Is there an explicit Dold-Thom theorem?

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.

• Closely related: Why the Dold-Thom theorem?, especially one of the answers – მამუკა ჯიბლაძე Sep 12 '18 at 21:23
• I have always wondered a similar question. Let $X$ be a finite-dimensional smooth manifold. It is my feeling that there should be an explicit map $\pi_n \text{Sym}^\infty(X) \to H_n X$ given by taking a map from $S^n$ and making it transverse to the strata with nonzero multiplicity. Then one should be able to explicitly build a pseudomanifold $\tilde S$ equipped with a branched covering map $\tilde S \to S^n$ and a map $\tilde S \to X$; the map $S^n \to \text{Sym}^\infty$ encodes the number of points in a preimage. A homotopy should give a cobordism through pseudomanifolds in $X$. – Mike Miller Sep 13 '18 at 1:26
• This proof actually works 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 '18 at 1:28
• Wow, if I had just followed that link I would have seen Ryan giving my desired argument (but with CW complexes). My mistake! – Mike Miller Sep 21 '18 at 21:41
• @fritz There's a misspelling in Mike's comment (probably auto-correct) that makes it a lot harder to figure out what he's talking about unless you're familiar with Heegaard Floer homology (as in my aside to the original question). He means "Lipshitz", as in Robert Lipshitz; the relevant paper of Lipshitz is "A cylindrical reformulation of Heegaard-Floer homology" at msp.org/gt/2006/10-2/p09.xhtml, especially section 2. – dvitek Mar 24 '19 at 14:23

• Maybe I'm missing something, but I'm not sure how helpful this is for my specific question about the homotopy groups of $\mathrm{Sym}^nX$. I agree that the paper of Dold you mention solves the problem for homology (and a later paper of Puppe, MR0216497, solves the problem for cohomology). Your comment about the factorization of the Hurewicz map seems far more useful: what do you mean by the "stable range" of that filtration? – dvitek Sep 13 '18 at 17:53