nLab group homotopy

Tentative

If group cohomology studies morphism between Eilenberg-MacLane objects /spaces $[K(A, n), K(B, m)]$, usually in the case $n = 1$, one might imagine group homotopy would study $[M(A, n), M(B, m)]$, especially for $m = 1$, where the $M(A, n)$ are Moore spaces. (Or should we take co-Moore spaces?)

One difference is that $K(G, 1)$ is defined for nonabelian $G$, whereas $M(G, 1)$ is only defined for abelian $G$.

David Corfield: An asymmetry in the structure and co-structure of the circle? $S^1$ is both $K(\mathbb{Z}, 1)$ and $M(\mathbb{Z}, 1)$, and supports an abelian group structure but a nonabelian cogroup structure, so first cohomology must be abelian, but not necessarily first homotopy.

Urs Schreiber: might it be more an asymmetry in the definition: since Moore space involves homology which is (in the sense described there) the abelian version of homotopy.

David Corfield: some people, e.g., Hatcher, talk of homotopy with coefficients as probing spaces with co-Moore spaces, based on cohomology, so I’m not sure if your point is right. The ordinary homotopy groups come from maps out of co-Moore spaces for $G = \mathbb{Z}$, i.e., the spheres. These are all cogroups, but not abelian in the case of the 1-sphere. Cohomology groups come from mapping into E-M spaces. In the case of $K(\mathbb{Z}, 1)$ we again have the circle, but this time it supports an abelian group. I guess $K(\mathbb{Z}, 1)$ comes from forming loop spaces from above, higher $n$, so will be abelian, whereas $M(\mathbb{Z}, 1)$ comes from suspending from below, and only goes abelian at the second stage.

According to Baues, the groups $[M(A, n), M(B, m)]$ are not at all understood. This is clear, since for $A = B = \mathbb{Z}$, these are the homotopy groups of the spheres.

Need to say something about homology decomposition and homotopy decomposition (see Baues, Homotopy Types. Homology decomposition fails to be natural, but Baues introduces ‘boundary invariants’ which are natural.