Let $H$ be an $(\infty,1)$-topos (seen as a generalization of the homotopy category of spaces).

You can define the suspension of an object $X$ as the (homotopy) pushout of $*\leftarrow X \to *$, hence you can define inductively the spheres $\mathbb{S}^n$ (the sphere of dimension $-1$ is the initial object of $H$ and the sphere of dimension $n+1$ is the suspension of the sphere of dimension $n$).

You can also define the loop spaces of a pointed object as the (homotopy) pullback of $*\to X \leftarrow *$. It will be itself pointed (because there is an obvious commutative diagram with a $1$ instead of $\Omega{}X$, so there is (I think) an arrow between this $1$ and $\Omega{}X$).

Then, given two integers $n, k$, you can define $\pi_k(\mathbb{S}^n)$ as the set of connected components (global elements up to homotopy) of the $k$-fold loop space of the $n$-sphere (I don’t know if this definition is one of the two described in the nlab)

Is there a natural group structure on $\pi_k(\mathbb{S}^n)$?

Is there something known about these groups in general?

For example,

  • Are they completely known for some $H$?
  • Is it always true that $\pi_k(\mathbb{S}^n)$ is trivial for $k<n$ and isomorphic to $\mathbb{Z}$ for $k=n$?
  • Are they isomorphic (or related in some way) to the usual homotopy groups of spheres?


What if you assume that $H$ is a cohesive $(\infty,1)$-topos? (see here for the nLab page)

  • 1
    $\begingroup$ A brief comment: if you take the collection of global elements of $\Omega^k\mathbb{S}^n$, then this naturally wants to form the 0-cells of an $\infty$-groupoid. Unless you take connected components it will not be a group, compare the case of $\Omega^k S^n$ in $Top$ - it is only an $A_\infty$-space, not a group. $\endgroup$
    – David Roberts
    Commented Oct 24, 2011 at 22:32
  • $\begingroup$ Thanks, I guess I wanted to take the collection of global elements up to homotopy, but for some reason I did not write the "up to homotopy" part. $\endgroup$ Commented Oct 24, 2011 at 23:08
  • 7
    $\begingroup$ I think the fact that $\pi _k (S^n)$ is a group should follow from general nonsense: $\Omega^k X$ is an $E_k$-object in $H$, so $\pi _0$ of it will be a group if $k>0$ (and abelian if $k>1$). I also think there would be a counterexample to your second point if you take $H$ to be sheaves of spaces over a sphere $S^k$. There will be a non-trivial global section of the constant sheaf with fibres $S^k$ (which should be the k-sphere object $\mathbb S^k$ in this category), so that $\pi _0(\mathbb S^k )=\mathbb Z$. $\endgroup$ Commented Oct 24, 2011 at 23:30
  • 6
    $\begingroup$ If the (ordinary) category of sets counts as an $\infty$-topos, then the spheres are all points, and the homotopy groups of those are all known. $\endgroup$ Commented Oct 24, 2011 at 23:35

5 Answers 5

  • If $H$ is the terminal category (=sheaves on the empty space), then $\pi_k^HS^n$ (notation for homotopy groups of "spheres" in $H$) is known!

  • The slice category $H=\mathrm{Spaces}/B$ is an $(\infty,1)$-topos. The homotopy groups of spheres in this setting amount to the homotopy groups of the space $\mathrm{map}(B,S^n)$ of unbased maps (with basepoint at a constant map $B\to S^n$). This shows that $\pi_k^HS^n$ need not be trivial if $k<n$. This also provides non-trivial examples in which $\pi_k^HS^n$ is isomorphic to the "usual" homotopy groups of spheres (e.g., if $B=BG$ for $G$ a finite group, by Miller's theorem.)

  • If $f: H\to H'$ is a geometric morphism, then the pullback functor $f^*: H'\to H$ induces a homomorphism $\pi_k^{H'}(S^n)\to \pi_k^{H}(S^n)$. In particular, if $H$ has a point (a geometric morphism $\mathrm{Spaces}\to H$), then $\pi_kS^n$ is a summand of $\pi_k^HS^n$.

Edit. As I understand it, if $H$ is cohesive, then $p^\*: \mathrm{Spaces} \to H$ is supposed to be fully faithful, where $p:H\to\mathrm{Spaces}$ is the unique geometric morphism. Spheres are in the image of $p^\*$, so it ought to follow that that $\pi_k^H S^n = \pi_kS^n$. The only example of cohesive topos I understand is $H=s\mathrm{Spaces}$ (simplicial spaces), and it is certainly true in this case.

  • $\begingroup$ Thank you very much! Are there natural conditions to put on $H$ in order to be able to say something more about the homotopy groups of spheres? For example, what happen if we ask that $H$ is a cohesive $(\infty,1)$-topos? (I know almost nothing about higher topos theory, but looking at the page in the nlab, it seems to be a good restriction. In particular, I think that the example of your second point is not cohesive) $\endgroup$ Commented Oct 27, 2011 at 20:45
  • $\begingroup$ (unless $B$ is contractible) $\endgroup$ Commented Oct 27, 2011 at 20:50
  • $\begingroup$ @Charles the page ncatlab.org/nlab/show/cohesive+%28infinity,1%29-topos gives as an example the cohesive (oo,1)-topos of smooth oo-groupoids. $\endgroup$
    – David Roberts
    Commented Oct 28, 2011 at 1:00
  • 4
    $\begingroup$ @David: Yes, but that is not an example I feel I understand! $\endgroup$ Commented Oct 28, 2011 at 19:04
  • $\begingroup$ Indeed, in a cohesive oo-topos the "categorical" homotopy groups of those spheres discussed here (the "discrete spheres") are the same as those in ooGrpd = Top. But the point is that there also geometric spheres, for which it is different. For instance in Sh_oo(Mfd) there is the geometric n-sphere S^n in Mfd --yoneda--> Sh_oo(Mfd). This being a 0-stack it has trivial categorical homotopy groups. But the cohesive oo-topos also knows how to compute the expected homotopy groups from it, the "geometric homotopy groups". This isn't mysterious:you all know this phenomon from motives/A1-homotopy $\endgroup$ Commented May 4, 2012 at 21:11

Because Charles has already answered (incredibly nicely!) your other questions, I'll just answer your question about a natural group structure. My answer is really just an elaboration on Sam's comment.

