We know that if $\mathcal C$ is a combinatorial monoidal model category such that all objects are cofibrant and the class of weak equivalences is stable under filtered colimits, then $\mathsf{Cat}_{\mathcal C}$, the category of all (small) $\mathcal C$-enriched categories, has the categorical model structure. My question is:

Is the homotopy category of $\mathsf{Cat}_{\mathcal C}$, with respect to the categorical model structure, the category of all (small) $\operatorname{Ho}\mathcal C$-enriched categories, $\mathsf{Cat}_{\operatorname{Ho}\mathcal C}$?

In other words, does every $\operatorname{Ho}\mathcal C$-enriched category arises from a $\mathcal C$-enriched category?

(I am asking this question for the following reason. We know that $(\infty,1)$-categories are schematically categories enriched over the category of homotopy types, which is equivalent to $\operatorname{Ho}\mathsf{SSet}$. If the above question is not true for $\mathcal C=\mathsf{SSet}$, then there exists a $\operatorname{Ho}\mathsf{SSet}$-enriched category that does not arise from a simplicial category. This seems weird, since the theory of simplicial categories is a well-accepted model for $(\infty,1)$-categories. Therefore I expect the answer to the above question to be ''Yes'', at least in the case where $\mathcal C=\mathsf{SSet}$. But I am not able to give a proof, nor can I find any references to this question.)

  • $\begingroup$ You also need the model category to have a colimit-preserving tensor product satisfying Quillen's pushout-tensor axiom, or else this makes no sense. $\endgroup$ Commented Nov 18, 2019 at 12:59
  • $\begingroup$ Also, I think the answer is no, since iirc there are monoids in the stable homotopy category that don't lift to E_1-algebras. This gives a counterexample to your question when you take such a homotopy monoid as a one-object category enriched in the stable homotopy category. $\endgroup$ Commented Nov 18, 2019 at 13:02
  • $\begingroup$ @HarryGindi Thans for your example in the second comment. In the first comment, I suspect the definition of a monoidal model category has already included that the tensor product is a Quillen bifunctor? $\endgroup$
    – Frank Kong
    Commented Nov 18, 2019 at 13:11
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    $\begingroup$ To highlight what may be a significant misunderstanding (depending on what you mean by "schematically"), it is really not at all true that $(\infty,1)$-categories should be thought of as categories enriched over $\mathrm{Ho}\mathbf{SSet}$. $\endgroup$ Commented Nov 18, 2019 at 23:02
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    $\begingroup$ To expand on what Kevin is saying, the statement '∞-categories are canonically schematically enriched over homotopy types' refers to being enriched over the ∞-category of homotopy types, not the homotopy category. If you take the Dwyer-Kan localization of sSet rather than the classical Gabriel-Zisman localization, this is a theorem (more or less equivalent to the proof of Quillen equivalence between quasicategories and categories enriched in Kan complexes). $\endgroup$ Commented Nov 19, 2019 at 12:41

3 Answers 3


A recipe for a counterexample:

Let $(X,e:\Delta^0\to X,m:X\times X\to X)$ be monoid object in the homotopy category of spaces $h\mathcal{S}$ (that is, an H-monoid). Note that this is a property of the triple. Then we can form an $h\mathcal{S}$-enriched category $\mathbf{B}X$ with set of objects $\{\ast\}$ and such that $\mathbf{B}X(\ast,\ast)=X,$ with composition given by $$m:X\times X \to X.$$ and unit given by $e:\Delta^0 \to X$ the basepoint of $X$.

In general, an H-monoid structure $(X,e,m)$ only specifies a canonical $A_2$-algebra structure on $X$ with the added stipulation that there exists an $A_3$-structure extending the $A_2$-structure, and there exist many examples (none of which I know!) of $A_2$-structures admitting (possibly many) extensions to an $A_3$ structure, none of which extend to an $A_\infty$-structure. Unfortunately, even having asked some experts, it's not so easy to come up with an example of such a space.

But it is a theorem that every $A_\infty$-space is equivalent to an honest monoid in spaces.

So it suffices to find any H-monoid $(X,e,m)$ in spaces that doesn't admit a lift to an $A_\infty$-monoid, then take $\mathbf{B}X$. This gives an example of an $h\mathcal{S}$-enriched category that cannot lift to an honest $\mathcal{S}$-enriched category, because if it did, it would necessarily specify an $A_\infty$-structure on $X$ lifting $(X,e,m)$.

