# stable (?) model category of simplicial monoids

If $\mathcal{C}$ is the category of commutative unitary monoids, one can endow the category of simplicial objects in $\mathcal{C}$, $s\mathcal{C}$, with the structure of a cofibrantly generated model category, in which fibrations are those maps inducing fibrations of the underlying simplicial sets (on the other hand, weak equivalences are possibly not precisely those maps inducing weak equivalences on underlying simplicial sets, as commutative monoids are not always Kan fibrant).

Is the homotopy category $\text{Ho} \mathcal({C})$ triangulated?

Perhaps it is pre-triangulated, at best, but the suspension functor is not necessarily an equivalence?

• You haven't given enough information to specify your model category- you can't give us only the fibrations, and from your parenthetical it sounds like you aren't just lifting the model structure from simplicial sets (otherwise equivalences would be detected below, independent of fibrancy.) – Dylan Wilson Aug 26 '17 at 23:19
• And I doubt that any natural choice of model structure that deserves to be called 'a homotopy theory of commutative monoids' would give you an additive homotopy category, let alone a triangulated one. And for sure you wouldn't be able to desuspend- which is already part of the definition of "pre-triangulated". I think what you'll get for any reasonable choice is a semi-additive homotopy category, and not really anything more. – Dylan Wilson Aug 26 '17 at 23:21
• (I mean, you could cheat and put a group completion in the definition of weak equivalence, and then you'd get simplicial abelian groups...and that'd give you the connective part of a triangulated category with a t-structure, which is probably the best you can get.) – Dylan Wilson Aug 26 '17 at 23:54
• $R$ a ring. In $R$-algebras and $R$-modules, effective epimorphisms are precisely the surjections on underlying sets, so one can put the model structures on rings given by declaring a map of rings $f$ to be "blah" iff for any projective $P$, $\text{Hom}(P, f)$ is "blah" as a map of simplicial sets. Then I'd show weak equivalences can be detected down below, as follows. – user97068 Aug 27 '17 at 0:00
• Any $R$-module is an abelian group. Any simplicial set is cofibrant. Any simplicial $R$-module, resp. $R$-algebra, $X$, has a cofibrant underlying simplicial set, because simplicial abelian groups are Kan fibrant. Then $X$ is fibrant-cofibrant. So map that induces weak equivalence on underlying simplicial sets, induces homotopy equivalence on underlying simplicial sets. Since homotopy equivalences are closed under arbitrary products, and since the forgetful functor to sets commutes with products, we're done because we can test "being a weak equivalence" on $P$ free. Converse true too and clear – user97068 Aug 27 '17 at 0:01