TITLE: An abstract nonsense proof of the Hurewicz theorem
QUESTION [27 upvotes]: 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?

REPLY [6 votes]: One approach is to replace the free abelian group functor with the free commutative monoid functor, also known as the infinite symmetric product. 
Let $X$ be a pointed CW complex, and let $\mbox{SP}^\infty(X)$ be the infinite symmetric product. By Dold-Thom theorem, if $X$ is connected then the homotopy groups of $\mbox{SP}^\infty(X)$ are the homology groups of $X$. 
The advantage of $\mbox{SP}^\infty$ over the free abelian group functor is that it has a natural filtration, and a cell structure that can be analyzed. Given Dold-Thom theorem, the Huriewicz theorem is equivalent to saying that if $X$ is $k-1$-connected, then the map $X\to \mbox{SP}^\infty(X)$ is $k+1$ connected (for simplicity let's assume $k>1$). This can be proved by analysing the cell structure on $\mbox{SP}^\infty(X)$. You can find this  in chapter 6 of the book of Aguilar, Gitler and Prieto.