# Why the Dold-Thom theorem?

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.

• Sort of intuitive heuristics behind the fact (for me) is that (a) SP(X) is the free commutative topological monoid on X (of sorts); (b) connected topological monoids possess homotopy inverses, so it is actually 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. – მამუკა ჯიბლაძე Oct 7 '14 at 17:53
• @მამუკა ჯიბლაძე, I agree that is the right perspective, but you need a little bit more. $\mathbb Z[X]$ is a simplicial abelian group. There are two things you can do with it: turn it into a chain complex and take homology; or turn it into a space and take homotopy groups. You need to know that they correspond. This is sometimes included in Dold-Kan. – Ben Wieland Oct 7 '14 at 19:27
• @BenWieland yes, I agree. On the other hand I only mentioned the simplicial context as a sort of analogy, although it is also true that one might switch to it via, say, singular simplices and then relate Sing(SP(X)) to Z[Sing(X)]... – მამუკა ჯიბლაძე Oct 7 '14 at 19:47
• For intuition, I think a key is the first case of X a sphere. If one uses the model of a cube relative to its boundary, then the free abelian group on X is essentially the iterated bar construction, and so a model for an Eilenberg-MacClane space. – Dev Sinha Oct 15 '14 at 2:37
• @მამუკაჯიბლაძე: why is SP(X) only a free commutative topological monoid "of sorts"? Why the "of sorts"? – Bruno Stonek Mar 7 '16 at 15:37

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.

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!

• Hm. I think the kicker in the whole thing is Exercise 3, which is basically that Abelian Groups are an Abelian Category (that should be intuitive!) --- what's novel is that this part still works if our Abelian groups have interesting (but not too wild) topology. Or I could say "trust me, for homotopy theory, this is pretty darn intuitive", but that wouldn't be nice. – Jesse C. McKeown Oct 7 '14 at 22:06
• There is (or ought to be) a slogan: "homology is abelianized homotopy"; the earliest inkling of this was Poincaré's formula $H_1 \simeq (\pi_1 / [\pi_1,\pi_1])$. And so compare/contrast (as I'm sure you have) $SP^\infty X \simeq Z[X]$ and $J_\infty X \simeq \Omega \Sigma X$; but that's not close enough to being an argument. – Jesse C. McKeown Oct 7 '14 at 22:11
• @JesseC.McKeown commenting to your first comment :D Indeed the crux is the fact that abelianizing turns cofibration sequences into fibration sequences and so connects homology of the source with homotopy of the target. This also works for other homology theories, giving things like $\tilde E_*X\cong\pi_*(E\wedge X)$ (although one has to stabilize $X$ into $\Sigma^\infty X$, and weaken "abelian" to "triangulated"...) – მამუკა ჯიბლაძე Oct 8 '14 at 4:31
• @მამუკაჯიბლაძე I'm having a hard time with the German, but Dold-Thom's Satz 6.10.III seems to say that the inclusion $SP(X)\to \mathbb{Z}[X]$ is a weak homotopy equivalence for $X$ a "countable simplicial complex". I'm guessing that one can replace the "simplicial" hypothesis by CW, but what about "countable"? Also, what is the reason behind the fact that most expositions on the Dold-Thom theorem focus on the SP construction instead of the $\mathbb{Z}[-]$ one? The way I see it, the one with $\mathbb{Z}$ is better because we don't need X to be connected... – Bruno Stonek Mar 13 '16 at 11:26
• The $SP$ construction is (we have to admit) simpler; it's easy enough to describe its approximants $SP^{(n)}(X) = (X^n) /_{strict} (\Sigma_n)$ and their inclusions, while building up enough continuous strict algebra to really nail down the exercises may be more daunting. Incidentally, an over-emphasis on $\mathbb{Z}[-]$ will give misleading ideas about the correct maps between these things. There are $E^\infty$-maps $\mathbb{Z}[\sph^2]\to\mathbb{Z}[\sph^n]$ that are nontrivial, and therefore not strict abelian! – Jesse C. McKeown Mar 14 '16 at 14:57

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.

