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The ordinary homotopy groups of a space $X$ are the homotopy groups of the corresponding singular simplicial set $Sing(X)$. The ordinary homology groups of $X$ are the homotopy groups of the simplicial set $F(Sing(X))$, where $F$ is the functor that replaces each set $Sing(X)_n$ with the free abelian group on that set. There should obviously be some nice property of $F$ that makes the Hurewicz theorem work, but all the proofs I can find in the literature do things entirely by hand at the level of checking individual maps and homotopies to $X$. Is this nice property known? Is there a satisfying answer?

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    $\begingroup$ Pick a simplicial group model $G$ for the loop space of $X$, then the Hurewicz homomorphism is realized as that induced on $G\to G/[G,G]$. This is in many places, e. g. May $\endgroup$ Oct 10, 2017 at 20:59
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    $\begingroup$ @მამუკაჯიბლაძე that's a more than slightly ambiguous reference, tbh. $\endgroup$
    – David Roberts
    Oct 10, 2017 at 22:55
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    $\begingroup$ @anton I'd be interested to see the simple abstract proof, so that I can work out whether that's what I'm asking. $\endgroup$ Oct 10, 2017 at 23:51
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    $\begingroup$ @DavidRoberts You are right, sorry, I meant "Simplicial objects in algebraic topology", Theorem 26.9 (on page 121) $\endgroup$ Oct 11, 2017 at 4:55
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    $\begingroup$ @OscarRandal-Williams Why hide it - it turns out I actually don't know. I thought I knew but it turned out I never did. $\endgroup$ Oct 12, 2017 at 9:26

4 Answers 4

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I’d argue that it boils down to the generator $S^n\to K(\mathbb{Z},n)$ being an $(n+1)$-equivalence.

More detail: If you take the represented version of homology, it is given by $$ H_n(X;\mathbb{Z}) \cong [ S^{n+t} , X\wedge K(\mathbb{Z}, t)] $$ for $t$ large. Then the Hurewicz map is the map induced by the generator $g:S^t \to K(\mathbb{Z}, t)$: $$ \mathrm{H}: [ S^{n+t} , X\wedge S^k] \xrightarrow{(\mathrm{id}_X\wedge g)_*} [ S^{n+t} , X\wedge K(\mathbb{Z}, t)]. $$ If $X$ is $(n-1)$-connected then since $g$ is a $(t+1)$-equivalence, the map $(\mathrm{id}_X\wedge g)_*$ is an isomorphism.

Even MORE detail: It is true that the existence of $K(\mathbb{Z},n)$ requires genuine work. But it does not require the Hurewicz theorem. In fact, you only need to produce an $(n-1)$-connected space $X$ with $\pi_n(X) \cong \mathbb{Z}$ (then you can take a Postnikov section to get $K(\mathbb{Z},n)$). And there is an obvious candidate: $S^n$.

So the question becomes, how can you show $\pi_n(S^n) \cong \mathbb{Z}$ for all $n$ (the connectivity is easy as a result of $S^n = * \cup D^n$ and the cellular approximation theorem). We can prove it for $n = 1$ using covering spaces, and use Freudenthal to get it for higher $n$ (it's a little tricky for $n= 2$, but luckily $S^1$ is an H-space).

What about proving Freudenthal? The suspension $\Sigma: \pi_n(X) \to \pi_{n+1}(\Sigma X)$ is equivalent to the map induced by $\sigma: X\to \Omega\Sigma X$, adjoint to $\mathrm{id}_{\Sigma X}$, and its connectivity can be estimated using the James construction.

What about the James construction? This lovely paper

Fantham, Peter; James, Ioan(4-OX); Mather, Michael On the reduced product construction. (English summary) Canad. Math. Bull. 39 (1996), no. 4, 385–389.

gives a proof based only on Mather's second cube theorem, which itself depends on some point-set topology due to Str{\o}m (no relation).

