It's probably a really naive question, but I didn't find any references. Given a category $V$ and we know that it is equivalent to the category $\mathbf{Vect}(X)$ of smooth vector bundles over a smooth manifold $X$, under what assumptions could we determine $X$? Or is this generally hopeless?



Reconstructing base manifold from its category of smooth vector bundles
Yes, it is possible to recover the manifold through the following steps:


* [Smooth Serre-Swan theorem](https://ncatlab.org/nlab/show/smooth+Serre-Swan+theorem): $\mathbf{Vect}(X)$ is equivalent to the category of finitely generated projective modules over $C^\infty(X)$.
* If $R$ is any commutative ring, then $R$ is canonically isomorphic to the ring of natural endomorphisms of the identity functor on the category of finitely generated projective modules. Thus, we obtain $C^\infty(X)$ as the ring of natural endomorphisms of $\mathrm{id}\_{\mathbf{Vect}(X)}$.
* Finally, it is [well-known](https://math.stackexchange.com/a/1765040/745634) that a manifold can be recovered from its ring of smooth functions. More precisely, the functor $C^\infty$ from manifolds to the opposite category of commutative rings is fully faithful.