• Could you elaborate in your answer about how $\pi_*(F(X))$ relates to reduced K-homology? – Chris Gerig Oct 9 '14 at 20:38
• @ChrisGerig well it is a combination of several propositions whose proofs sort of interleave - (1) that one may (if the basepoint is nice enough) replace $F(X)$ by $Hom(C(X),\mathscr K)$, the latter being the algebra of compact operators on a Hilbert space; (2) that $F(S^n)$ is the $(n-1)$-connected cover of the representing space for $KO^n$; (3) that for sufficiently nice subspaces $Y\subset X$, the map $F(X)\to F(X/Y)$ is a quasifibration with fibre $F(Y)$ (this last step is the analog of the Exercise 3 from Jesse C. McKeown's answer). Btw $kO$ is the connective $K$-theory, I'll correct. – მამუკა ჯიბლაძე Oct 9 '14 at 22:38
• So I think this response shifts my question onto another isomorphism, and my question is then: "Why is $\pi_*(F(X))\cong \widetilde{kO}_*(X)$ true, intuitively?" I see a proof of the isomorphism, but I think it skirts my question. – Chris Gerig Oct 23 '14 at 18:47
• @Chris: you probably won't like this answer, but the way the answer almost has to go is that $F(X)$ is a model for the free $ko$-module spectrum on $X$ (give or take some fuss about basepoints). This is, more or less by definition, the spectrum whose homotopy groups are the $ko$-groups of $X$. – Qiaochu Yuan Oct 23 '14 at 19:10
• Other than that, I don't know what kind of answer you're looking for. An intuition about why two things are isomorphic necessarily involves an intuition about the two things; do you have an intuition for what ko-homology, as opposed to ko-cohomology, is, other than what I wrote down above? (Because I don't, and I'd love one!) – Qiaochu Yuan Oct 23 '14 at 19:45

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 $X$.
• Thank you for this summary! All that spectra stuff starts to make some sense. It's extremely difficult to learn this stuff as there are very few sources about this and fewer that treat the "technical" stuff that goes into proving this beautiful theory. I'm finally convinced though at least in the case of connective ring spectra, close cousins of simplicial rings, that the theory starts looks rather reasonable (as opposed to unjustly beautiful). Would you say in general that the tools that go into constructing and working with general spectra are refinements of simplicial techniques? – Saal Hardali Feb 12 '16 at 22:51

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?

• I am confused. Could you elaborate in your answer on forming that new CW-complex? I don't see how to put together the union of copies of k-cells with varying k -- what are the attaching maps? I would think this affects whether the resulting "complex" could even have a fundamental class. – Chris Gerig Oct 13 '14 at 0:40
• Give me a few days. My inbox has become thick in the past few... – Ryan Budney Oct 14 '14 at 22:13
• I suppose the philosophy of this answer is that the homology of a space allows for very stratified singular objects in the space to `detect holes'. Homotopy isn't nearly as flexible, but if you replace the space $X$ by $SP(X)$ you are replacing the space by one where homotopy allows for similarly stratified objects. – Ryan Budney Oct 21 '14 at 22:59
• I'm sorry I cant' wrap my head around this yet, even for $i=1$. In the first paragraph, $k$ can't vary across the cells of $S^i$, right? And I don't see how the "subspace" of $S^i\times X$ come together to form a CW-complex. I also didn't understand the last sentence; how does the map $S_f\to S^i$ relate to "lifted maps to $X^k$"? And how do I ultimately get an element of homology of $X$? – Chris Gerig Oct 22 '14 at 22:41
• From cell to cell $k$ can vary. But on the interior of each cell I'm saying the lift has to be unique up to the action of $\Sigma_k$. – Ryan Budney Oct 23 '14 at 3:39