Let $X$ is be smooth projective curve over an algebraically closed field $k$.
Let $S$ be a scheme and let $S \xrightarrow[]{t}\Bbb A^1$ be a global function. Let $\mathcal C_S$ be the category whose objects are pairs $(\mathcal E, \nabla )$ where $\mathcal E$ is a degree zero vector bundle on $X_S = X\times S$ and $\nabla$ is a connection on $\mathcal E$, and whose morphisms are isomorphisms commuting with the connections.
Prove that
- $\mathcal C_S$ with the "obvious" inverse image functor is a fibered catgory over the category $\operatorname{Aff}/\Bbb A^1$.
My attempt:
- I think that the obvious inverse image functor should take an object $(\mathcal E, \nabla ) \in \mathcal C_S$ as input and outputs an affine scheme over $\Bbb A^1$. The problem here is that neither $X$ nor $S$ are affine, so I thought rather about $\operatorname{Spec(\mathcal E(X\times S))} $, or even $\operatorname{Spec(\mathcal E(X))} $ but it doesn't seem right. I don't know what would be this inverse image functor?
Thank you for your help.
The problem is a bit ambiguously stated, but I think it should probably be interpreted so that $\mathcal C_S$ is the fiber of the fibered category over $S$, which must be restricted to be affine. Then we have the fibered category $(S\to \mathbb A^1)\mapsto \mathcal C_S$, and then you can start to think about an inverse image functor induced by a map $T\to S$ of affines over $\mathbb A^1$. Note that your attempt is about trying to construct a fiber functor, or the forgetful functor, from $\mathcal C_S$ to $\mathrm{Aff}/\mathbb A^1$, which is not the same as the inverse image functor.