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There is a sequence of forgetful functors $\mathbf{Ab} \to \mathbf{Grp}\to \mathbf{Set}_*$. Further for a pointed topological space $(X,x_0)$ it is $\pi_2(X,x_0) \in \mathbf{Ab}$ and $\pi_1(X,x_0) \in \mathbf{Grp}$ and $\pi_0(X,x_0) \in \mathbf{Set}_*$. Can this sequence be extended on the left side, i.e. are there strictly "better" or smaller categories $\mathbf{C}_n, n\geq 0$ and a sequence of forgetful functors $$\dots \to \mathbf{C}_n \to \dots \to \mathbf{C}_2 = \mathbf{Ab} \to \mathbf{C}_1 = \mathbf{Grp} \to \mathbf{C}_0 = \mathbf{Set}_*$$ such that $\pi_n(X,x_0) \in \mathbf{C}_n$ for all $n \geq 0$? What would be for example the additional (to abelian) special property that all $\pi_3(X,x_0)$ have (like $\pi_1$ are groups and $\pi_2$ have the additional property "abelian")? Can it be described explicitly in a nice way?

psl2Z
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    See Eilenberg MacLane Spaces. You can set $\pi_3$ to any abelian group you like. – lulu Dec 15 '25 at 22:18
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    @lulu I don't see how that comment is related to the question. – Ben Steffan Dec 15 '25 at 22:18
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    The question is "which abelian groups are $\pi_3$'s", and @lulu's answer is "all of them". – Naïm Camille Favier Dec 15 '25 at 22:20
  • @BenSteffan What do you mean? By Eilenberg-Maclane theory you find a space with $\pi_3\simeq G$ for any abelian $G$. Thus there are no restrictions on the $G$ which might occur as $\pi_3$ other than that they must be abelain. – lulu Dec 15 '25 at 22:22
  • @NaïmCamilleFavier That's not the question, no. The question is whether $\pi_n$ factors through some sufficiently nice functor $C \to \mathrm{Ab}$. – Ben Steffan Dec 15 '25 at 22:27
  • @lulu Okay, fine. The way it is written ("property") that comment actually answers the question. Ignore my bickering. – Ben Steffan Dec 15 '25 at 22:31
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    The higher homotopy groups are $\pi_1X$-modules (i.e. modules over the group ring $\mathbb{Z}[\pi_1X]$). Off the top of my head, I don't know anything about the realisation problem in this case. – Tyrone Dec 16 '25 at 13:17
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    "trivial and uninteresting" is in the eye of the beholder#/media/File:Beholder_(D&D).JPG). If all you care about is whether there's some juicy categories there, I suppose $\textbf{Ab}$ seems trivial and uninteresting (except don't forget the $\pi_1 X$ module structure mentioned by @Tyrone, and other variations in the answers). But if you care about topology, the values of these groups, and the patterns in those values, this remains one of the richest open topics in topology. – Lee Mosher Dec 16 '25 at 15:23
  • This may be bickering about terminology, but is the map from Ab to Grp actually a forgetful functor? It is clearly functorial but I don't see anything that it forgets. Its image inside Grp still has all the structure and information of Ab. – quarague Dec 17 '25 at 11:59
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    @quarague It forgets the property of being abelian: see https://ncatlab.org/nlab/show/stuff%2C+structure%2C+property#more_examples – Naïm Camille Favier Dec 17 '25 at 12:44
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    @Tyrone The realization problem with prescribed $\pi_1$-actions is also solvable. This is somewhere in G.W. Whitehead's book IIRC, but the proof is essentially the same as in the plain realization problem. – Thorgott Dec 18 '25 at 23:30

2 Answers2

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As mentioned in the comments, once you quotient by homotopy and look at the mere group $\pi_n X$, these are all abelian groups with no restrictions once $n \geq 2$ so that it doesn't look like these categories are getting more restrictive. But the pattern that you've noticed actually does exist and continue! The caveat is that you have to remember more information!

Recall that one can build a (pointed) mapping space $[X,Y]$ whose points are continuous (pointed) maps $X \to Y$. Then $\pi_0 [X,Y]$ is the set of connected components of this space, which recovers exactly the set of (pointed) maps $X \to Y$ up to homotopy. Then we can think of $\pi_n X = \pi_0 [S^n, X]$ as the homotopy classes of maps from the $n$-sphere into $X$ (as an easy exercise, check this for $\pi_0$, $\pi_1$, and $\pi_2$). You might imagine that when we take connected components of this space we lose a lot of information, and you would be completely correct! It turns out that the category of sets is just not "big" enough (in a homotopical sense) to see much of the interesting structure present in the space $[S^n,X]$. But if we stay in the ($\infty$-)category of spaces, then we can see much more!

In particular, we find that $[S^n, X]$ is an $E_n$-algebra in spaces! This is a homotopical notion of "group" that gets progressively more commutative as $n$ gets large. In particular:

  • $E_0$-spaces are just pointed spaces
  • $E_1$-spaces are like nonabelian groups up to homotopy
  • $E_2$-spaces are like abelian groups that "remember" if you move two elements around each other in 2d
  • $E_3$-spaces are like abelian groups that "remember" if you move two elements around each other in 3d
  • ...

