In higher category theory, it is common that a weak structure cannot be strictified in all directions simultaneously. For instance, a monoidal category is not (in general) equivalent to one that is both strict and skeletal, and a tricategory is not (in general) equivalent to one whose units and interchange law are both strict.

Now a model category can be regarded as a particular sort of strictification of an $(\infty,1)$-category. From this perspective, all sorts of questions along the above lines suggest themselves. For concreteness, I'll ask a particular one:

Does there exist a locally presentable $(\infty,1)$-category which (provably) cannot be presented by a model category in which all objects are both fibrant and cofibrant?

But I would be interested in answers to any similar question.

  • 5
    $\begingroup$ Can the $(\infty,1)$-category of $\infty$-groupoids be presented by a model category in which all objects are fibrant-cofibrant? $\endgroup$ – Guillaume Brunerie Apr 29 '12 at 17:33
  • 1
    $\begingroup$ Guillaume: I don't know! I kind of suspect not, but I have no idea how to prove it. $\endgroup$ – Mike Shulman Apr 30 '12 at 3:40
  • $\begingroup$ Under mild conditions at least we can switch from all objects being cofibrant to all objects being fibrant: ncatlab.org/nlab/show/… But do we have many examples at all of model categories where all objects are fibrant and cofibrant? $\endgroup$ – Urs Schreiber May 1 '12 at 11:26
  • $\begingroup$ @Mike: I think that a tricategory is not in general equivalent to one whose interchange law is strict (regardless of whether units are weak or not). The way you phrased things gives me the impression that you believe that there is a strictification result that says that any tricategory is equivalent to one whose interchange law is strict. Do you believe that such a strictification result exists? If yes, do you have a reference? (I strongly believe that such a strictification result does not exist) $\endgroup$ – André Henriques May 1 '12 at 14:44
  • 1
    $\begingroup$ @Urs: no, not many at all! $\endgroup$ – Mike Shulman May 1 '12 at 19:53

Unfortunately I don't have an answer to the actual question. If you ask for more than just a model category I think there are examples:

There is no combinatorial symmetric monoidal simplicial model category $\mathcal{S}$ such that $\mathbf{Ho}(\mathcal{S})$ is the stable homotopy category and every object of $\mathcal{S}$ is both fibrant and cofibrant.

Otherwise the forgetful functor $\mathcal{S} \rightarrow \mathbf{sSet}$ given by homming out of the unit object gives a symmetric monoidal Quillen adjunction which seems to have "too many good properties." Here is a sketch:

As a right Quillen adjoint it commutes with the formation of loop spaces (up to equivalence), and it preserves all weak equivalences, hence it should factor through the zero-space functor from $\Omega$-prespectra to $\mathbf{sSet}$.

Since everything is fibrant, it should be possible to transfer the model structure on $\mathcal{S}$ to commutative algebras in $\mathcal{S}$ (use combinatorial for this).

The above two facts taken together contradict Remark 11.2 of the paper

May, J. P. What precisely are $E_{\infty}$ ring spaces and $E_{\infty}$ ring spectra? New topological contexts for Galois theory and algebraic geometry (BIRS 2008), 215–282

where this is deduced from a result due to Lewis (Theorem 11.1 in the above paper).

  • $\begingroup$ Thanks! I hadn't thought of it, but you're right that Lewis' no-go theorem is definitely in this vein. $\endgroup$ – Mike Shulman May 1 '12 at 19:46
  • $\begingroup$ Now that I think about it, combinatorial is probably not needed here because we already assume that the unit is both fibrant and cofibrant. In any case, given the above I would be even more surprised if there were a model category of spectra (instead of spaces) such that all objects are fibrant and cofibrant. $\endgroup$ – Daniel Schäppi May 2 '12 at 1:16

Here is another answer that involves adding extra properties. If we have a model category which

  • is locally cartesian closed, as a category (such as if it is a presheaf category)
  • has its cofibrations being the monomorphisms (hence in particular all objects are cofibrant)
  • is right proper (such as if all objects are fibrant)

then pullback along a fibration $g\colon A\to B$ preserves both cofibrations and acyclic cofibrations, and so the adjunction $g^* \dashv \Pi_g$ is Quillen. Therefore, the $(\infty,1)$-category presented by this model category is locally cartesian closed.

Thus, an $(\infty,1)$-category which is not locally cartesian closed cannot be presented by a model category with all three of the above properties.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.