The Stacks project

4.36 Presheaves of categories

In this section we compare the notion of fibred categories with the closely related notion of a “presheaf of categories”. The basic construction is explained in the following example.

Example 4.36.1. Let $\mathcal{C}$ be a category. Suppose that $F : \mathcal{C}^{opp} \to \textit{Cat}$ is a functor to the $2$-category of categories, see Definition 4.29.5. For $f : V \to U$ in $\mathcal{C}$ we will suggestively write $F(f) = f^\ast $ for the functor from $F(U)$ to $F(V)$. From this we can construct a fibred category $\mathcal{S}_ F$ over $\mathcal{C}$ as follows. Define

\[ \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ F) = \{ (U, x) \mid U\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}), x\in \mathop{\mathrm{Ob}}\nolimits (F(U))\} . \]

For $(U, x), (V, y) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ F)$ we define

\begin{align*} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_ F}((V, y), (U, x)) & = \{ (f, \phi ) \mid f \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(V, U), \phi \in \mathop{\mathrm{Mor}}\nolimits _{F(V)}(y, f^\ast x)\} \\ & = \coprod \nolimits _{f \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(V, U)} \mathop{\mathrm{Mor}}\nolimits _{F(V)}(y, f^\ast x) \end{align*}

In order to define composition we use that $g^\ast \circ f^\ast = (f \circ g)^\ast $ for a pair of composable morphisms of $\mathcal{C}$ (by definition of a functor into a $2$-category). Namely, we define the composition of $\psi : z \to g^\ast y$ and $ \phi : y \to f^\ast x$ to be $ g^\ast (\phi ) \circ \psi $. The functor $p_ F : \mathcal{S}_ F \to \mathcal{C}$ is given by the rule $(U, x) \mapsto U$. Let us check that this is indeed a fibred category. Given $f: V \to U$ in $\mathcal{C}$ and $(U, x)$ a lift of $U$, then we claim $(f, \text{id}_{f^\ast x}): (V, {f^\ast x}) \to (U, x)$ is a strongly cartesian lift of $f$. We have to show a $h$ in the diagram on the left determines $(h, \nu )$ on the right:

\[ \xymatrix{ V \ar[r]^ f & U & (V, f^*x) \ar[r]^{(f, \text{id}_{f^*x})} & (U, x) \\ W \ar@{-->}[u]^ h \ar[ru]_ g & & (W, z) \ar@{-->}[u]^{(h, \nu )} \ar[ru]_{(g, \psi )} & } \]

Just take $\nu = \psi $ which works because $f \circ h = g$ and hence $g^*x = h^*f^*x$. Moreover, this is the only lift making the diagram (on the right) commute.

Definition 4.36.2. Let $\mathcal{C}$ be a category. Suppose that $F : \mathcal{C}^{opp} \to \textit{Cat}$ is a functor to the $2$-category of categories. We will write $p_ F : \mathcal{S}_ F \to \mathcal{C}$ for the fibred category constructed in Example 4.36.1. A split fibred category is a fibred category isomorphic (!) over $\mathcal{C}$ to one of these categories $\mathcal{S}_ F$.

Lemma 4.36.3. Let $\mathcal{C}$ be a category. Let $\mathcal{S}$ be a fibred category over $\mathcal{C}$. Then $\mathcal{S}$ is split if and only if for some choice of pullbacks (see Definition 4.33.6) the pullback functors $(f \circ g)^*$ and $g^* \circ f^*$ are equal.

Proof. This is immediate from the definitions. $\square$

Lemma 4.36.4. Let $ p : \mathcal{S} \to \mathcal{C}$ be a fibred category. There exists a contravariant functor $F : \mathcal{C} \to \textit{Cat}$ such that $\mathcal{S}$ is equivalent to $\mathcal{S}_ F$ in the $2$-category of fibred categories over $\mathcal{C}$. In other words, every fibred category is equivalent to a split one.

Proof. Let us make a choice of pullbacks (see Definition 4.33.6). By Lemma 4.33.7 we get pullback functors $f^*$ for every morphism $f$ of $\mathcal{C}$.