Since $\mathsf{Set}$ is a 1-category, everything $>1$ looks the same, which explains why truncating $[S^n,X]$ to $\mathsf{Set}$ by taking $\pi_0$ sees that $\pi_0 X$ is a pointed set, $\pi_1 X$ is a group, and then all higher $\pi_{\geq 2} X$ are "just" abelian groups.

If we instead found a way to truncate to $\mathsf{Cat}$, the 2-category of categories, then we could see slightly further! Now an $E_0$-algebra is a pointed category, an $E_1$-algebra is a monoidal category, an $E_2$-algebra is a braided monoidal category and only for $E_{\geq 3}$-algebras do we stabilize to symmetric monoidal categories.

Since spaces form an $\infty$-category, here we can see all the gradations of $E_n$-structure, and there's lots of interesting things to say about this! Though doing so would require some serious homotopy theory.


I hope this helps ^_^

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    Can you recommend any sources that provide a more detailed account of this story (apart from Lurie, of course)? – GeometriaDifferenziale Dec 17 '25 at 00:32
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    Hm... There's a nice lecture here explaining the basics. Another good talk here draws nice pictures in the $E_2$-case, and explains the connection to braided categories. There's another nice set of three lectures here. If you want to se a lot of this worked out carefully with details, but in a more approachable level of generality than Lurie, you might try Claudia Scheimbauer's thesis – Chris Grossack Dec 17 '25 at 19:27
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    NB: I learned a lot of this stuff "by osmosis" while trying to learn about factorization homology and TQFTs, which is pretty clearly reflected in the references I could quickly think of. It's very possible that a real homotopy theorist would have better references for all this. For instance, I think there's some discussion in various chapters of the 2020 Handbook of Homotopy Theory (especially Ch 19). And of course $E_n$-algebras were originally developed for "recognition principles", and the original references for that are likely to be good, even though I haven't personally read them. – Chris Grossack Dec 17 '25 at 19:36
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    I thought a bit harder about it, and you might also be interested in Chapter III (and some of the exercises in Chapter IV) of Hiro Lee Tanaka's notes here – Chris Grossack Dec 18 '25 at 15:07
  • I learned a lot of this stuff "by osmosis" while trying to learn about factorization homology and TQFTs, which is pretty clearly reflected in the references I could quickly think of.

    Don't worry, that's actually a bonus.

    – GeometriaDifferenziale Dec 19 '25 at 12:31
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I will provide two different interpretations of an answer to this question. The first was already provided by @lulu in the comments and says that there there is no additional property of an abelian group that obstructs it occurring as a homotopy group $\pi_n$ with $n\ge2$ of a pointed space $X$ (even better, for any group $G$ and abelian groups $A_n,\,n\ge2$, there is a pointed space $X$ with $\pi_1(X)\cong G$ and $\pi_n(X)=A_n$ for $n\ge2$).

The second is that this question is the "shadow" of a question with a more positive answer. Namely, there is a sequence of forgetful functors (which, however, are not faithful!) $$ \dotsc\rightarrow\mathrm{Mon}^n(\mathbf{Spc})\rightarrow\dotsc\rightarrow\mathrm{Mon}(\mathbf{Spc})\rightarrow\mathbf{Spc}_{\ast}, $$ where the $n$-th term is the $\infty$-category of $n$-fold monoid objects in the $\infty$-category $\mathbf{Spc}$ of spaces (also known as $\mathbb{E}_n$-algebras). The $n$-fold loop space functor $\mathbf{Spc}_{\ast}\rightarrow\mathbf{Spc}_{\ast}$ has a canonical lift through the $n$-th term of this sequence (the $n$ compatible monoid structures coming from composition of loops in the $n$ directions of $I^n/\partial I^n\cong S^n$), but not in general through the $(n+1)$-th term. The $n$-th homotopy group functor is obtained by applying $\pi_0$ to this functor, which (since it is a cartesian monoidal functor) induces a functor $\mathrm{Mon}^n(\mathbf{Spc})\rightarrow\mathrm{Mon}^n(\mathbf{Set})$. However, $n$-fold monoid objects in $\mathbf{Set}$ are either monoids for $n=1$ or commutative monoids for $n\ge2$ (by the Eckmann-Hilton argument), so you cannot see this interesting & subtle higher structure at the level of homotopy classes!

Thorgott
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    What is the relationship between the $\mathbb{E}_n$-monoids in your answer and the $\mathbb{E}_n$-algebras in Chris's answer? Are they essentially the same thing? – Naïm Camille Favier Dec 17 '25 at 23:05
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    @NaïmCamilleFavier I should've probably also just said "$\mathbb{E}_n$-algebras" (and edited this), but yes, they're the same thing. [To be slightly more precise, the category of $n$-fold monoid objects and the category of $\mathbb{E}_n$-algebras in $\mathbf{Spc}$ are a priori different things. That they are equivalent is a Theorem.] – Thorgott Dec 17 '25 at 23:49