The Dold-Thom theorem tells us that we can recover the reduced homology of a pointed space $(X,x)$ via taking homotopy groups of the symmetric product:

$$\pi_i(\mathrm{Sym}^{\infty}(X,x)) \cong H_i(X,x;\mathbb{Z}).$$

The usual proof of this theorem generally involves verifying that the left-hand side satisfies the axioms of a (reduced) homology theory, and as such is pretty abstract. The only application I know is constructing Eilenberg-MacLane spaces as infinite symmetric products of Moore spaces. (There's also a factorization-homology proof due to Bandklayder, which is perhaps a bit more useful for my question but is definitely outside my wheelhouse.)

I'm wondering if there are any explicit convergence results: we have inclusions $$\mathrm{Sym}^0X \hookrightarrow \mathrm{Sym}^1X \hookrightarrow \mathrm{Sym}^2X \hookrightarrow \cdots,$$ so how far do we have to go in this sequence before the homotopy groups stabilize?

To make this question explicit, suppose that $X$ is a finite simplicial complex, such that $X$ has at most $n$ nondegenerate simplices in each dimension up to $d$. **Is there a function $f(i,n,d)$ such that for all such $X$ we have**

$$\pi_i(\mathrm{Sym}^{f(i,n,d)}(X,x)) \cong H_i(X,x;\mathbb{Z})?$$

Part of my difficulties in thinking about this problem are that I don't know how to do many computations of $\mathrm{Sym}^n X.$ I know that $\mathrm{Sym}^n(S^1) \simeq S^1$ and that we know a little bit about the algebraic topology of $\mathrm{Sym}^g(\Sigma_g)$, where $\Sigma_g$ is the Riemann surface of genus $g$ (this perhaps betrays my exposure to Heegaard-Floer homology).

In general, I believe one can compute a simplicial decomposition of $\mathrm{Sym}^nX$, though it's pretty complicated to write down and I'd appreciate a reference to automated computations. But I don't know how to compute the homotopy groups of an arbitrary abstract simplicial complex in an algorithmic fashion, so this has been pretty inaccessible to me.

actuallyworks when $X$ is a surface, and indeed Lipschitz' cylindrical picture makes this clear (and motivated the question for me, but it's definitely intrinsically interesting). But I never worked out the details in general. – Mike Miller Sep 13 at 1:28