Following Spanier's book on algebraic topology chapter $1$, section $6$ about suspensions, I'm wondering about the following questions:

1) We know that $S^n$ is an $H$-cogroup for all $n\geq1$ because $S^n = S(S^{n-1})$ where $S$ is the "reduced suspension" functor, which has been proven to take values in the category of $H$-cogroups. Is it also true that $T^n = S^1 \times \dots \times S^1$ is an $H$-cogroup? If so, how does one show this?

2) Suppose the answer to the above question is yes (which I intuitively think it is), then is there an $H$-cogroup isomorphism $S^n \to T^n$? Or at least an $H$-cogroup homomorphism? If so, then there would be a natural transformation between the functors $\pi_{S^n}$ and $\pi_{T^n}$, or even a natural equivalence if the map $S^n \to T^n$ is isomorphic (but it cannot be a homeomorphism. I'm a bit confused here, can it they be $H$-cogroup isomorphic but not homeomorphic?)

3) What I'm really after is a correspondence between $\pi_{S^n}(X)$ and $\pi_{T^n}(X)$ for given $X$ and $n\geq1$. The two are group-isomorphic, right? Is the above the right way to show this? If not, what is?

Note on notation: $\pi_Q$ is a covariant functor from $\mathsf{hTop}_{\bullet}$ to $\mathsf{Grp}$ which sends $X\in \operatorname{Obj}(\mathsf{hTop}_{\bullet})$ to the homotopy classes of basepoint-preserving maps $\operatorname{hom}_{\mathsf{hTop}_{\bullet}}(Q,X)$ and a morphism $[f]\in \operatorname{hom}_{\mathsf{hTop}_{\bullet}}(X,Y)$ to $$([g]\mapsto[f\circ g])\in \operatorname{hom}_{\mathsf{hTop}_{\bullet}}(\operatorname{hom}_{\mathsf{hTop}_{\bullet}}(Q,X), \operatorname{hom}_{\mathsf{hTop}_{\bullet}}(Q,Y)).$$

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    $\begingroup$ $\pi_k T^n$ is trivial for all $k>1$. There are lots of homotopically nontrivial maps between tori. $\endgroup$ – user98602 Aug 29 '15 at 23:38
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    $\begingroup$ Wait, why is $\pi_{S^n}(X)$ isomorphic to $\pi_{T^n}(X)$? Last I remember, $\pi_{S^2}(S^1)$ was trivial, but $\pi_{T^2}(S^1)$ is non-trivial. $\endgroup$ – Dan Rust Aug 29 '15 at 23:39
  • $\begingroup$ I guess my pretense was wrong, and I'm glad to have found out! $\endgroup$ – PPR Aug 29 '15 at 23:42
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    $\begingroup$ I believe that the only connected manifold cogroups are spheres, and that this is proved somewhere on this site. $\endgroup$ – user98602 Aug 29 '15 at 23:46
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    $\begingroup$ $S^n$ and $T^n$ are not even homotopy equivalent (except when $s = 1$; they can be distinguished by $H_1$), so there's no hope for them to be equivalent as anything more structured. $\endgroup$ – Qiaochu Yuan Aug 30 '15 at 0:00

1) As is shown here, if $M$ is a compact connected $n$-dimensional manifold which is a co-$H$-space, then $M$ is simply connected and has the same integral homology as $S^n$ (i.e. $M$ is a $\mathbb{Z}$-homology sphere). By a version of Whitehead's Theorem (see Hatcher's Algebraic Topology, Corollary $4.33$), $M$ has the same homotopy groups as $S^n$ (i.e. $M$ is a homotopy sphere), so by the generalized Poincaré conjecture, $M$ is homeomorphic to $S^n$; note, as $M$ was assumed to be connected, $n > 0$.

As every $H$-cogroup is a co-$H$-space (because every group is a monoid), we see that the only compact connected orientable manifolds which could possibly be $H$-cogroups are the spheres $S^n$ for $n > 0$. As $n > 0$, $S^n = \Sigma S^{n-1}$, so we see that they are in fact $H$-cogroups (and hence co-$H$-spaces).

In summary, if $M$ is a compact connected manifold, then:

  • $M$ is a co-$H$-space if and only if $M$ is homeomorphic to $S^n$ for some $n > 0$.
  • $M$ is an $H$-cogroup if and only if $M$ is homeomorphic to $S^n$ for some $n > 0$.

In particular, there is no compact connected manifold which is a co-$H$-space but not an $H$-cogroup.

By the above, we see that $T^n$ is not an $H$-cogroup for $n > 1$. For $n = 1$, we see that $T^1 = S^1$ is an $H$-cogroup.

2) As $T^n$ is not an $H$-cogroup for $n > 1$, there is no $H$-cogroup isomorphism or homomorphism $S^n \to T^n$. For $n = 1$, $S^1$ and $T^1$ are homeomorphic (via the identity).

3) As $T^n$ is not an $H$-cogroup for $n > 1$, $\pi_{T^n}(X)$ is not naturally a group in general; in particular, it cannot be isomorphic as a group to $\pi_{S^n}(X)$.

Although $\pi_{T^2}(X)$ is not naturally a group is general, the set $[T^2, X]_*$ of pointed homotopy classes of maps $T^2 \to X$ can be computed in terms of $\pi_1(X)$ and Whitehead products, see this MathOverflow question which asks about maps from the product of two arbitrary spheres. I don't know if the argument there can be generalised to $T^n$.

  • $\begingroup$ Would you agree that if someone has difficulties distinguishing $S^n$ and $T^n$ then this answer might not prove that useful? (It is good for reference, however.) I think one should address the fact the OP is failing to distinguish $T^n$ and $S^n$. $\endgroup$ – Pedro Tamaroff Jul 18 '16 at 3:40
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    $\begingroup$ @PedroTamaroff: I don't think the OP has difficulties distinguishing these two spaces, I'm not sure why you think they do. $\endgroup$ – Michael Albanese Jul 18 '16 at 5:16
  • $\begingroup$ It seems you haven't read the question with enough care, then. $\endgroup$ – Pedro Tamaroff Jul 18 '16 at 5:31
  • $\begingroup$ @MichaelAlbanese, thank you very much, now I have some order in my confusion about 1) and 2). What I was really after is 3) which I will now look into. $\endgroup$ – PPR Jul 18 '16 at 10:24

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