Edit: Kevin Carlson's comment below gives an example of such spaces from Stasheff and Adams and a source, so check it out!

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    $\begingroup$ It may not be easy to come up with examples, but Stasheff did it (citing some help from Adams) in the paper introducing $A_n$ spaces. According to Theorem 17 of Homotopy Associativity of H-spaces, I, any odd-dimensional Moore space of $\mathbb Z$ localized away from $p$ admits an $A_{p-1}$ structure but no $A_p$ structure. I have never heard of any other examples, though I hope more are known to somebody! $\endgroup$ Commented Nov 18, 2019 at 17:50
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    $\begingroup$ @KevinCarlson Assuming that $p$ must be a prime, that gives an example of a space that admits a (say) $A_4$-structure (hence also an $A_3$-structure) but no $A_5$-structure (hence no $A_\infty$-structure). Is there an example of a space admitting an $A_3$-structure but no $A_4$-structure? $\endgroup$ Commented Nov 19, 2019 at 21:22
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    $\begingroup$ @MikeShulman I believe these are all the examples in the paper, so that one gets $A_4$-but-no-$A_5$ and $A_2$-but-no-$A_3$, but misses $A_3$-but-no-$A_4$. It would certainly be nice to have examples at every level of coherence. $\endgroup$ Commented Nov 20, 2019 at 0:28
  • $\begingroup$ @KevinCarlson Okay. In particular it would be nice to know whether there is an example when $\mathcal{C} = \rm Gpd$ rather than $\rm SSet$, which would necessarily be $A_3$-but-not-$A_4$ since an $A_4$ groupoid is already $A_\infty$. $\endgroup$ Commented Nov 20, 2019 at 6:27
  • $\begingroup$ @MikeShulman Yes, that would be nice. It seems to me that the category whose objects are rooted binary trees and whose morphisms are sequences of moves generated by the associator is an $A_3$ groupoid which manifestly doesn’t satisfy the pentagon identity, so at least an example of an $A_3$ structure not extending to an $A_4$. $\endgroup$ Commented Nov 21, 2019 at 16:00

Summarizing a discussion in the comments on Harry's answer, we can consider the case $\mathcal C=\mathrm{Gpd}$, with the canonical model structure: weak equivalences are just the equivalences of groupoids, cofibrations are functors injective on objects, and fibrations are isofibrations.

A $\mathcal C$-enriched category with one object is then precisely a strictly monoidal groupoid. In contrast, a $\mathrm{Ho}(\mathcal C)$-enriched category with one object is essentially the same thing as an $A_3$-monoidal groupoid-we get a tensor product, which is certainly associative up to non-specified isomorphism, but there is no reason why the associators should satisfy the pentagon identity, and indeed they need not.

For a counterexample, consider the $A_3$-monoidal groupoid $G$ freely generated by a single object $x$. For simplicity, let's also make it strictly unital. Then the objects of $G$ are given by rooted binary trees, corresponding to parenthesizations of finite strings of $x$'es. The morphisms are freely generated by the associator isomorphism between the two rooted binary trees of three leaves, together with the functoriality of the tensor product $\otimes:G\times G\to G$ which fuses two trees by adding a new root to their disjoint union.

It is in fact possible to describe the morphisms in $G$ without reference to the tensor product: they are sequences of elementary moves, where an elementary move out of a binary tree $T$ is given by taking a node $N$ which is the left child of its parent $P$ and replacing $P$'s right child with $N$, $P$'s left child with $N$'s left child, $N$'s left child with its right child, and $N$'s right child with $P$'s right child. Under this description, one can see that the two different paths from $((xx)x)x$ to $x(x(xx))$ around the pentagon are not equal in $G$. (For comparison, the free strictly unital $A_4$-monoidal groupoid on $x$ has exactly one morphism between two binary trees whenever there exists any such sequence of elementary moves transforming one into the other.)

Thus $G$ gives an $A_3$-monoidal groupoid which does not arise from an $A_4$-monoidal one, equivalently, not from a strictly monoidal one.

EDIT: As Harry notes, this is not quite a counterexample to your question, as it's not clear that it's impossible to make a different choice of the associators that would extend to $A_4$. It seems unlikely, but also that it would be complicated to make an elementary argument against it.