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    $\begingroup$ How so? Is it because $S^n$ is a Moore space? That the Eilenberg-Mac Lane space represents on cohomology? $\endgroup$
    – David Roberts
    Oct 11, 2017 at 5:08
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    $\begingroup$ What is your definition of $K(\mathbb{Z},n)$? The various usual definitions are not obviously equivalent without the Hurewicz Theorem. Similarly, many of the standard facts about connectivity of spectra and maps implicitly use the Hurewicz Theorem, so it is far from clear that you have a non-circular argument. $\endgroup$ Oct 11, 2017 at 18:05
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    $\begingroup$ Without the Hurewicz theorem how do you know $K(\mathbb{Z},n)$ exists? $\endgroup$ Oct 11, 2017 at 18:05
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    $\begingroup$ @Oscar: I don't know what construction you have in mind that requires the Hurewicz theorem, but all that I think I need to construct $K(\mathbb{Z}, n)$ is a model for the classifying space functor $B$ which takes topological abelian groups to topological abelian groups, which IIRC was written down by Segal, applied $n$ times to $\mathbb{Z}$. Verifying that this space has the correct homotopy groups just requires knowing that $\Omega B G \cong G$. But then I guess verifying that it represents (co)homology requires more. $\endgroup$ Oct 11, 2017 at 18:09
  • $\begingroup$ More detail on Eilenberg-Mac Lane spaces added. $\endgroup$
    – Jeff Strom
    Oct 12, 2017 at 12:50
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Here is a sketch of the proof, some details filled below. All categories are $(\infty,1)$-categories and all functors are $(\infty,1)$-functors unless specified otherwise. The notion of a topological abelian group can be defined within higher topos theory. The category of abelian groups is equivalent to the category of cartesian functors from a certain representing category $\mathrm{\mathcal{T}Ab}$ with finite products to the topos. The 1-truncation of $\mathrm{\mathcal{T}Ab}$ is equivalent to the Lawvere theory of abelian groups, and its models in any 1-truncated topos is the classic category of abelian groups. The important thing here is that the free abelian groups functor is defined naturally for any cocomplete category with finite products as a colimit of finite powers of the generating object, thus it is preserved by any functor which preserves colimits and finite products. In particular, it it preserved by the $n$-truncation functor $\tau_{\leqslant n}:\mathrm{Space}\to\mathrm{Space}_{\leqslant n}$, since $\mathrm{Space}$ is a topos (HTT, lemma 6.5.1.2, also on nLab). Hurewicz theorem requires working not with $\mathrm{Space}$ itself, but with the category of pointed and, more generally, $k$-connected spaces $\mathrm{Space}^{>k}$. The truncation functor on these categories also preserves finite products and colimits. The free abelian group functor on pointed spaces acts as $\left(X,\ast\right)\mapsto F\left(X\right)/F\left(\ast\right)$, i.e. as reduced homology. Note that it's just a specialization of the general definition. $F$ maps the subcategory $\mathrm{Space}^{>k}\hookrightarrow\mathrm{Space}^{>-1}$ to itself, since $\mathrm{Space^{>k}}$ is closed under finite products and small colimits. This means that for any $\left(n-1\right)$-connected space $X$ the canonical morphism $X\to FX$ under truncation induces a morphism $\tau_{\leqslant n}X\to F\tau_{\leqslant n}X$, where $\tau_{\leqslant n}X$ lives in the category of $\left(n-1\right)$-connected $n$-truncated spaces $\mathrm{Space}_{\leqslant n}^{>n-1}$ and $F$ is the free abelian group functor for $\mathrm{Space}_{\leqslant n}^{>n-1}$. The loop–deloop adjunction induces an equivalence between the categories $\mathrm{Space}^{>k}$ and $E_{k+1}\mathrm{Grp}\left(\mathrm{Space}\right)$ — $E_{k+1}$-monoidal group objects in spaces, thus $\mathrm{Space}_{\leqslant n}^{>n-1}\simeq E_{n}\mathrm{Grp}\left(\mathrm{Set}\right)$. For $n=0$ $E_{n}\mathrm{Grp}\left(\mathrm{Set}\right)\simeq\mathrm{Set}_{\bullet}$ and the Hurewicz moprhism is the inclusion of a pointed set into its free abelian group (note that the marked point maps to $0$). For $n=1$ $E_{n}\mathrm{Grp}\left(\mathrm{Set}\right)\simeq\mathrm{Grp}\left(\mathrm{Set}\right)$, the abelian groups in the $1$-category of groups are just classical abelian groups and the Hurewicz morphism is the abelianization of the group $\pi_{1}\left(X\right)$. For $n>1$ the sequence of $E_{n}$-monoids stabilizes, all $E_{n}\mathrm{Grp}\left(\mathrm{Set}\right)\simeq\mathrm{Ab}\left(\mathrm{Set}\right)$. The category of abelian groups in the $1$-category of abelian groups is the $1$-category of abelian groups itself, thus both the forgetful functor and the free abelian group functor are identity and the Hurewicz morphism is the identity as well, QED.

