The homotopy category of the category of enriched categories

Question

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.)

Answer 1

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.

Answer 2

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!

Answer 3

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.