I've been trying to learn some of the basic language of infinity-category theory (in the sense of Lurie), and in particular, to understand which basic statements in (1-)categories have analogues in the infinity-categorical setting. Specifically, I'm trying to understand how equivalences behave in infinity categories, motivated by the following question.

Vague Question: in an infinity-category $\mathcal C$, to what extent is it true that a quasi-inverse to a morphism $f$, if it exists, is unique up to a contractible space of choices?

Here is how I believe this question should be made precise. If we let $I$ denote the nerve of the undirected interval category (the category with two objects and exactly one morphism between each pair of objects), then there is an inclusion $\Delta^1\hookrightarrow I$ coming from the inclusion of the directed interval category in the undirected interval category. Thus we have a restriction map $$\mathcal C^I\rightarrow\mathcal C^{\Delta^1}$$ of simplicial sets (in fact a functor of infinity-categories which is a fibration for the Joyal model structure), where informally the left-hand side is the infinity-category of quasi-inverse pairs $(g,h)$ of morphisms in $\mathcal C$ (with a family of homotopies realising their quasi-inverseness), the right-hand side is the infinity-category of arrows in $\mathcal C$, and the functor takes a pair $(g,h)$ to its first component $g$.

Precise Question: if $f$ is a morphism in an infinity-category $\mathcal C$, is it true that the fibre of the map $\mathcal C^I\rightarrow\mathcal C^{\Delta^1}$ over the point $f$ of $\mathcal C^{\Delta^1}$ is either empty or a contractible Kan complex?

I am interested both in answers to the precise question and the vague questions above, especially if there are alternative precise formulations which make the answer clearer.


The condition that $f$ admit a quasi-inverse should be be equivalent to the assertion that it be an isomorphism in the homotopy category of $\mathcal C$. This is asserted in this nlab page, and something similar in the language of topological categories is proved in the setting of topological categories in Proposition of Higher Topos Theory.


2 Answers 2


A possibly simpler way of proving what you are after is using marked simplicial set.

Recall that marked simplicial sets are pairs $(X,S)$ where $X$ is a simplicial set and $S\subseteq X_1$ is a set of 1-simplices of $X$ containing all degenerate 1-simplices. If $X$ is a simplicial set we will denote the minimal and maximal marking by $X^\flat$ and $X^\sharp$ respectively

If $X,Y$ are two marked simplicial sets, we can form additional simplicial sets $\mathrm{Map}^\flat(X,Y)$ and $\mathrm{Map}^\sharp(X,Y)$ by $$ \mathrm{Hom}_{\mathrm{sSet}}(K,\mathrm{Map}^\flat(X,Y))=\mathrm{Hom}_{\mathrm{sSet}^+}(K^\flat\times X,Y)\,,$$ $$ \mathrm{Hom}_{\mathrm{sSet}}(K,\mathrm{Map}^\sharp(X,Y))=\mathrm{Hom}_{\mathrm{sSet}^+}(K^\sharp\times X,Y)\,.$$

In HTT. a simplicial model structure is constructed on the category of marked simplicial sets such that the fibrant objects are precisely the $\infty$-categories with the equivalences marked.

The important part here will be that

  • For any anodyne morphism of simplicial sets $A\to B$ the map $A^\sharp\to B^\sharp$ is a marked trivial cofibration. This is because of the definition of marked trivial cofibration (HTT. and the fact that for any simplicial set $A$ and every $\infty$-category $C$ $$ \mathrm{Map}^\sharp(A^\sharp,C)=\mathrm{Map}(A,\mathrm{Core}(C))$$

  • If $f:X\to Y$ is a marked trivial cofibration and $C$ is a fibrant object (i.e. an $\infty$-category with the equivalences marked), then the map $$f^*:\mathrm{Map}^\flat(B,C)\to \mathrm{Map}^\flat(A,C)$$ is a trivial fibration (HTT.

Then we can factorize the arrow you want to study as $$C^J=\mathrm{Map}^\flat(J^\sharp,C)\to \mathrm{Map}^\flat((\Delta^1)^\sharp, C)\to \mathrm{Map}^\flat((\Delta^1)^\flat,C)=C^{\Delta^1}$$ The first arrow is a trivial fibration and the second is precisely the inclusion of the subcategory of $C^{\Delta^1}$ spanned by the equivalences.