The trickiest part of the theorem is to actually state it: we need to define the free abelian group functor $F$ as a functor on the category of spaces (here and below all categories are $(\infty,1)$-categories and all functors are $(\infty,1)$-functors unless specified otherwise) and show that it can be constructed via finite products and small colimits. This isn't obvious since the natural homotopization of abelian monoids is the category of $E_{\infty}$ monoids, which is the category of algebras over the operad of abelian groups $\mathrm{Comm}$. Thus we will proceed in several steps: first introduce the notion of $E_{\infty}$-groups which are the same as connective spectra. The abelian groups are $H\mathbb{Z}$-modules in the category of spectra (by Dold–Kan correspondence the category of simplicial abelian groups is equivalent to the category of non-negative chain complexes, which are equivalent to connective $H\mathbb{Z}$-module spectra). Here $H\mathbb{Z}$ is an $E_{\infty}$-ring spectrum which is the 0-truncation of the sphere spectrum. The free abelian group functor in the category of spectra is thus $X\mapsto H\mathbb{Z}\otimes X$, by the universality of left adjoints the free abelian group functor in the category of spaces is $X\mapsto\Omega^{\infty}(H\mathbb{Z}\otimes\Sigma^{\infty}X_{+})$. This is the composition of three freeness functors: the free $E_{\infty}$-monoid one, followed by the group completion of a monoid, followed by the free $H\mathbb{Z}$-algebra one which is the smash product of spectra. The first two ones are of the required type (products and colimits) as free algebra functors. The third one is also of this type since the smash product aka $E_{\infty}$-tensor product can be represented by the Day convolution, which involves only finite products and colimits, of the corresponding functors on the representing Lawvere theory of $E_{\infty}$-groups. Note that if we model this theory in $1$-categories rather than $\infty$-categories, then the $E_{\infty}$-monoid structure reduces to the abelian monoid one, the tensor product is just the ordinary tensor product over $\mathbb{Z}$ and the whole composition equals to the free abelian group functor.

