**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.



Why the Dold-Thom theorem?
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!