As Sam pointed out, there is in fact an $E_k$ structure on the space of all maps $\{ * \to \Omega^k S^n\}$; I'll illustrate this in a moment. But then by the usual Eckman-Hilton argument (or drawing pictures), $\pi_0$ of this space will have the structure of a group for $k\geq 1$, and of a commutative group for $k \geq 2$. The fact that we're taking $S^n$ is not so important here, it's true for any object $X$.

In what follows, I'll let $H(A,B)$ denote the space of morphisms from $A$ to $B$ in the $\infty$-category $H$. In particular, when $A=\ast$ we get the space of global elements of $B$.

By the universal property of pullbacks, a map $\ast \to \Omega X$ is the same as a homotopy coherent map from $\ast$ to the diagram $D := \ast \to X \leftarrow \ast$. Without loss of generality we assume that the two maps $\ast \to X$ in $D$ are the same map. We choose this to be the base point in the space of maps $H(\ast,X)$.

Then a map from $\ast$ to the diagram $D$ is precisely a loop in the Hom-space $H(\ast,X)$. (This is the only key observation--it follows easily from the definition of the Hom Kan complex in an $\infty$-category, if you like.) In other words, $$ H(\ast,\Omega X) \cong \Omega H(\ast,X) $$ where $\Omega$ in the right hand side actually means based loop space, in the usual sense of topology. By induction, the space of global elements of $\Omega^k X$ has the structure of a $k$-fold loop space. And we're finished.


If it may be forgiven to resurrect a very old question, it's worth pointing out that these are not the "homotopy groups of spheres" that appear in synthetic homotopy theory / homotopy type theory. Those are internal group objects in the $\infty$-topos, not external "ordinary" groups. So in particular it is still true that "$\pi_1(S^1)=\mathbb{Z}$" in the terminal category, because both "$\pi_1(S^1)$" and "$\mathbb{Z}$" denote a (or rather the) object of the terminal category and hence are equal --- it's irrelevant that in that case this internal $\mathbb{Z}$ doesn't have the external $\mathbb{Z}$ as its set of global elements. Similarly, in $\infty Gpd^A$ for a set $A$, we have "$\pi_1(S^1)=\mathbb{Z}$", but the internal object $\mathbb{Z}$ has the external group $\mathbb{Z}^A$ as its set of global elements.

There is a conjecture that internally all the homotopy groups of spheres are always isomorphic to the classical ones. This is still open (indeed the meaning of "always" has yet to be formulated precisely), but it seems to be true for all Grothendieck $\infty$-toposes, because inverse image functors preserve spheres, loop spaces, truncations, and the construction of specific finitely presented abelian groups.


In the ordinary homotopy category, the groups $A_k^n=\pi_k(S^n)/(\text{odd torsion})$ are linked by the EHP exact sequence $$ A_k^n \xrightarrow{E} A_{k+1}^{n+1} \xrightarrow{H} A_{k+1}^{2n+1} \xrightarrow{P} A^n_{k-1} $$ It is not hard to show that if $A$ and $B$ are two bigraded groups which each have an EHP sequence, and $f\colon A\to B$ is a morphism that is compatible with the EHP maps, and $f\colon A^1_1\to B^1_1$ is an isomorphism, then $f\colon A\to B$ is an isomorphism. (Perhaps we need the edge condition $A_k^1=B_k^1=0$ for $k>1$ as well.) There is a similar but more complex story for odd primes. Thus, a central question is whether one can define an EHP sequence in any $(\infty,1)$ topos. A closely related question is whether one can define an EHP sequence in any model of homotopy type theory; I am not sure about the exact relationship between these two formulations. Key ingredients include

  • The James construction
  • Homology groups
  • Some general theory of fibrations
  • The theorem of Whitehead, that homology equivalences between simply connected spaces induce isomorphisms of homotopy groups.

I believe that the first three of these are available in homotopy type theory, but the last is not. However, one only needs some specific cases of the Whitehead theorem, and these may hold even if the general result does not.


In any $\infty$-topos, the loop space $\Omega S^1$ of the circle is the free $\infty$-group object on the terminal object. Indeed, there are equivalences natural in a $\infty$-group $G$: \begin{align} \mathrm{Grp}(\Omega S^1, G) &\simeq \mathrm{Map}_*(S^1, \mathbf{B}G) \\ &= \mathrm{Map}_*(\Sigma S^0, \mathbf{B}G) \\ &\simeq \mathrm{Map}_*(S^0, \Omega\mathbf{B}G) \\ & \simeq \mathrm{Map}_*(S^0, G) \\ & \simeq \mathrm{Map}(\mathrm{pt}, G) \end{align} Hence $\pi_1(S^1)$ is the free group on $\mathrm{pt}$ and, for $k > 1$, the group $\pi_k(S^1)$ is trivial.


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