Some other trivial applications of the above technique: for any abelian group $A$ the Hurewicz morphism for $A$-homology on the lowest nontrivial homotopy group is the composition of abelianization and the map $\pi_{n}\left(X\right)\to A\otimes\pi_{n}\left(X\right)$. Less trivially, let us consider the free group functor, which is equivalent to $\Omega\Sigma$. The natural transformation $X\to\Omega\Sigma X$ is described by the Freudenthal's theorem: it is $\left(2n-2\right)$-connected if $X$ is $\left(n-1\right)$-connected. After looping and truncating this is equivalent to the statement that on the category of $\left(n-2\right)$-truncated $E_{n}$-groups the free group functor is an equivalence. This in turn is equivalent to the statement that the categories of models of $E_{n+k}$-groups in $n$-categories are equivalent for $k\geqslant1$, which is a generalization of the claim that $E_{k}\mathrm{Grp}\left(\mathrm{Set}\right)\simeq\mathrm{Ab}$ for $k\geqslant2$ (in fact we don't need the structure of a group, only a monoid). The structure of $E_{n+k}$-monoid is given by $\left(n+k\right)$ commuting (necessarily equal) associtive multiplications, which amounts to giving a series of commuting $\left(n+k\right)$-dimensional cubes which have the muplication maps on edges, the commutativity conditions on 2-faces etc for all operation arities. The highest degree condition corresponds to the cube itself and is $\left(n+k\right)$-dimensional. In an $n$-category it reduces to a relation, i.e. if the corresponding $\left(n+k\right)$-dimensional paths exists then they are unique and no relation of higher degree between them is possible, thus any $E_{n+1}$-monoid is automatically an $E_{n+k}$-monoid.

As usual in category theory, we have moved the burden of work from the proof of the statement to the definitions, which reduces the actual proof to some purely formal statements. One could probably say that it requires a perverted state of mind to claim that the proof above is "simple", however I still make this claim: it involves only general facts about higher algebras and higher categories. In a world where children study homotopy theory instead of set theory all statements above would be most natural and can be verified mentally.

I think I'll state it concisely: Hurewicz theorem is the statement that for $(n-1)$-connected spaces the $n$-truncation of the morphism into abelianization is an abelianization of $\pi_n$. The truncation commutes with free algebra functor by abstract nonsense and the free algebra is $(n-1)$-connected, thus we study ablianization on the category of $(n-1)$-connected $n$-truncated spaces, which is equivalent to $Set_*$, $Grp$ or $Ab$ depending on $n$, where abelianization is obvious, QED.

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    $\begingroup$ As someone who likes and uses $\infty$-categories, this answer terrifies me. If I am correctly interpreting your insistence on using only native $\infty$-categorical concepts (which means you aren't allowed to cheat and get the symmetric monoidal structure on $Sp$ using a model structure or anything) you have used all of the following things to prove the Hurewicz theorem: the fact that spectra has the structure of symmetric monoidal $\infty$-category, the definition and basic properties of various $\infty$-operads, $\infty$-categorical Day convolution, the fact that algebras over $\endgroup$ Oct 12, 2017 at 17:29
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    $\begingroup$ (contd) varius $\infty$-operads have nice explicit descriptions in the 1-categorical setting, the fact that grouplike $E_k$-spaces are equivalent to $k$-fold loop spaces (which includes checking somehow that $\Omega^k$ preserves sifted colimits as a functor on connected enough spaces- a fact which seems dangerously close to the Hurewicz theorem already), the claim that the sphere spectrum (and I mean like.. THE sphere spectrum) is the universal $E_{\infty}$-group (including defining what that means), which is roughly the Barrat-Priddy-Quillen theorem, and the ($\infty$-)Dold-Kan correspondence $\endgroup$ Oct 12, 2017 at 17:33
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    $\begingroup$ I find this absolutely ridiculous. $\endgroup$ Oct 12, 2017 at 17:33
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    $\begingroup$ @DylanWilson Well, the OP has asked for an abstract nonsense proof. I have delivered on both points, didn't I? More seriously, the essential part of the proof is in the last paragraph. Most of the rest is required for the comparison to classic singular homology. The proof is trivial if you can define homology $\infty$-categorically in a simple way, e.g. if you know some simple description of the abelian operad (I don't). I consider Day convolution and operads to be a trivial matter. You're not terrified by Lawvere theories or coends, are you? (..cont) $\endgroup$ Oct 12, 2017 at 17:56
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    $\begingroup$ Finally, I describe a general approach to such algebraic questions. It may be an overkill for Hurewicz, but it is more powerful. E.g. I show that it proves Freudenthal, there are probably other interesting algebras which would be amenable to the method. $\endgroup$ Oct 12, 2017 at 18:05
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Here are some thoughts; I don't know if they'll add up to a satisfying answer. Let $X$ be $(n-1)$-connected. We have a Hurewicz map $\pi_n(X) \to H_n(X)$ given by applying $H_n$ to maps $S^n \to X$ and we want to know that it's an isomorphism (or abelianization if $n = 1$). Applying $\text{Hom}(-, A)$ for an arbitrary abelian group $A$, the Yoneda lemma shows that this is equivalent to knowing that the corresponding map

