Why the Dold-Thom theorem?

Question

Dold-Thom Theorem: $$\pi_i(SP(X))\cong\tilde{H}_i(X)$$

It's pretty miraculous, no? I've seen its proof, where you show that the composition of the functors on the left-side satisfies the axioms of a homology theory. I've also seen many uses of the theorem, to explain features about Eilenberg-MacLane spaces and other (categorical) phenomena which relate homotopy and homology. But,

Is there an intuitive reason (geometrically?) why it's true?
Is the Dold-Thom theorem to be expected? Why would one come to think of this?

It is very intuitive and clear in low degrees, but the geometry might stop after this. The $i=0$ case is the connectedness of $X$. The $i=1$ case is the ability to lift and commute loops, when analyzing the compositions $\pi_1(X)\to \pi_1(X)^d\to \pi_1(\text{Sym}^d(X))\to H_1(\text{Sym}^d(X))\to H_1(X)$. Perhaps I can argue similarly in higher degrees when $X$ is a closed Riemann surface. This is clear for the sphere, since $\text{Sym}^d(\mathbb{C}P^1)\approx\mathbb{C}P^d$ and $\pi_i(\mathbb{C}P^\infty)\cong\tilde{H}_i(\mathbb{C}P^1)$. Note that this is also clear in the 1-dimensional case when $X\simeq S^1$, as $\text{Sym}^d(\mathbb{C}-\lbrace0\rbrace)\approx \mathbb{C}^{d-1}\times(\mathbb{C}-\lbrace 0\rbrace)$. The content of the theorem is reduced to low degrees for these simple examples.

Answer 1

Imposing some reasonable conditions on our spaces (I think semilocally-simply-connected ought to do), one works through

Exercise 1 $\mathbb{Z}[X]$, the free topological $\mathbb{Z}$-module continuously generated by a convenient space $X$ is an $E^\infty$ space; the maps $\mathbb{Z}[X] \to \mathbb{Z}[Y]$ induced by $ X \to Y \to \mathbb{Z}[Y]$ make this construction continuously functorial; these induced maps are again $E^\infty$ maps.

Exercise 2 a weak homotopy equivalence of spaces $X \simeq X'$ induces a weak homotopy equivalence of $\mathbb{Z}$-modules.

Exercise 3 For a cofibration $X \to Y$, there is a pullback square $$ \begin{array}{c} \mathbb{Z}[X] & \to & \mathbb{Z}[Y] \\ \downarrow & & \downarrow \\ \mathbb{Z} & \to & \mathbb{Z}[Y/X] \end{array}$$

Exercise 4 $\pi_0 \mathbb{Z}[*] \simeq \mathbb{Z}$; otherwise $\pi_k \mathbb{Z}[*] \simeq 0$.

Exercise 5 the functor $X\mapsto \mathbb{Z}[X]$ preserves colimits of sequences of cofibrations.

Corollary We have verified that the functors $\pi_k \mathbb{Z}[X]$ satisfy the Eilenberg-Steenrod axioms for ordinary homology.

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Using the natural map $\mathbb{Z}[X] \to \mathbb{Z}$, write $\tilde{\mathbb{Z}}[X]$ for its kernel. To complete the exercises, მამუკა ჯიბლაძე's cogent remark explains why the natural map $SP^\infty X \to \tilde{\mathbb{Z}}[X]$ is an equivalence for connected $X$.
Here is the remark: Some intuitive heuristics behind the fact are (a) $SP(X)$ is the free commutative topological monoid on $X$ (of sorts), (b) connected topological monoids possess homotopy inverses, so it is actually a free topological abelian group on $X$ (again sort of), (c) homology of $X$ is more or less the same as homotopy of the free topological abelian group on $X$. All this is almost rigorous in the simplicial context, where $\tilde H_*(X)=\pi_*\mathbb{Z}[X]$ more or less by definition.

But yes, it really is magical!

Answer 2

Maybe Segal's fascinating extension of this fact to the K-homology adds some intuitive understanding of what happens underneath. Unfortunately I only was able to find a paid version of the text ("K-homology theory and algebraic K-theory"), it is in the book "K-theory and operator algebras" (Springer LNM 575, pp 113-127)

In Segal's setup $X$ is any compact Hausdorff space with a basepoint. He takes its Gelfand dual $C(X)$ (continuous real-valued functions on $X$ vanishing at the basepoint, a (unitless) C*-algebra). Recall that Gelfand duality recovers $X$ from $C(X)$ as the spectrum of the latter. That is, every C*-homomorphism $C(X)\to\mathbb R$ has form $f\mapsto f(x)$ for some (fixed) $x\in X$.

