The simpler question is to study the 2-groupoid $\mathrm{Aut}(\mathsf{Cat})$ of all autoequivalences of the category of (small) categories (i.e. the groupoid of equivalences of categories $\mathsf{Cat} \to \mathsf{Cat}$). This is equivalent to $\mathbb{Z}/2$ considered as a locally discrete 2-groupoid: the terminal object $\mathbb{1}$ is fixed up to unique isomorphism, and the arrow category $\mathbb{2}$ is the unique minimal generator (i.e. it is a generator and no proper subobject is a generator) so also fixed up to isomorphism. Since every category is functorially a colimit of copies of $\mathbb{2}$, once we decide whether our autoequivalence fixes or swaps the two maps $\mathbb{1} \to \mathbb{2}$ we've determined the autoequivalence up to a canonical isomorphism. And of course the nontrivial autoequivalence is the "opposite category" functor $\mathcal{C} \mapsto \mathcal{C}^\mathrm{op}$.

This is a somewhat artificial question, since an autoequivalence of the category $\mathsf{Cat}$ must respect the notion of isomorphism of categories, whereas all we should really want to respect is equivalence of categories. One way to fix this would be to consider $\mathsf{Cat}$ as a 2-category, and ask about the 3-groupoid of autobiequivalences of $\mathsf{Cat}$. But this starts to involve heavy machinery, and the problem will only get worse if we want to go up the categorical ladder. Hence a more "homotopical" approach:

Question: What is the biequivalence type of the 2-groupoid $\mathrm{Aut}(\mathrm{ho}\mathsf{Cat})$ of autoequivalences of $\mathrm{ho}\mathsf{Cat}$, the category of small categories localized at the equivalences of categories? Equivalently, $\mathrm{ho}\mathsf{Cat}$ is the category of small categories modulo the congruence identifying naturally isomorphic functors. In particular, is $\mathcal{C} \mapsto \mathcal{C}^\mathrm{op}$ isomorphic to the identity (presumably the answer is no)? Are there any other autoequivalences, essentially (probably not, but you never know)?

A variant would be to ask about the 2-groupoid of functors $\mathsf{Cat} \to \mathsf{Cat}$ which preserve equivalences of categories and induce equivalences on $\mathrm{ho}\mathsf{Cat}$.

I'm aware that Toën showed in some sense that the $(\infty,1)$-category of $(\infty,1)$ categories has an automorphism group of $\mathbb{Z}/2$, and Barwick and Schommer-Pries generalized this to $(\infty,n)$-categories for larger $n$. This question is meant to be an easier / more elementary warmup to those questions.

The approach that worked for $\mathsf{Cat}$ certainly isn't going to work for $\mathrm{Ho}\mathsf{Cat}$: Freyd showed ("On the concreteness of certain categories." Symposia Mathematica. Vol. 4. 1968-9, pp. 431-56) that $\mathrm{Ho}\mathsf{Cat}$ is not concrete, so it certainly can't have a generating object, never mind a colimit-dense object.

By playing around, I've convinced myself at least that the indiscrete categories (the equivalence class of the terminal category) can be characterized in $\mathrm{Ho}\mathsf{Cat}$ by the fact that all functors into it are isomorphic (consider functors from the walking equalizer diagram to see that such a category is a preorder and functors from the terminal category to see that all objects are isomorphic). So $\mathrm{Ho}\mathsf{Cat}(\mathbb{1},-)$ tells us how many isomorphism classes of objects a category has, and what isomorphism classes of functors are doing to them. But that's about all I've got.