$$\text{Hom}(H_n(X), A) \to \text{Hom}(\pi_n(X), A)$$

is always an isomorphism. Now, by the universal coefficient theorem $\text{Hom}(H_n(X), A)$ can be identified with $H^n(X, A)$, so this map can be thought of as the map

$$H^n(X, A) \to \text{Hom}(\pi_n(X), A)$$

given by applying $\pi_n$ to maps $X \to B^n A$, once we've shown that the classifying space $B^n A \cong K(A, n)$ exists and represents cohomology (I don't know off the top of my head whether there's a clean way to do this that avoids the Hurewicz theorem).

The nice thing about having massaged the statement to this form is that now it is entirely a statement about computing homotopy classes of maps and we can hope for a reasonably conceptual $\infty$-categorical way to understand, if not non-circularly prove, it. My understanding would use the following: there is a functor $\tau_{\le n}$ sending a space to its $n$-truncation, which is left adjoint to the inclusion from $n$-truncated spaces (spaces with vanishing $\pi_k, k \ge n+1$) into spaces, and which correspondingly has the universal property that

$$\text{Map}(\tau_{\le n} X, Y) \cong \text{Map}(X, Y)$$

for any $n$-truncated space $Y$. Now, $B^n A$ is $n$-truncated, so setting $Y = B^n A$ and $X$ as above gives

$$\text{Map}(\tau_{\le n} X, B^n A) \cong \text{Map}(X, B^n A)$$

or, restated in terms of cohomology,

$$H^n(\tau_{\le n} X, A) \cong H^n(X, A).$$

In other words, $H^n$ only depends on the $n$-truncation of $X$. This is useful because the assumption that $X$ is $(n-1)$-connected implies that $\tau_{\le n} X$ has only a single nonzero homotopy group, and hence it can be identified with $B^n \pi_n(X)$. Now we've reduced to showing that

$$\text{Map}(B^n \pi_n(X), B^n A) \cong \text{Hom}(\pi_n(X), A)$$

which at least morally follows from taking loop spaces $n$ times and using that abelian groups fully faithfully embed into $n$-fold loop spaces, $n \ge 1$ (for $n = 1$ and $\pi_1$ nonabelian, the same statement for groups). I also don't know whether there's a clean proof of this that avoids the Hurewicz theorem!

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One approach is to replace the free abelian group functor with the free commutative monoid functor, also known as the infinite symmetric product.

Let $X$ be a pointed CW complex, and let $\mbox{SP}^\infty(X)$ be the infinite symmetric product. By Dold-Thom theorem, if $X$ is connected then the homotopy groups of $\mbox{SP}^\infty(X)$ are the homology groups of $X$.

The advantage of $\mbox{SP}^\infty$ over the free abelian group functor is that it has a natural filtration, and a cell structure that can be analyzed. Given Dold-Thom theorem, the Huriewicz theorem is equivalent to saying that if $X$ is $k-1$-connected, then the map $X\to \mbox{SP}^\infty(X)$ is $k+1$ connected (for simplicity let's assume $k>1$). This can be proved by analysing the cell structure on $\mbox{SP}^\infty(X)$. You can find this in chapter 6 of the book of Aguilar, Gitler and Prieto.

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