Loading [MathJax]/extensions/tex2jax.js

The Stacks project

Remark 4.41.3. We can use the $2$-Yoneda lemma to give an alternative proof of Lemma 4.37.3. Let $p : \mathcal{S} \to \mathcal{C}$ be a category fibred in groupoids. We define a contravariant functor $F$ from $\mathcal{C}$ to the category of groupoids as follows: for $U\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ let

\[ F(U) = \mathop{\mathrm{Mor}}\nolimits _{\textit{Cat}/\mathcal{C}}(\mathcal{C}/U, \mathcal{S}). \]

If $f : U \to V$ the induced functor $\mathcal{C}/U \to \mathcal{C}/V$ induces the morphism $F(f) : F(V) \to F(U)$. Clearly $F$ is a functor. Let $\mathcal{S}'$ be the associated category fibred in groupoids from Example 4.37.1. There is an obvious functor $G : \mathcal{S}' \to \mathcal{S}$ over $\mathcal{C}$ given by taking the pair $(U, x)$, where $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $x \in F(U)$, to $x(\text{id}_ U) \in \mathcal{S}$. Now Lemma 4.41.2 implies that for each $U$,

\[ G_ U : \mathcal{S}'_ U = F(U)= \mathop{\mathrm{Mor}}\nolimits _{\textit{Cat}/\mathcal{C}}(\mathcal{C}/U, \mathcal{S}) \to \mathcal{S}_ U \]

is an equivalence, and thus $G$ is an equivalence between $\mathcal{S}$ and $\mathcal{S}'$ by Lemma 4.35.9.


Comments (0)


Your email address will not be published.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.