Let $\mathfrak S$ be the category of schemes. My goal is just to visualize the $2$-morphisms in the $2$-category consisting of the following objects: categories over $\mathfrak S$, i.e. $$ \textrm{ categories } X, \textrm{ together with a (covariant) functor } X\to \mathfrak S. $$ (I'm following Fantechi's paper "Stacks for everybody", where at some point I get lost.)

So, suppose we have two "$0$-objects" (functors) $X\to\mathfrak S$ and $Y\to\mathfrak S$; then we have a category $\textrm{hom}_\mathfrak S(X,Y)$ whose objects are the commutative triangles that one expects (and that I am unable to draw here). My trouble is in understanding the morphisms in this category, i.e. the so called "$2$-morphisms". Such morphisms should of course be natural transformations $\phi\to\psi$ of functors $\phi,\psi:X\to Y$ over $\mathfrak S$. It is exactly this condition "over $\mathfrak S$" that I cannot translate. More precisely, in the paper it is written that a morphism in $\textrm{hom}_\mathfrak S(X,Y)$ is $$ \textrm{"a natural transformation over the identity functor on } \mathfrak S" $$ Can anyone explain to me the part "over the identity functor"?


  • $\begingroup$ I think there are two simplifying cases that make this easier to visualize. Before I type them both up I was wondering if you've worked either out yet. Much simpler: consider $\mathfrak{S}$ to be the Zariski site on a fixed variety $X/k$. Then objects are opens $U\hookrightarrow X$ and a good sample category would be vector bundles (or even just line bundles!). You can translate everything to topological language then. The other example is to go back to $\mathfrak{S}$ as schemes, but then consider $X$ and $Y$ to be varieties as categories over $\mathfrak{S}$ and see what happens. $\endgroup$
    – Matt
    Feb 1 '13 at 22:32
  • $\begingroup$ 1-st simplification. Let me call $\mathcal L$ the category of line bundles on $X$. I suppose we have two $\mathfrak S$- functors $\phi,\psi:\mathcal L\to\mathcal L$ (say). (I tried explicitly with $\phi=1,\psi=-^\vee$.) But still I can't figure what I have to impose on a natural transformation $\phi\to\psi$ to translate the compatibility with $\mathfrak S$. My only guess is to "add" to the square of the natural transformation the morphisms from the bundles to $X$ and impose commutativity. But it is a very special case of bundles that we have morphisms between objects of $\mathcal L$ and $X$... $\endgroup$
    – Brenin
    Feb 1 '13 at 23:31
  • $\begingroup$ 2-nd simplification. I have to think about it a little bit more, because I feel like I don't capture the meaning of considering a scheme $X$ as a category over $\mathfrak S$. Does it mean the following? to look at the identity functor $1_\mathfrak S$ and at the fiber over the object $X$, which is the subcategory consisting of the single object $X$ and has as morphisms the isomorphisms $Z\to X$ in $\mathfrak S$. $\endgroup$
    – Brenin
    Feb 1 '13 at 23:43
  • $\begingroup$ I would call this "simplification" with line bundles on varieties rather a "complication". As always in basic category theory, only few data is available, and finding the correct notion of morphism means just unwinding the whole structure and demanding the morphism to be compatible with this structure. The correct definition should satisfy coherence, i.e. all meaningful diagrams should commute. In particular, any two parallel morphisms should be equal. $\endgroup$ Feb 2 '13 at 14:26
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    $\begingroup$ Thus, if $X,Y$ are fibered categories over a category $C$, and $F,G : X \to Y$ are functors over $C$, a natural transformation $\eta : F \to G$ should also be over $C$: For $x \in X$, lying over $c \in C$, we have the morphism $\eta(x) : F(x) \to G(x)$ in $Y$, which projects down to some endomorphism of $c$. There is also the identity of $c$. Of course we require that these two morphisms coincide. $\endgroup$ Feb 2 '13 at 14:29

Let $\mathcal{C}$ be a category. The standard definition of 2-morphism in $(\mathfrak{Cat} \downarrow \mathcal{C})$ is this: given objects $P : \mathcal{D} \to \mathcal{C}$ and $Q : \mathcal{E} \to \mathcal{C}$ and functors $F, G : \mathcal{D} \to \mathcal{E}$ such that $Q F = P = Q G$, a 2-morphism $\alpha : F \Rightarrow G$ is a natural transformation $\alpha : F \Rightarrow G$ such that, for all objects $d$ in $\mathcal{D}$, the morphism $Q (\alpha_d) = \textrm{id}_{P d}$. This is what is meant by "a natural transformation over the identity functor". Another name for it is "vertical natural transformation", because the morphisms $\alpha_d$ are "vertical" in the sense of being contained in the fibre of $Q$ over $P d$.


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