Hard as I tried, I couldn't find a proof of Remark in Higher Topos Theory, or prove it myself. It seems to need an explicit formulation for the unstraightening functor, so my question is: is an explicit expression known for the unstraightening? Anyway, is it possible to obtain Remark without having one?

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    $\begingroup$ You better write clearly what is the question.. I guess users may not prefer to go to that page and see what It is.. you can atleast say what the remark is.. $\endgroup$ May 24, 2018 at 15:58

2 Answers 2


Wow, I have always thought that unstraightening has to be easier than straightening, but I've never actually looked at Lurie's treatment before, so I'm surprised to realize he defines straightening directly while defining unstraightening as its right adjoint.

Anyway, I'm pretty sure that Remark follows from the description of straightening over a point given between Remarks and, by applying the functoriality properties of Proposition After all, in Remark, we are calculating the following quantity:

$Un_\phi \mathcal F \times_S \{s\} = Map_S(\{s\} \to S, Un_\phi \mathcal F) = Nat_{\mathcal C}(St_\phi(\{s\} \to S), \mathcal F)$.

where we have $\phi: \mathfrak C[S]^\mathrm{op} \to \mathcal C$, $s \in S$, and $\mathcal F: \mathcal C \to \mathsf{sSet}$. Note that

$St_\phi(\{s\} \to S) = St_\phi((\{s\} \to S)_!(\{s\}=\{s\})) \\ \qquad ~~~\qquad = St_{\phi \circ \mathfrak C(\{s\} \to S)}(\{s\}) = (\phi(s))_! \circ St_{\{s\}}(\{s\})$

where in the last two steps, we have applied Prop and Prop respectively. We have also written $\phi(s)$ for $\phi \circ \mathfrak{C}(\{s\} \to S)$, and sometimes written $\{s\}$ for the identity map $\{s\} = \{s\}$.

Now we calculuate:

$Nat_{\mathcal C}((\phi(s))_! \circ St_{\{s\}}(\{s\}),\mathcal F) = Nat_{\{s\}}(St_{\{s\}}(\{s\}), \mathcal F(\phi(s))) \\ \qquad \qquad \qquad \qquad \qquad = Nat_{\{s\}}(|\{s\}|_{Q_\bullet}, \mathcal F(\phi(s))) = Map_{\{s\}}(\{s\}, Sing_{Q_\bullet} \mathcal F(\phi(s))) \\ \qquad \qquad \qquad \qquad \qquad = Sing_{Q_\bullet} \mathcal F(\phi(s))$

as desired. Here we have applied adjointness, then the description of straightening over a point from between and, then adjointness again, and finally evaluated at a point.

The upshot is that we are getting an explicit partial description of the unstraightening functor, in the sense that we are seeing what its fibers are.

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    $\begingroup$ It is little disturbing to see these numbers $,,$ with out mentioning what they actually are. Atleast you could have clearly said what it is as you have enough experience here. $\endgroup$ May 24, 2018 at 16:58
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    $\begingroup$ Here is a glossary: (the remark in question) calculates the fiber $Unst_\phi \mathcal F \times_S \{s\}$; the paragraph between and caluclates straightening over a point in terms of $Q_\bullet$. is a pair of functoriality properties for straightening. I only need formal properties of the things I haven't defined, so I refer to HTT (which is freely available from Lurie's website) for more details. I agree that numerical HTT references are unpleasant to look at, but at least they are precise, and one must learn to live with them in higher category theory today. $\endgroup$
    – Tim Campion
    May 24, 2018 at 17:10
  • $\begingroup$ @TimCampion Hi Tim! Is the description in p. 164 of Rezk's notes on quasicategories (a link to that page for convenience) the one sought in this question? $\endgroup$
    – Emily
    Dec 6, 2020 at 3:43
  • $\begingroup$ (P.S. Sorry for pinging you if it isn't; I've just been learning a bit about un/straightening and am currently quite confused!) $\endgroup$
    – Emily
    Dec 6, 2020 at 3:43
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    $\begingroup$ By the way, since leaving this comment here I stumbled upon these really cool slides of Alexander Campbell, which describe (forthcoming work) on how to factor the straightening-unstraightening adjunction into a composition of an equivalence and two adjunctions. These give a nice description of unstraightening as $\mathrm{Un}_{A}\cong A_{/(-)}\circ(\eta_{A},\mathrm{id})^{*}\circ\mathrm{N}\circ\text{collage}$ (in Alexander's notation, see there). $\endgroup$
    – Emily
    Jan 29, 2021 at 19:42