We construct a new category $\mathcal{S}'$ as follows. The objects of $\mathcal{S}'$ are pairs $(x, f)$ consisting of a morphism $f : V \to U$ of $\mathcal{C}$ and an object $x$ of $\mathcal{S}$ over $U$, i.e., $x\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$. The functor $p' : \mathcal{S}' \to \mathcal{C}$ will map the pair $(x, f)$ to the source of the morphism $f$, in other words $p'(x, f : V\to U) = V$. A morphism $\varphi : (x_1, f_1: V_1 \to U_1) \to (x_2, f_2 : V_2 \to U_2)$ is given by a pair $(\varphi , g)$ consisting of a morphism $g : V_1 \to V_2$ and a morphism $\varphi : f_1^\ast x_1 \to f_2^\ast x_2$ with $p(\varphi ) = g$. It is no problem to define the composition law: $(\varphi , g) \circ (\psi , h) = (\varphi \circ \psi , g\circ h)$ for any pair of composable morphisms. There is a natural functor $\mathcal{S} \to \mathcal{S}'$ which simply maps $x$ over $U$ to the pair $(x, \text{id}_ U)$.

At this point we need to check that $p'$ makes $\mathcal{S}'$ into a fibred category over $\mathcal{C}$, and we need to check that $\mathcal{S} \to \mathcal{S}'$ is an equivalence of categories over $\mathcal{C}$ which maps strongly cartesian morphisms to strongly cartesian morphisms. We omit the verifications.

Finally, we can define pullback functors on $\mathcal{S}'$ by setting $g^\ast (x, f) = (x, f \circ g)$ on objects if $g : V' \to V$ and $f : V \to U$. On morphisms $(\varphi , \text{id}_ V) : (x_1, f_1) \to (x_2, f_2)$ between morphisms in $\mathcal{S}'_ V$ we set $g^\ast (\varphi , \text{id}_ V) = (g^\ast \varphi , \text{id}_{V'})$ where we use the unique identifications $g^\ast f_ i^\ast x_ i = (f_ i \circ g)^\ast x_ i$ from Lemma 4.33.7 to think of $g^\ast \varphi $ as a morphism from $(f_1 \circ g)^\ast x_1$ to $(f_2 \circ g)^\ast x_2$. Clearly, these pullback functors $g^\ast $ have the property that $g_1^\ast \circ g_2^\ast = (g_2\circ g_1)^\ast $, in other words $\mathcal{S}'$ is split as desired. $\square$

Comments (5)

Comment #7087 by Rakesh Pawar on

We have these functors between 2-categories

(1) Do these functors form equivalences as 2-categories or (2) these are equivalences of corresponding 1-categories, by which I mean, on , the set of isomorphism classes of objects in the category ?

I have the same question about the corresponding statement of Presheaves in Groupoids and the category of fibered categories in groupoids.

Comment #7090 by on

@#7087. In order to say what an equivalence is between 2-categories, one needs to work in the 3-category of all 2-categories, which is something I want to avoid doing at all cost! It is quite possible that your option (2) is OK, but I haven't checked all the details.

Comment #7091 by Rakesh Pawar on

Thanks for your comment.

Is it possible to say it is equivalence of 2-categories as in (1) as follows: The functors which induce equivalence of categories, i.e. for all objects , the functor induced by , is essentially surjective and fully faithful.

I suppose this will avoid talking about 3-categories in practice. I guess my question is ''is the notion of equivalence of 2-categories, (in the 3-category of 2-categories) synonymous to the above naive formulation of an equivalence of 2-categories?''

Comment #7093 by on

Dear Rakesh Pawar, two quick things. First, I bet your question at the end of #7091 has a negative answer. Second, now I am getting worried that even your option (2) in #7087 may not be OK in the specific example of the text (but I haven't had coffee yet today). It has been suggested to me by experts that it doesn't pay to think (eplicitly) about "truncated" higher categories (above 2-categories) and if needed one should directly jump to -categories (in which case you can read all about these things on kerodon. Anecdotally, Fulton asked for a precise formulation of the folklore "the 2-category of algebraic stacks is equivalent to the 2-category of presentations" where algebraic geometers know what it means but if you want to spell it out formally you almost immediately get a headache and give up.

Comment #7094 by Rakesh Pawar on

If (2) in #7087 is not OK in these particular situations, and if the question at the end of #7091 has a negative answer, then there is no hope for me ever to check if these are equivalences. :-) Thanks for sharing the insight and the anecdote. I will pass it around, whenever 2-categories will crop up around me. :-)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02XU. Beware of the difference between the letter 'O' and the digit '0'.