  • $\begingroup$ Here is an attempted (but not careful) argument. You might be able to determine this by choosing a skeleton for ho(Cat) (as a 1-category): start with the subcategory of skeletal 1-categories, where isomorphic objects are equal and natural equivalences of categories are also isomorphisms; then choose one representative for each isomorphism class of skeletal category. I suspect that the argument that you give calculates the self-equivalences of sk(ho(Cat)) as Z/2. $\endgroup$ Nov 12, 2015 at 18:11
  • $\begingroup$ The trouble is that the skeleton of ho(Cat) still looks quite different from the skeleton of Cat, because isomorphic functors have been identified. For example, instead of containing the category of groups as a subcategory, we now have groups-and-conjugacy-classes-of-functions as a subcategory. More to the point, the characterization I used of the arrow category $\mathbb{2}$ (as a minimal generator) no longer applies to ho(Cat), skeletal or no: ho(Cat) doesn't even have a generator. Plus, colimits in ho(Cat) are hopeless, so that part of the argument isn't going to carry over either. $\endgroup$
    – Tim Campion
    Nov 12, 2015 at 19:13
  • 3
    $\begingroup$ Here are some thoughts: Since the equivalence class of the terminal category is preserved, we can show that the collection of categories equivalent to posets is preserved. (They are the categories P such that $hoCat(D,P) \to Set( hoCat(pt, D), hoCat(pt, P))$ is injective for all cats D). Now hoPoset = Poset only has $\mathbb{Z}/2$ many automorphisms. WLOG we can assume our functor is the identity on Poset. So then the free-walking arrow is preserved, identically. From this it follows that the class of groupoids is preserved (as well as the connected groupoids which I will call "groups"). cont $\endgroup$ Nov 12, 2015 at 21:03
  • 3
    $\begingroup$ Then from this we see that the class of what Clark and I call the "gaunt" categories is preserved since those are the categories such that $hoCat(G,C) = hoCat(pt,C)$ for all groups G. Again we have hoGaunt = Gaunt, has only two automorphisms and we may assume WLOG that it is the identity automorphism on Gaunt. Using this I think I can show that for any category C we have that C and F(C) have equivalent presentations (free categories on directed graphs are gaunt). So I think any such functor must be the identity on objects of hoCat. It could a priori still be interesting on morphisms? $\endgroup$ Nov 12, 2015 at 21:09
  • 1
    $\begingroup$ @ChrisSchommer-Pries Thank you, this is really neat! What I don't understand is the "presentation" bit. Every category is a coequalizer of a fork between two free categories: the fork must be fixed by the autoequivalence, but why should the coequalizer of the fork be fixed? If the coequalizer in $\mathsf{Cat}$ descends to a coequalizer in $\mathrm{ho}\mathsf{Cat}$, then it follows -- is that the idea? $\endgroup$
    – Tim Campion
    Nov 12, 2015 at 22:15

1 Answer 1


$\newcommand{\hoCat}{\mathrm{ho}\mathsf{Cat}} \newcommand{\Cat}{\mathsf{Cat}} \newcommand{\Gaunt}{\mathsf{Gaunt}} \newcommand{\hoGaunt}{\mathrm{ho}\mathsf{Gaunt}} \newcommand{\Free}{\mathsf{Free}} \newcommand{\hoFree}{\mathrm{ho}\mathsf{Free}} \newcommand{\comp}{\mathrm{comp}} \newcommand{\inv}{^{-1}} \require{AMScd}$

Chris Schommer-Pries' argument in the comments, can be beefed up to show that indeed, essentially the only autoequivalences of $\hoCat$ are the identity and taking the opposite category.

Summary. Use Chris's argument to show that an autoequivalence $F: \hoCat \to \hoCat$ restricts to the identity on the free categories $\Free = \hoFree$ (possibly after composing with the known autoequivalence $(-)^\mathrm{op}: \hoCat \to \hoCat$). As Chris then argues, $\mathsf{Free}$ forms a (large) regular generator in $\hoCat$, and it follows that $F$ is essentially the identity on objects. We just need the slightly stronger statement that $\Free$ is a (large) dense generator to show that $F$ is also the identity on morphisms. It turns out that the $\Cat$-functoriality of the presentations of categories is enough to guarantee this, even though these presentations are not actually functorial on $\hoCat$. In fact, this argument generalizes to show that if the category of algebras for a monad is modded out by a congruence relation, then the free algebras still form a dense generator in the resulting category $\widetilde{\mathcal{C}^T}$ if the canonical coequalizer diagrams descend.


  • As I said in the question statement, the terminal category $\mathrm{pt}$ remains terminal in $\hoCat$, so it is preserved by any autoequivalence. My argument in the question statement is overly complicated: it's clear that when modding out by a congruence, the terminal object remains terminal.

  • As Chris argues, the subcategory $\mathrm{ho}\mathsf{Pre}$ of preorders is preserved by any autoequivalence because it is the class of categories $P$ with $\hoCat(D,P) \to \mathsf{Set}(\hoCat(\mathrm{pt},D),\hoCat(\mathrm{pt},P))$ injective for every category $D$. One direction is clear, the other comes from taking $D$ to be the walking equalizer diagram. Now, since $\mathrm{ho}\mathsf{Pre} = \mathrm{ho}\mathsf{Pos} = \mathsf{Pos}$, our autoequivalence of $\hoCat$ restricts to either the identity or opposite on $\mathsf{Pos}$ by the same argument I gave in the question statement characterizing the autoequivalences of $\Cat$. After composing with the "opposite" autoequivalence of $\hoCat$, we can assume our autoequivalence is the identity on $\mathsf{Pos}$.

