On the one hand, in their paper Simplicial structures on model categories and functors, Rezk, Schwede and Shipley proved that a simplicial model category structure on a given model category is unique up to simplicial Quillen equivalence.

On the other hand, we know that every model category can be simplicially enriched (for instance taking a cosimplicial resolution of the source or a simplicial one of the target), but in a way which does not give in every case a honest simplicial model category, but simplicial mapping spaces with good homotopy invariance properties and a composition defined up to homotopy (see for instance the category of chain complexes over a field).

Now, let us consider a model category $M$. Suppose that $M$ is equipped with two simplicial mapping space functors $Map(-,-)$ and $\underline{Map}(-,-)$, both homotopy invariant under weak equivalences of a cofibrant source or a fibrant target, and with composition defined at least up to homotopy (so they both induce mapping spaces with a well defined composition on the homotopy category $Ho(M)$ of $M$). Suppose moreover that we have $\pi_0Map(X,Y)\cong[X,Y]\cong\pi_0\underline{Map}(X,Y)$ where $X$ is a cofibrant object of $M$, $Y$ a fibrant object and $[-,-]$ denotes the set of homotopy classes. Do we then have $\pi_nMap(X,Y)\cong\pi_n\underline{Map}(X,Y)$ for every $n$ ?

A more general idea underlying my question is that I wonder if a result similar to the result of Rezk, Schwede and Shipley, in a weaker version, could hold under weaker assumptions (especially, when axiom SM7 of simplicial model categories is not fulfilled anymore).

  • $\begingroup$ Welcome to MathOverflow! I enjoyed your paper on PROPs from last year $\endgroup$ Jul 8, 2013 at 17:01
  • $\begingroup$ Hovey's book highlights some of the difficulties when SM7 fails. In particular, not knowing the answer to Conjecture 5.6.6 seems to cause problems, and on page 144 he mentions that Ho(M) is a central Ho(sSet)-algebra (with the action given by framing) iff conjecture 5.6.6 holds. So what you're asking may be equivalent to this old open problem but I'm not sure. Perhaps someone can find a proof of unicity which doesn't make use of this centrality $\endgroup$ Jul 8, 2013 at 17:09
  • $\begingroup$ Thank you for your welcome and your indications ! It seems to me quite strange and unlikely that a model category could give rise to two non-equivalent higher homotopy theories, precisely two non-equivalent $(\infty,1)$-categories with the same underlying homotopy category, but I neither see a proof of unicity up to homotopy nor a counter-example. $\endgroup$ Jul 8, 2013 at 21:17
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    $\begingroup$ Surely, you need to impose more conditions on your homotopy mapping spaces. For now all your requirements are satisfied by homotopy discrete mapping spaces with prescribed $\pi_0$. $\endgroup$ Jul 9, 2013 at 5:38
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    $\begingroup$ @David: Hovey's coherence conjectures (numbered 5.6.6 and 5.7.5 in his book on model categories) are proved in my paper "Propriétés universelles et extensions de Kan dérivées", Theory and Applications of Categories 20 (2008), 605-649. Scherer and Chacholski also proved it independently in their paper "Representations of spaces", Algebraic & Geometric Topology 8 (2008) 245–278. The paper of Scherer and Chacholski also gives an alternative to my answer below. $\endgroup$ Jul 18, 2013 at 20:21

1 Answer 1


If, given any fixed cofibrant object $A$, there is a funtor $map(A,-)$ from $M$ to simplicial sets which preserves weak equivalences between fibrant objects and commutes with homotopy limits up to canonical weak equivalences (e.g. $map(A,-)$ is a right Quillen functor), and such that $\pi_0(map(A,X))=[A,X]$ (functorially) for any fibrant object $X$, then, for any fibrant object $X$, we have a canonical isomorphism $map(A,X)\simeq Map(A,X)$ in the homotopy category of simplicial sets, where $Map(A,X)$ denotes the usual simplicial enrichment. This is because both $map(A,X)$ and $Map(A,X)$ must represent the same presheaf. For (a more precise statement and) a proof, see for instance Remark 6.14 in this paper (but this is in fact a very formal and easy fact which can be found in many ways and disguises in the literature).

  • $\begingroup$ Beautiful answer! $\endgroup$ Jul 15, 2013 at 6:22

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