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Let $B$ be a commutative ring with unity.

By Theorem 4 from Chapter 2 Section 4 of 'Homotopical Algebra' we can impose a model structure on $sMod_{B}$ by declaring a map of simplicial $B$-modules $f:M \rightarrow N$ to be a fibration (resp. a weak equivalence) if $\underline{Hom}(P,M) \rightarrow \underline{Hom}(P,N)$ is a fibration (resp. a weak equivalence) for all projective $B$-modules $P$. Call this model structure MS1

Alternatively, using the forgetful/free adjunction between $sMod_{B}$ and $sSet$, by Theorem 5.1 of Goerss-Jardine, we have another model structure on $sMod_{B}$. Call this model structure MS2.

Now since $B$ is a projective $B$-module, and $\underline{Hom}(B,M) \cong M$ as simplicial sets, it follows that $f:M \rightarrow N$ a fibration/weak equivalence in MS1 implies $f:M \rightarrow N$ is a fibration/weak equivalence in MS2.

Do these model structures coincide? If not, can we exhibit an example highlighting their nonequivalence?

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Yes, MS1=MS2. In MS2, fibrations and weak equivalences are created by $hom(B,-)$. But since fibrations and weak equivalences of simplicial sets are preserved under (arbitrary) products, then you can equivalently create these by the functors $hom(B^{\oplus S},-)$ for all sets $S$. Then, any projective module $P$ is a retract of a free module $B^{\oplus S}$, and thereafter the corepresented functor $hom(P,-)$ is a retract of $hom(B^{\oplus S},-)$. Since fibrations and weak equivalences of simplicial sets are also stable under retracts, it follows that any fibration or weak equivalence in MS2 is necessarily also such in MS1.

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  • $\begingroup$ Why exactly are weak equivalences of simplicial sets preserved under arbitrary products? It's clear that they're preserved by finite products, but I can't think of an argument for arbitrary ones. $\endgroup$ Commented Jul 15, 2018 at 0:18
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    $\begingroup$ Hmm, yeah they're definitely not in general. But they are among Kan complexes, since these are then computing the derived values (via the Quillen adjunction sSet <==> Fun(S,sSet) for an indexing set S). So luckily we're good here, because simplicial abelian groups are all Kan complexes. Whew, that was a close one. Nice catch! $\endgroup$ Commented Jul 16, 2018 at 2:27
  • $\begingroup$ See the comments here for a counterexample to weak equivalences being preserved by arbitrary products: mathoverflow.net/questions/181188/… $\endgroup$ Commented Jul 16, 2018 at 2:29
  • $\begingroup$ Marc Hoyois linked me the same post in the chat room! It’s a good example :) $\endgroup$ Commented Jul 16, 2018 at 2:30
  • $\begingroup$ Ah, sure enough. Yeah, it feels like exactly the minimal example that leverages non-Kan-ness -- something about there not being the composite paths that there ought to be, and so things go screwy when the number of subdivisions of the interval that are needed to represent some path in each factor is unbounded (which is only possible with an infinite product). $\endgroup$ Commented Jul 16, 2018 at 2:37

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