  • $\begingroup$ This is a very neat argument! In your final line, when you assert that $\mathrm{Map}^\flat(J^\sharp,C)\rightarrow\mathrm{Map}^\flat((\Delta^1)^\sharp,C)$ is a trivial fibration, you mean for the Joyal model structure, right? $\endgroup$ Jul 2, 2018 at 15:42
  • $\begingroup$ @AlexanderBetts Trivial fibrations for the Kan or the Joyal model structure are the same thing (since the cofibrations are the same). They are Kan fibrations with contractible fibers $\endgroup$ Jul 2, 2018 at 16:14
  • $\begingroup$ @AlexanderBetts Essentially the result you want can be stated as saying that $(\Delta^1)^\sharp\to I^\sharp$ is a marked anodyne morphism. I am pretty sure there must be a more direct proof of that, but I don't see it. $\endgroup$ Jul 2, 2018 at 16:53
  • 1
    $\begingroup$ The map $j : (Δ^1)^{\sharp} \to J^{\sharp}$ is (marked anodyne because it is) a transfinite composition of pushouts of maps of the form HTT. followed by a pushout of a map of the form HTT. $\endgroup$ Jul 23, 2018 at 10:37
  • $\begingroup$ @DanielGerigk I suspected as much, but I don't quite see how to get the sequence of pushouts. Would you like to write a little more details (either in comments or in a new answer)? $\endgroup$ Jul 23, 2018 at 11:08

If $C$ is a category, its core is the subgroupoid consisting of the isomorphisms of $C$. This generalizes; if $\mathcal{C}$ is an $\infty$-category, we define its core is the $\infty$-subgroupoid consisting only of equivalences.

If $\mathcal{C}$ is a quasi-category, define $\mathrm{Core}(\mathcal{C})$ by the following pullback of simplicial sets:

$$ \require{AMScd} \begin{CD} \mathrm{Core}(\mathcal{C}) @>>> \mathcal{C} \\ @VVV @VVV \\ \mathbf{N}(\mathrm{Core}(\mathrm{h}\mathcal{C})) @>>> \mathbf{N}(\mathrm{h}\mathcal{C}) \end{CD} $$ The bottom horizontal map is an inner fibration because it is a functor between (nerves of) categories (introduction to HTT section 2.3). I think you can show the right vertical map is an inner fibration as well. Therefore, $\mathrm{Core}(\mathcal{C})$ is a quasi-category. (and furthermore, this diagram computes a pullback in the $\infty$-category of $\infty$-categories)

HTT propositions and together imply that $\mathrm{Core}(\mathcal{C})$ is the maximal Kan subcomplex of $\mathcal{C}$. And by construction, we can see a morphism of $\mathcal{C}$ is in $\mathrm{Core}(\mathcal{C})$ if and only if it is an isomorphism in the homotopy category.

(the remarks following the proof of state this construction is actually right adjoint to the inclusion $\mathbf{Kan} \to \mathbf{QuasiCat}$)

Since $\Delta^1 \to I$ is a Kan equivalence, it follows that there is an equivalence of $\infty$-groupoids $$ \mathrm{Core}(\mathcal{C})^I \to \mathrm{Core}(\mathcal{C})^{\Delta^1} $$ and the objects of $\mathrm{Core}(\mathcal{C})^{\Delta^1} \subseteq \mathcal{C}^{\Delta^1}$ are precisely functors mapping the arrow of $\Delta^1$ to an isomorphism of $\mathrm{h}\mathcal{C}$.

  • $\begingroup$ I feel like I'm missing something in the above argument, but I can't put my finger on it. $\endgroup$
    – user13113
    Jul 2, 2018 at 14:16
  • $\begingroup$ I think this line of argument can be completed by observing/checking that the obvious square with vertices $\mathrm{Core}(\mathcal C)^I$, $\mathrm{Core}(\mathcal C)^{\Delta^1}$, $\mathcal C^I$ and $\mathcal C^{\Delta^1}$ is a pullback of simplicial sets. Thus if $f$ lies in $\mathrm{Core}(\mathcal C)$, the fibre of $\mathcal C^I\rightarrow\mathcal C^{\Delta^1}$ over $f$ is the same as the fibre of $\mathrm{Core}(\mathcal C)^I\rightarrow\mathrm{Core}(\mathcal C)^{\Delta^1}$ over $f$. It is thus a contractible Kan complex as the latter map is a trivial Kan fibration. $\endgroup$ Jul 3, 2018 at 8:25
  • $\begingroup$ For clarity, I feel like this argument doesn't come to a precise conclusion at the end -- the statement I think I can extract from the last line is that an equivalence $f$ can be viewed as an object of $\mathrm{Core}(\mathcal C)^{\Delta^1}$, and the fibre of $\mathrm{Core}(\mathcal C)^I\rightarrow\mathrm{Core}(\mathcal C)^{\Delta^1}$ over $f$ is a contractible Kan complex. This definitely answers the vague version of the question, but a little extra argument is needed to answer the precise version (e.g. as in my above comment). $\endgroup$ Jul 3, 2018 at 11:41
  • $\begingroup$ P.S. As I'm new to MO, I'm not sure what the accepted thing to do is: should I add an extra argument to this answer to link back to the precise question, or should I write this in a separate answer? I'm happy to follow Hurkyl's lead on this, and do whichever they would prefer! $\endgroup$ Jul 3, 2018 at 11:48

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