Segal considers $$ F(X):=\bigcup_{n\geqslant0}\mathrm{Hom}_{\textrm{Algebras}}(C(X),\mathrm{Mat}_{n\times n}\mathbb R), $$ "a kind of non-abelian spectrum of $C(X)$" (union is wrt embedding $n$-matrices into $n+1$-matrices via $A\mapsto\left(\begin{smallmatrix}A&0\\0&0\end{smallmatrix}\right))$.

What matters is that an element of $F(X)$ can be viewed as a "finitely supported family of real f.d. vector spaces $V_x$ indexed by points of $x$"; moreover there is a natural topology on $F(X)$ such that (a) if points $x_1$ and $x_2$ are moved towards each other to coincide in $x$ then $V_x$ becomes identified with the resulting limit of $V_{x_1}\oplus V_{x_2}$; (b) if a point $x$ moves towards the basepoint it just falls out of the picture.

Thus a point of $F(X)$ is like a nonnegative linear combination of points of $X$, except that mutliplicities of points are, instead of natural numbers, finite-dimensional real vector spaces.

It turns out that $\pi_*(F(X))$ is isomorphic to $\widetilde{\mathrm{kO}}_*(X)$ (reduced connective $K$-homology of $X$).

Furthermore, there is a natural map $F(X)\to SP(X)$ sending "$V_1x_1+...+V_kx_k$" to $\dim(V_1)x_1+...+\dim(V_k)x_k$ and the induced map of homotopy groups $\widetilde{\mathrm{kO}}_*(X)\to\pi_*(SP(X))$ is the one you've just guessed.

Answer 3

This won't involve any geometry, but here is a model-independent description of the situation as I understand it. I will not prove anything. The very short summary is that

The infinite symmetric product and singular chains are both models of the free $\mathbb{Z}$-module spectrum on a space $X$, where $\mathbb{Z}$ is regarded as a ring spectrum, and "homotopy groups of the free $\mathbb{Z}$-module spectrum on $X$" is a model-independent definition of the (ordinary) homology groups of $X$.

First, here's the simplest version of a general definition. Let $X$ be a set and let $R$ be a commutative ring. Then the $R$-homology of $X$ is equivalently one of the following $R$-modules:

• the $R$-module $R[X]$ of formal $R$-linear combinations of elements of $X$,
• the direct sum $\displaystyle \bigoplus_{x \in X} R$, or equivalently the colimit of the constant diagram $X \ni x \mapsto R \in \text{Mod}(R)$,
• the value on $X$ of the left adjoint to the forgetful functor $\text{Mod}(R) \to \text{Set}$,
• the value on $X$ of the unique cocontinuous functor $\text{Set} \to \text{Mod}(R)$ sending $\text{pt}$ to $R$.

Intuitively, $R$-homology is the canonical covariant way to linearize a set $X$ into an $R$-module: above I've just given four different ways of saying "the free $R$-module on $X$."

Moreover, the forgetful functor $\text{Mod}(R) \to \text{Set}$ factors through abelian groups, and hence its left adjoint also factors through abelian groups in the other direction as the composite

$$R[-] : \text{Set} \xrightarrow{\mathbb{Z}[-]} \text{Ab} \cong \text{Mod}(\mathbb{Z}) \xrightarrow{R \otimes (-)} \text{Mod}(R)$$

which is just a fancy way of saying that we can write

$$R_0(X) \cong R[X] \cong R \otimes \mathbb{Z}[X].$$

This is "universal coefficients for sets": it says that to understand the free $R$-module on a set it suffices to understand the free $\mathbb{Z}$-module / abelian group on a set.

The last description of $R$-homology above reflects "Eilenberg-Steenrod for sets," which says that $\text{Set}$ is the free cocomplete category on a point.

Now suppose we want to linearize, not sets, but spaces, by which I mean (weak) homotopy types / $\infty$-groupoids. So let $X$ be a space and let $R$ be an $E_{\infty}$-ring spectrum. Then the $R$-homology of $X$ is equivalently (the homotopy groups of) one of the following $R$-module spectra, where $\text{Mod}(R)$ denotes the $(\infty, 1)$-category of $R$-module spectra (and "functor" means "$(\infty, 1)$-functor"):

• the smash product $R \wedge \Sigma^{\infty}_{+} X$, where $\Sigma^{\infty}_{+} X$ is the suspension spectrum of $X$ with a disjoint basepoint,
• the homotopy / $(\infty, 1)$-colimit of the constant diagram $X \ni x \mapsto R \in \text{Mod}(R)$,
• the value on $X$ of the $(\infty, 1)$-left adjoint to the forgetful functor $\text{Mod}(R) \to \text{Space}$,
• the value on $X$ of the unique homotopy cocontinuous functor $\text{Space} \to \text{Mod}(R)$ sending $\text{pt}$ to $R$.

