The ordinary homotopy groups of a space $X$ are the homotopy groups of the corresponding singular simplicial set $Sing(X)$. The ordinary homology groups of $X$ are the homotopy groups of the simplicial set $F(Sing(X))$, where $F$ is the functor that replaces each set $Sing(X)_n$ with the free abelian group on that set. There should obviously be some nice property of $F$ that makes the Hurewicz theorem work, but all the proofs I can find in the literature do things entirely by hand at the level of checking individual maps and homotopies to $X$. Is this nice property known? Is there a satisfying answer?
[Math] An abstract nonsense proof of the Hurewicz theorem
homotopy-theorysimplicial-stuff
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There are maps $|Sing(X)| \to |\underline{Sing}(X)| \to X$ which realize to weak homotopy equivalences. The inclusion of the n-skeleton $|Sing(X)|^{(n)}| \to |Sing(X)|$ is n-connected, because this is always true for CW-complexes, and so the map $|Sing(X)|^{(n)} \to X$ is n-connected. You can't really do any better than this estimate because the n-skeleton has zero homology groups in degrees above n.
The simplicial space $\underline{Sing}(X)$ contains the sub-simplicial space of constant simplices $\Delta^n \to X$. This is homeomorphic to $X$ itself and so, if we write $cX$ for the constant simplicial space with value $X$, we get a map $cX \to \underline{Sing}(X)$. This inclusion $X \to \underline{map}(\Delta^n,X)$ is a homotopy equivalence because the simplex is contractible, so this map of simplicial spaces is levelwise a weak equivalence. The geometric realization of $cX$ is $X$ itself, and so is its n-skeleton for all n. An excision argument (which takes some work) will show that the same is true for the simplicial space (at least under good conditions), and so each of the skeleta of $|\underline{Sing}(X)|$ is homotopy equivalent to $X$.
Therefore the map on n-skeleta is n-connected. I realize that this is the "compare with $X$" game that you mentioned, but my point is that because the simplicial space $\underline{Sing}(X)$ is homotopically constant the comparison $|Sing(X)| \to |\underline{Sing}(X)|$ really is comparing with $X$. From the point of view of homotopy theory it's not arising from good levelwise structure of the map at all, but comes from the simplicial assemblage.
I don't know whether this helps your generalization.
So the short answer is that there is not such a model structure. The difficulty arises in trying to show that the class of weak equivalences has all of the necessary properties; in particular, even two-of-three does not hold for the naive definition. The first difficulty arises even before that: on ordinary simplicial sets we can arrange for a model of every set that is "minimal" on the $\pi_0$-level, meaning that every connected component has exactly one $0$-simplex. In simplicial commutative monoids we can no longer do this. However, we could assume that in order to be a weak equivalence we need to be a $\pi_*$-isomorphism when choosing any (coherently chosen) basepoints.
For the purposes of our discussion we are going to assume that $\pi_*$-equivalences use the model $S^n = \Delta^n/\partial\Delta^n$. (This is the model that most closely mimicks the boundary maps in the Dold-Kan correspondence.) Now let $X = S^2$, and let $Y$ be $S^2$ with an extra $0$-simplex connected by a $1$-simplex to the original basepoint. (So it looks like a balloon on a string.) We define a map $X\rightarrow Y$ to be the inclusion of $S^2$ in the obvious manner, and a map $Y\rightarrow X$ to be collapsing the extra $1$-simplex back down. Then the composition of these two maps is the identity on $X$, so obviously a weak equivalence. The map $X\rightarrow Y$ is also a weak equivalence, because adding the "string" can't add any new homotopy groups to $X$. However, the map $Y\rightarrow X$ is not a weak equivalence, as $\pi_2Y$ based at the extra point is a one-point set but $\pi_2X$ at its image is a two-point set.
The problem arose because in order to show that $\pi_*$ was invariant of basepoint in the usual Kan complex model we needed to be able to "pull back" simplices along paths in the simplicial set, which used the Kan condition. The new model does not have such a condition, and thus we can't necessarily pull things back.
Another observation along these lines. Take any connected simplicial set $X$, and let $Y$ be $X$ with a "string" added to it at any basepoint. Then $*\rightarrow Y$ (including into the new point) is a weak equivalence, and $X\rightarrow Y$ (including into itself) is a weak equivalence. Thus in the homotopy category, $X$ is isomorphic to a point (and thus the homotopy category is just the category of sets) ... which is presumably not desired.
-- The Bourbon seminar
Best Answer
I’d argue that it boils down to the generator $S^n\to K(\mathbb{Z},n)$ being an $(n+1)$-equivalence.
More detail: If you take the represented version of homology, it is given by $$ H_n(X;\mathbb{Z}) \cong [ S^{n+t} , X\wedge K(\mathbb{Z}, t)] $$ for $t$ large. Then the Hurewicz map is the map induced by the generator $g:S^t \to K(\mathbb{Z}, t)$: $$ \mathrm{H}: [ S^{n+t} , X\wedge S^k] \xrightarrow{(\mathrm{id}_X\wedge g)_*} [ S^{n+t} , X\wedge K(\mathbb{Z}, t)]. $$ If $X$ is $(n-1)$-connected then since $g$ is a $(t+1)$-equivalence, the map $(\mathrm{id}_X\wedge g)_*$ is an isomorphism.
Even MORE detail: It is true that the existence of $K(\mathbb{Z},n)$ requires genuine work. But it does not require the Hurewicz theorem. In fact, you only need to produce an $(n-1)$-connected space $X$ with $\pi_n(X) \cong \mathbb{Z}$ (then you can take a Postnikov section to get $K(\mathbb{Z},n)$). And there is an obvious candidate: $S^n$.
So the question becomes, how can you show $\pi_n(S^n) \cong \mathbb{Z}$ for all $n$ (the connectivity is easy as a result of $S^n = * \cup D^n$ and the cellular approximation theorem). We can prove it for $n = 1$ using covering spaces, and use Freudenthal to get it for higher $n$ (it's a little tricky for $n= 2$, but luckily $S^1$ is an H-space).
What about proving Freudenthal? The suspension $\Sigma: \pi_n(X) \to \pi_{n+1}(\Sigma X)$ is equivalent to the map induced by $\sigma: X\to \Omega\Sigma X$, adjoint to $\mathrm{id}_{\Sigma X}$, and its connectivity can be estimated using the James construction.
What about the James construction? This lovely paper
gives a proof based only on Mather's second cube theorem, which itself depends on some point-set topology due to Str{\o}m (no relation).