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If $\pi : F \to N$ is a fiber bundle, and $\phi : M \to N$ (here, $M$ and $N$ are manifolds), then the standard way to define the pullback bundle of $F$ by $\phi$ is $$\phi^* F := \{(m,f) \in M \times F \mid \phi(m) = \pi(f)\},$$ written as a submanifold of $M \times F$.

If additionally, $\psi : L \to M$ (where $L$ is also a manifold), then $\psi^* \phi^* F$ is a submanifold of $L \times \phi^* F$ which is a submanifold of $L \times M \times F$, and $\psi^* \phi^* F$ is naturally isomorphic to $(\phi \circ \psi)^* F$, which is a submanifold of $L \times F$. One can show that the notion of pullback is a contravariant functor (with a minor caveat).

In being super-careful about the types of domain/codomain in various bundle computations in some of my research, I'm finding that, while it is a natural way to construct a pullback bundle, this "tacking on" of the new basespace and identifying pullback bundles using the above isomorphism tends to get in the way of really rigorously typed computations (i.e. ensuring that domain/codomain are always exactly matched, and making no implicit identifications of spaces).

QUESTION: Is there an alternate construction for pullback bundles which doesn't require the identification of spaces (in the manner described above) in order for pullback to be a contravariant functor?

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My thought is that the submanifold-of-direct-product construction could still be used as "hidden implementation details", over which an equivalence relation is imposed to identify the isomorphic spaces, and then somehow translate into this scheme the natural morphisms/maps one gets from the fact that the pullback bundle is defined as a submanifold of a direct product (e.g. restrictions of projections onto each factor, etc). –  Victor Dods Jul 19 '12 at 23:15
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I am not sure I fully understand your problem. Are you bothered by the fact that $(\phi \circ \psi)^* F$ and $\psi^* \phi^* F$, are merely naturally isomorphic and not equal in terms of the standard construction? You probably know that the pullback bundle is unique up to unique isomorphism, if you define it in the categorical sense. In that sense the problem 'shouldn't' matter, but I can imagine that in explicit computations you really need to use a construction. Perhaps you can give an example of this 'getting in the way' of using the isomorphism? –  wildildildlife Jul 19 '12 at 23:54
    
Identifying the naturally isomorphic pullback bundles in the category of bundles seems like the way to go, but I can't think of how to do this in the category of bundles -- it seems like one must talk only about the subcategory of pullback bundles (assuming this is actually a subcategory); put an equivalence relation on this category which relates isomorphic bundles in the sense described above. OR, is there a more natural way to do this in the full category of bundles? –  Victor Dods Jul 20 '12 at 0:31
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Congratulations! You have discovered the notion of a pseudofunctor. There is in general no way of "rectifying" a pseudofunctor to be a genuine functor. –  Zhen Lin Jul 20 '12 at 0:42
    
@ZhenLin Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Lord_Farin Jun 15 '13 at 13:51

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