MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $[X,S^3]$ be the set of homotopy classes of maps from the pointed CW complex X to the pointed 3-sphere. The group structure on S^3 endows this set with a group structure. Is this group ever noncommutative?

share|cite|improve this question
up vote 10 down vote accepted

Yes, by Yoneda's lemma. To generalize a bit, you have a group object $G$ in some category with finite products, and you ask whether the functor it represents is pointwise abelian. This is the same as asking whether two natural transformations $[-,G\times G]=[-,G]\times[-,G]\to[-,G]$ are equal (one being $(f,g)\mapsto f\cdot g$ and the other being $(f,g)\mapsto g\cdot f$). By Yoneda, this is this case iff the representing maps $G\times G\to G$ are equal. That is, the group $[X,G]$ is always abelian iff the multiplication map on $G$ is itself commutative. In the case of $G=S^3$ in the homotopy category of spaces, the group operation is not (homotopy-)commutative, though I don't know an easy proof of that off the top of my head.

If you want an explicit example of an $X$ that makes $[X,G]$ nonabelian, you just have to unravel the proof of Yoneda's lemma. That means you set $X=G\times G$ and consider the identity map $G\times G\to G\times G$, which corresponds the two projection maps $p_0,p_1\in[G\times G,G]$ when you identify $[G\times G,G\times G]=[G\times G,G]\times [G\times G,G]$. So in your case, the group $[S^3\times S^3, S^3]$ is nonabelian, and in particular the two projection maps do not commute.

share|cite|improve this answer
Eric, if $[S^3 \times S^3,S^3]$ is a noncommutative group, which noncommutative group is it? (It looks like an extension of a subgroup of $[S^3$ wedge $S^3,S^3] = \mathbf{Z}+\mathbf{Z}$ by a quotient of $\pi_1([S^5,S^3]) = \mathbf{Z}/12$ to me.) Which element is $gh$ and which element is $hg$? – ya-tayr Mar 14 '14 at 2:56
That's a great question that I hope someone else can answer! – Eric Wofsey Mar 14 '14 at 2:57
That $S^3$ is not homotopy commutative was apparently proved in [Samelson,H.,"Groups and wpaces of loops, Comm.Math.Helv.,28,278-87 (1954)] – Mariano Suárez-Alvarez Mar 14 '14 at 5:22
@SpecialUnit2, why a subgroup of $Z\oplus Z$ and not the whole thing, and why a quotient of $Z/12$ and not the group itself? – Mariano Suárez-Alvarez Mar 14 '14 at 5:52
@MarianoSuárez-Alvarez, he was considering a certain long exact sequence from which it was not a priori clear that the corresponding maps are surjective and injective. They are, however, consider my answer to… – Achim Krause Mar 14 '14 at 6:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.