Here is a proposed easier example, coming from $\mathcal C=\mathrm{Cat}$ instead of $\mathrm{Gpd}$, again with the canonical model structure. We let $C$ be the free $A_3$-monoidal category generated by a co-pointed object. So as compared to $G$ from above, $C$ contains a map $p:x\to I$ contracting a leaf which freely generates $C$ from $G$ under $\otimes$ and naturality with respect to the elementary moves. I'll refer to any tensor product of $p$'s and identities as a projection. Call the canonical associators, given by elementary moves, $\alpha$; I claim there is no alternative choice $\beta$ of associators for an $A_3$-structure on $C$, so that $C$'s underlying $\mathrm{Ho}(\mathrm{Cat})$-enriched category arises from no monoidal category.

To show that $C$ admits no $A_4$-monoidal structure, note that $\beta_{x,x,x}:(xx)x\to x(xx)$ is uniquely determined as it was in $G$-we've added no new morphisms between isomorphic trees. Furthermore, while morphisms between trees with different numbers of leaves cannot be uniquely written as strings of elementary moves and projections, the only relation on them is naturality of projections with respect to elementary moves. The induced equivalence relation respects lengths of strings of elementary moves and projections, so that any morphism of $C$ has a well defined length given by the length of any representing string of elementary moves and projections.

Then to show for instance that $\beta_{xx,x,x}=\alpha_{xx,x,x}$, we can consider the equality $$(px)(xx)\circ \beta_{xx,x,x}=\beta_{x,x,x}\circ ((px)x)x:((xx)x)x\to x(xx).$$ This implies that $\beta_{xx,x,x}$ must be an elementary move, by computing lengths. And there is a unique elementary move $((xx)x)x\to (xx)(xx)$, namely $\alpha_{xx,x,x}$. Similarly one shows all the associators between four-leaf trees are the canonical ones, so that the pentagon cannot commute.

If I haven't missed anything in this example, it seems as if its nerve should give rise to an $A_3$-but-not-$A_4$ space. I can't tell if it can be made to give such a groupoid, though.

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    $\begingroup$ A comment: The thing you need is that the multiplication and unit (A_2 structure) don't support any A_4 structure but do support an A_3 structure. I'm not sure if this is true in this case, since we might connive to pick isos such that the diagram commutes by some application of the axiom of choice. Can you show that this isn't possible? The Stasheff example establishes this by showing that an A_4 structure exists on a space but no A_5 structure exists whatsoever, not just that no A_5 structure exists extending that A_4 structure. $\endgroup$ Commented Nov 21, 2019 at 22:15
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    $\begingroup$ @HarryGindi Yes, thanks, I let myself slip into talking about a slightly different problem. $\endgroup$ Commented Nov 22, 2019 at 20:21
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    $\begingroup$ @HarryGindi I think adding projections lets us avoid your concern; see my edit. $\endgroup$ Commented Nov 22, 2019 at 21:25

There's a different obstacle; the change-of-enrichment functor $\mathsf{Cat}_{\mathcal C} \to \mathsf{Cat}_{\operatorname{Ho}\mathcal C}$ doesn't invert the weak equivalences! Instead, this is a relative functor, mapping the weak equivalences of the former category into the weak equivalences of the latter category.

One can, however, ask whether the map $\operatorname{Ho}(\mathsf{Cat}_{\mathcal C}) \to \operatorname{Ho}(\mathsf{Cat}_{\operatorname{Ho}\mathcal C})$ is an equivalence, and if not, whether there is a map $\mathsf{Cat}_{\operatorname{Ho}\mathcal C} \to \operatorname{Ho}(\mathsf{Cat}_{\mathcal C})$ or the other way around. I don't have an answer; I've recently become interested in this and related questions myself.

If $\mathcal{C}$ models $\mathcal{G}\!pd_\infty$, I mildly expect $\mathsf{Cat}_{\operatorname{Ho}\mathcal C}$ to be the homotopy category of the full subcategory of $(\mathsf{Set} \downarrow \mathcal{C}\!at_\infty)$ spanned by the essentially surjective functors; i.e. the $\infty$-category of small $\infty$-categories equipped with the choice of a strict 'set of objects'. However, I haven't been able to convince myself that this ought to work out.

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    $\begingroup$ As Harry's answer indicates, an object of $\mathbf{Cat}_{\mathrm{Ho} \mathcal C}$ does not induce an $\infty$-category, regardless of what you're going for with the "set of objects" idea. $\endgroup$ Commented Nov 19, 2019 at 4:54

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