This is an old question, but it seems worthwhile to give the full explicit description of unstraightening, for convenient reference. I'll do this for contravariant unstraightening, and trying to match the notation used in HTT 2.2.1 (i.e., resisting the temptation to replace $\mathcal{C}$ with $\mathcal{C}^{\mathrm{op}}$ everywhere).

Fix functors $\phi\colon \mathfrak{C}[S]\to \mathcal{C}^{\mathrm{op}}$ and $F\colon \mathcal{C}\to \mathrm{Set}_\Delta$ of simplicial categories. I will describe the $n$-simplices of $\mathrm{Un}_\phi F$, which is a simplicial set mapping to $S$.

Given a map $s\colon \Delta^n\to S$ let $\phi_s= \phi\circ \mathfrak{C}[s]$, a functor $\mathfrak{C}[\Delta^n]\to \mathcal{C}^{\mathrm{op}}$. Then there is a bijective correspondence between $n$-simplices of $\mathrm{Un}_\phi F$, and pairs $(s,g)$, where $s\colon \Delta^n\to S$ is a map of simplicial sets, and $$ g\colon D^n\to F\circ \phi_s^{\mathrm{op}} $$ is a map of simplicial functors $\mathfrak{C}[\Delta^n]^{\mathrm{op}}\to \mathrm{Set}_\Delta$.

Here $D^n$ is a particular functor $\mathfrak{C}[\Delta^n]^{\mathrm{op}} \to \mathrm{Set}_\Delta$, defined as follows. Consider the simplicial category $\mathfrak{C}[(\Delta^n)^\rhd]\approx \mathfrak{C}[\Delta^{n+1}]$, which contains $\mathfrak{C}[\Delta^n]$ as a subcategory. Then $D^n$ is the functor represented by the object corresponding to the cone point $v$ of $(\Delta^n)^\rhd$. When you unwind this, you get that $D^n(x)$ is isomorphic to the nerve of a poset: $$ D^n(x) \approx N\bigl\{ S\subseteq \{x,x+1,\dots, n\} \;\bigm|\; x\in S \bigr\} \approx (\Delta^1)^{n-x}. $$ (In fact, $D^n$ is nothing other than the straightening of $(\mathrm{id}\colon \Delta^n\to \Delta^n)$ along $\phi=\mathrm{id}\colon \mathfrak{C}[\Delta^n]^{\mathrm{op}}\to \mathfrak{C}[\Delta^n]^{\mathrm{op}}$.)

In HTT, we are interested in $s\colon \Delta^n\to \{s_0\}\to S$ which factor through a single vertex in $S$, which maps under $\phi$ to some object $C$ in $\mathcal{C}$. In this case $F\circ \phi_s^{\mathrm{op}}$ is a constant simplicial functor $\mathfrak{C}[\Delta^n]^{\mathrm{op}}\to \mathrm{Set}_\Delta$ with value $F(C)$. So natural transformations $D^n\to F\circ \phi_s^{\mathrm{op}}$ are the same as maps of simplicial sets $Q^n\to F(C)$, where $Q^n$ is the enriched left Kan extension of $D^n$ along $\mathfrak{C}[\Delta^n]^{\mathrm{op}}\to \mathfrak{C}[\Delta^0]^{\mathrm{op}}=*$. Unwinding this should yield Lurie's description of $Q^n$, which will be as a quotient of $D^n(0)\approx (\Delta^1)^n$.


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