Intuitively, $R$-homology is the canonical covariant way to linearize a space into an $R$-module spectrum: above I've just given four different ways of saying "the free $R$-module spectrum on $X$."

The first description above should be regarded as a direct generalization of the isomorphism $R[X] \cong R \otimes \mathbb{Z}[X]$ to spaces, except that $\mathbb{Z}$ has been replaced with the sphere spectrum $\mathbb{S}$. More precisely, the forgetful functor $\text{Mod}(R) \to \text{Space}$ factors through spectra, and hence its left adjoint also factors through spectra in the other direction as the composite

$$\text{Space} \xrightarrow{\Sigma^{\infty}_{+}(-)} \text{Sp} \cong \text{Mod}(\mathbb{S}) \xrightarrow{R \wedge (-)} \text{Mod}(R).$$

In particular $\Sigma^{\infty}_{+}$, being left adjoint to the forgetful functor from spectra to spaces, should be thought of as the "free spectrum" functor, and $R \wedge (-)$, being left adjoint to the forgetful functor from $R$-module spectra to spectra, should be thought of as the "free $R$-module spectrum (on a spectrum)" functor.

The last description of $R$-homology reflects an $(\infty, 1)$-categorical version of Eilenberg-Steenrod for spaces, which says that $\text{Space}$ is the free homotopy cocomplete $(\infty, 1)$-category on a point.

Now, at long last, ordinary homology is the homotopy groups of the free $\mathbb{Z}$-module spectrum:

$$H_{\bullet}(X, \mathbb{Z}) \cong \pi_{\bullet}(\mathbb{Z} \wedge \Sigma_{+}^{\infty} X).$$

Hopefully I've phrased things so it's clear that this story about linearizing spaces is a direct analogue of the story about linearizing sets, provided you are willing to accept (that various $(\infty, 1)$-categorical machinery works the way it ought to and) that the correct analogue of abelian groups in this setting is spectra.

Here are some more things that ought to be true and that connect this story back to more model-dependent considerations.

• By a suitable version of the stable Dold-Kan theorem, the $(\infty, 1)$-category of $\mathbb{Z}$-module spectra should be equivalent to the $(\infty, 1)$-category presented by unbounded chain complexes of $\mathbb{Z}$-modules. This should restrict to an equivalence between connective $\mathbb{Z}$-module spectra and connective chain complexes.
• By the usual Dold-Kan theorem, the category of connective chain complexes of abelian groups is equivalent to the category of simplicial abelian groups (and there should be model structures on both sides making this a Quillen equivalence presenting an equivalence of $(\infty, 1)$-categories, and so forth). This equivalence more or less sends singular chains on a topological space $X$ to the free simplicial abelian group on the singular simplicial set of $X$, and modulo technical details this gives rise to the relationship between singular homology and the homotopy groups of the free $\mathbb{Z}$-module spectrum on $X$, which is connective since $\mathbb{Z}$ and suspension spectra are connective.
• The analogue of the free simplicial abelian group on a simplicial set for topological spaces is the free topological abelian group; this is roughly what the infinite symmetric product attempts to be, and modulo technical details (in particular, niceness hypotheses on $X$) this gives rise to the relationship between the homotopy groups of the infinite symmetric product and the singular homology of $X$.

Answer 4

Here is a sketch of a direct argument, stated with a little more precision but still lacking in details. It is a proposed map $\pi_i SP(X) \to \tilde H_i(X)$.

Given a map $f : S^i \to SP(X)$, via a general position argument we can (up to a small homotopy of $f$) endow $S^i$ with a CW-structure such that the restriction of $f$ to any cell admits a lift to some $X^k$, moreover we will demand that the number of distinct points is constant on the interior of the cell, and no points are allowed to be mapped to the basepoint (again, on the interior). So each such cell comes with $k$ maps to $X$. Think of these maps from the cells to $X$ as subsets of $S^i \times X$.

We form a new CW-complex, $S_f$. It is a subspace of $S^i \times X$, and consists of the union of the graphs of all the above maps.

There is a projection map $S_f \to S^i$ and a this is how we define the fundamental class of $S_f$, it will be an element in $H_i S_f$. All the cells of $S_f$ are parametrized so that projection $S_f \to S^i$ preserves the characteristic maps. You weight your cells by how many times that projection occurs in the lifted map to $X^k$.

Does that sound more sensible now?