4

One way to prove that a topological group has abelian fundamental group is to point out that the two group operations are homomorphisms for each other and apply the Eckmann-Hilton argument. (Admittedly, it's just as easy to do directly.) The question arises: if $A$ is a cogroup object and $B$ a group object in some category $\mathcal C$, is the naturality of the comultiplication and multiplication enough to make $\hom_{\mathcal C}(A,B)$ an abelian group, i.e. enough to make E-H applicable? The answer to this is no: the other well known nontrivial cogroups are free groups with a choice of generating set, and for instance the group of homomorphisms $\mathbb{Z}\to G$ is isomorphic to $G$.

So, I'd like to know whether there's any good reason that the most obvious cogroups in homotopy topological spaces have a comultiplication that plays nicely with every multiplication. Is there an abstract condition that implies this?

EDIT: Just to point out that the same argument shows that the maps from any suspension into a topological group form an abelian group. So if it's special, it's something special to suspensions, and not just to the circle.

Kevin Carlson
  • 52,457
  • 4
  • 59
  • 113
  • 1
    Regarding your edit, note that $\hom(\Sigma X, Y) = \hom(S^1 \wedge X, Y) \cong \hom(X, \hom(S^1, Y)) = \hom(X, \Omega Y)$ which is a mapping space into an H-group. In fact, it's something special about any smash product with a cogroup: $\hom(X \wedge H, Y)$ will be a group for $H$ a well-behaved topological cogroup. – Najib Idrissi Sep 06 '14 at 07:26
  • What argument do you have in mind that is more direct than the Eckmann-Hilton argument? – Qiaochu Yuan Sep 06 '14 at 07:42
  • Just a comment to say that you have to be working in the category of pointed spaces and homotopies respecting base point. – Ronnie Brown Sep 06 '14 at 10:39
  • @Qiaochu Yuan Most of the argument is the same: $a.b=ae.be=ab,b.a=eb.ae=ab$, but the units for . and for juxtaposition are already known to coincide. – Kevin Carlson Sep 06 '14 at 18:00

1 Answers1

7

The answer to this is no

The answer is yes. Remember that a group is a group object in the category of groups iff it's abelian; this is also by Eckmann-Hilton.

Sketch: if $B$ is a group object, then $\text{Hom}(-, B)$ is a functor to groups. If $A$ is a cogroup object, with $\Delta : A \to A \sqcup A$ the comultiplication, then the composite

$$\text{Hom}(A \sqcup A, B) \cong \text{Hom}(A, B) \times \text{Hom}(A, B) \to \text{Hom}(A, B)$$

is a homomorphism with respect to the group structure imposed by $B$ by functoriality.

Qiaochu Yuan
  • 419,620
  • Ugh, of course. Thanks! – Kevin Carlson Sep 06 '14 at 18:02
  • I skipped a detail here which is to show that the isomorphism $\text{Hom}(A \sqcup A, B) \cong \text{Hom}(A, B) \times \text{Hom}(A, B)$ respects the group structure on each side imposed by $B$. I think this shouldn't be hard though. – Qiaochu Yuan Sep 06 '14 at 18:39