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?
1 Answer
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Yes, it is possible to recover the manifold through the following steps:
 Smooth SerreSwan 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 wellknown 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.

$\begingroup$ Thanks! Very straightforward, should have thought of it myself :( $\endgroup$– xuq01Feb 27, 2023 at 5:59