  • As Chris argues, in particular the walking arrow $\mathbb{2}$ is fixed. So the groupoids are also fixed: they are characterized as the categories $G$ such that every functor $\mathbb{2} \to G$ extends up to natural isomorphism along the functor $\mathbb{2} \to \mathrm{pt}$. The connected groupoids are also fixed because $\hoCat(\mathrm{pt},-)$ tells us how many isomorphism classes a category has. Note that we don't know the autoequivalence fixes each individual groupoid, but only that the class of groupoids is preserved, but this will be good enough for the next step.

  • As Chris argues, the gaunt categories $\hoGaunt$ are fixed because they are the categories such that every functor from a connected groupoid extends along the functor to $\mathrm{pt}$. We have $\Gaunt = \hoGaunt$, and again the argument given in the question statement for $\mathsf{Cat}$ shows that the autoequivalence restricts to the identity on $\Gaunt$. In particular, it's the identity on the subcategory $\Free$ of free categories.

  • As Chris argues, the $\Cat$ - coequalizer $P^2X \stackrel{\to}{\to} PX \overset{\comp_X}{\to} X$ descends to $\hoCat$ (in particular, $\Free$ is a regular generator in $\hoCat$). Here $P$ is the "path" comonad on $\Cat$, induced by the forgetful functor to graphs, and the coequalizer is the canonical one: one leg $\mu_X$ de-parenthesizes a path of paths, while the other $P\comp_X$ composes each individual path in a path of paths; the coequalizer $\comp_X$ composes a path in $X$. The reason the coequalizer descends is that if $F: PX \to Y$ coequalizes $\mu_X$ and $P\comp_X$ up to a natural isomorphism $\alpha$, then for every path $\gamma_1, \dots, \gamma_k$ in $X$ we have $F(\gamma_k \circ \cdots \circ \gamma_1) = \alpha \circ F(\gamma_k) \circ \cdots \circ F(\gamma_1) \circ \alpha\inv$. But in the unary case, this tells us that $F(\gamma_i) = \alpha \circ F(\gamma_i) \circ \alpha\inv$. So we can commute the $\alpha$ down to the end to get $F(\gamma_k \circ \cdots \circ \gamma_1) = F(\gamma_k) \circ \cdots \circ F(\gamma_1) \circ \alpha \circ \alpha\inv = F(\gamma_k) \circ \cdots \circ F(\gamma_1)$. That is $F$ already coequalizes the fork strictly, so it descends to a functor $X \to Y$. Uniqueness of the factorization in $\hoCat$ is easy to see.

  • Let $i_\Free: \Free \to \hoCat$ be the inclusion functor. Let $\alpha: \hoCat(i_\Free - , X) \implies \hoCat(i_\Free - , Y)$ be a natural transformation. To show that $\Free$ is dense in $\hoCat$, we just need to show that $\alpha$ is induced by some $\bar\alpha: X \to Y$ (we know that such a $\bar\alpha$ is unique because we've shown that $\Free$ is a regular generator, and hence a generator, in $\hoCat$). There is an obvious candidate: by naturality $\alpha_{PX}(\comp): PX \to Y$ coequalizes the two functors $P^2X \to PX$ up to natural isomorphism, so by the previous part it descends to a functor $\bar\alpha: X \to Y$. So much is true of any regular generator; the trick is to see that $\bar\alpha$ really induces the natural transformation $\alpha$, i.e. that $\bar\alpha \circ f \cong \alpha_A(f)$ for any $f: A \to X$ with $A \in \Free$, leading to the following diagram: $$ \begin{CD} PA @>\comp_A>> A;\\ @VPfVV @VfVV \searrow^{\alpha_A(f)} \\ PX @>\comp_X>> X @>\bar\alpha>> Y; \end{CD} $$ Since the square commutes and the outside commutes up to natural isomorphism and $\comp_A$ is epi in $\hoCat$, the triangle also commutes up to natural isomorphism as desired.

The principle used to analyze the auotequivalences of $\Cat$ in the question statement, and then $\mathsf{Pos}$, and then $\Gaunt$, and finally $\hoCat$, is that an equivalence (or any colimit-preserving functor) is determined up to natural isomorphism by its action on a dense generator. One way to see this is that a dense generator $\mathcal{A} \subseteq \mathcal{C}$ embeds $\mathcal{C}$ as a full subcategory of $\mathrm{Fun}(\mathcal{A}^\mathrm{op},\mathsf{Set})$ containing $\mathcal{A}$ as the representables. Each object is (canonically) a colimit of representables, and each morphism is (canonically) a colimit of maps between representables, so the value of the functor on them is determined by colimit preservation and where it sends the representables.


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.