $G=Aut(F)$-principal bundle together with (trivially) a left $G$-action on $F$ and
• a $G$-principal bundle over $B$, where $G$ is an arbitrary topological group, together with a topological space $F$ and a left $G$-action on $F$ determines a fiber bundle over $B$ with fiber $F$.

Can one formulate this correspondence "categorically", i.e. is there an equivalence of the category $FB$ to a product (?) of two categories, one encoding the "glueing structure" (the principal bundle) and one encoding the "fiber information" (the space $F$ with the action)? (In particular, what should be the analogue to a morphism of fiber bundles?)

(Certainly is not possible to get such a description precisely like above because one have to fix the topological group $G$ to say what a space $F$ with a $G$-action is and conversely one have to fix a space $F$ to say what a $G=Aut(F)$-principal bundle is, but maybe there is a way to formulate this in general.)

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The question is not precise enough: the category FB is not defined precisely (e.g. do you require local triviality, paracompact base, possibility to have different fibers over different connected components and so on). If I understood right your category allows that the typical fiber of domain and target "fiber bundle" is different. Any morphism among principal G-bundles is iso; if you would have locally trivial fibrations with arbitrary fiber, not every map among such would be induced by morphisms of the fibers, thus no cat-product. Maps of vector bundles might have a non-constant rank. –  Zoran Skoda Mar 5 '10 at 14:35
Thank you for your response. I have edited the question. –  veit79 Mar 5 '10 at 15:05
Something is very wrong about your construction. Consider the mobius bundle over the circle: the fibers are isomorphic as topological spaces to \RR or the open interval, and the gluing is more or less straight forward. Then the automorphism group of the fiber includes at least the translations, but the translations do not as as global endomorphisms of the bundle. More generally, the problem is this: knowing how a group G acts on F does not determine the action of G on some isomorph of F --- rather, you have to fix an isomorphism, but in your context this isomorphism depends on local patch. –  Theo Johnson-Freyd Mar 6 '10 at 22:41
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## 1 Answer

It is definitely not an easy read, but you might want to have a look at Peter May's Classifying Spaces and Fibrations monograph. If i recall correctly, he constructs classifying spaces for fibrations with fiber F in some very categorical fashion. So it seems like it might imply something like the result you are interested in, although i doubt it is exactly what you want. It has been a long time since i read through any of it (I never got through all of it). This seems like something you might want to look at, let me know if you are having a hard time finding a copy via email.

Also... I could be completely wrong.

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A photocopy of this book is available from Peter May's webpage: math.uchicago.edu/~may/BOOKS/Classifying.pdf –  Dan Ramras Mar 5 '10 at 20:55
thats right! It is one of only two of his memoirs up there. We all owe andy baker a thank you for saying something to peter that then led to his having all this stuff scanned in. –  Sean Tilson Mar 5 '10 at 21:24
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