5
$\begingroup$

In classical category theory, there is the notion of functor (co)fibered in groupoids. Furthermore, via Grothendieck construction we have an equivalence between pseudo functors into the category of groupoids and functors cofibered in groupoids.

This leads to my question: why is the following theorem which can be found in Lurie's HTT (Theorem 2.1.2.2) an $\infty$-categorical analogue?

Let $S$ be a simplicial set, $\mathcal{C}$ a simplicial category, and $\phi: \mathfrak{C}[S] \rightarrow \mathcal{C}^{\mathrm{op}}$ a simplicial functor. The straightening and unstraigntening functors determine a Quillen adjunction $$ \operatorname{St}_{\phi} : (\mathit{Set}_{\Delta})_{/S} \rightleftarrows \mathit{Set}_{\Delta}^{\mathcal{C}} \colon\operatorname{Un}_{\phi}$$ where $(\mathit{Set}_{\Delta})_{/S}$ is endowed with the contravariant model structure and $Set_{\Delta}^{\mathcal{C}}$ with the projective model structure. Furthermore, if $\phi$ is an equivalence of simplicial categories, then it is an Quillen equivalence.

I know that fibrant object in the category of objects over $S$ (with the covariant model structure) are right fibrations- the $\infty$-categorical analogue of functors fibered in groupoids- and that the projectively fibrant functors are pointwise Kan complexes -the $\infty$-categorical analogue of groupoid. How does the theorem link those to two togethers?

$\endgroup$
2
  • 1
    $\begingroup$ Take $S$ to be an $\infty$-category. Then the theorem produces an equivalence of $\infty$-categories between the $\infty$-category $\mathsf{RFib}(S)$ of right fibrations over $S$ and the $\infty$-category of functors $S^{op} \to \mathsf{Spaces}$. (Take $\mathcal{C} = \mathfrak{C}[S]^{op}$). So the "link" is that they are equivalent. $\endgroup$ Commented Feb 8, 2019 at 17:00
  • $\begingroup$ The projective model structure (and also the injective model structure...) on simplicial functors models the $\infty$-category of functors between both sides- see HTT.4.2.4.4. $\endgroup$ Commented Feb 8, 2019 at 17:03

2 Answers 2

1
$\begingroup$

The unstraightening functor $\mathrm{Un}_{\phi}$ is an analog of the Grothendieck construction that associates a category $\mathcal{C}_{\chi}$ to a functor $\chi$ as at the beginning of section 2.1.1 of Lurie's HTT. Here is a heuristic way to see why.

We follow the notations in HTT section 2.2.1. For simplicity we take $\mathcal{C}=\mathfrak{C}[S]^{op}$ and $\phi=\mathrm{id}$. Let $s$ be a vertex of $S$, then by the definition of $\mathrm{Un}_{\phi}$ as a right adjoint, for any object $Y$ of $\mathrm{Set}_{\Delta}^{\mathcal{C}}$ we have \begin{equation} \mathrm{Hom}_{(\mathrm{Set}_{\Delta})_{/S}}(\{s\},\mathrm{Un}_{\phi}Y) =\mathrm{Hom}_{\mathrm{Set}_{\Delta}^{\mathcal{C}}}(\mathrm{St}_{\phi}\{s\},Y). \end{equation} To unwind $\mathrm{St}_{\phi}\{s\}$, we look at the simplicial category $\mathcal{M}$ as the pushout of $\require{AMScd}$ \begin{CD} \mathfrak{C}[\{s\}] @>>> \mathfrak{C}[S] \\ @V V V\\ \mathfrak{C}[\{s\}^{\triangleright}] \end{CD} Unwinding the definition of a pushout of categories, we see that a morphism from a vertex of $S$ to the cone point $v$ of $\{s\}^{\triangleright}$ must factor through the vertex $s$. So a natural transformation from $\mathrm{St}_{\phi}\{s\}$ to $Y$ should correspond to choices maps from $\{s\}$ to $Y(s)$. In other words, it corresponds to a point in $|Y(s)|$. This fact is made precise in Remark 2.2.2.11 in HTT (see e.g. Explicit expression of the unstraightening functor).

Similarly, an edge $s\rightarrow t$ in $\mathrm{Un}_{\phi}Y$ over $S$ should correspond to the edge $s\rightarrow t$ in $S$ and an edge from a point in $|Y(s)|$ to a point $|Y(t)|$ covering $s \rightarrow t$.

So we recover the Grothendieck construction, and see how the cone point in the definition of the straightening functor plays its role.

$\endgroup$
0
$\begingroup$

As Xiaowen mentions, it is probably a good idea to look at the unstraightening functor for an intuition. And while Xiaowen's answer is nice, we can be even more explicit. For simplicity, I will assume $\phi=\operatorname{id}$. Let $F:\mathcal{C}^{\mathrm{op}}\to \mathrm{Set}_\Delta$ be a simplicial functor. Then:

  • A vertex of $\operatorname{Un}_{\phi}(F)$is a pair $(s,v)$, where $s$ is a vertex of $S$ and and $v$ is a vertex of $F(s)$.
  • An edge $(s,v)\to (s',v')$of $\operatorname{Un}_{\phi}(F)$ that lies over an edge $f:s\to s'$is an edge $v\to Ff(v')$ in $F(s)$, where we used the same notation $f$ to denote the morphism of $\mathfrak{C}[S]$ determined by $f$.
  • A 2-simplex of $\operatorname{Un}_{\phi}(F)$depicted as a left hand triangle, lying over a 2-simplex of $S$ as depicted on the right triangles is a diagram $\Delta^1\times \Delta^1\to F(s)$ depicted as square

Simplices of higher dimensions can be depicted similarly, using cubical diagrams of higher dimensions. A full account of this can be found, e.g., in [$\S$4.1, Ara24].

[Ara24] The Grothendieck Construction for ∞-Categories Fibered over Categorical Patterns, arXiv:2404.01025, 2024.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.