# Divisibility in homology/homotopy

I have a simply-connected CW-complex $F$ of finite-type, and I know that the imprimitivity of its particular integral homology is divisible by an odd prime $p$; that is, $$\forall n,\exists \delta, \forall x : H_n(F,\mathbb{Z}), \nabla x = x\otimes 1 + 1 \otimes x + p (\delta x)$$ for a very nice delta.

Can I already conclude that an equivalent complex can be chosen with cellular attaching maps also divisible by $p$? Or do I need to know something more about the complex?

• Is $x$ a chain or a homology class? Can you ask the question more precisely? – Tom Goodwillie Jun 9 '14 at 22:17
• Here, $x$ is meant to be a homology class, though for my particular complex it could as easily be a chain --- but then I'd have to check the comultiplication again. If mathoverflow will let me, I'll adjust the question. – Jesse C. McKeown Jun 9 '14 at 23:10
• I don't know what "the cellular attaching maps are divisible by $p$" means. The same space can have more than one cell structure. – Tom Goodwillie Jun 10 '14 at 0:20
• The cone of any element in the homotopy groups of spheres which is not detected by the Hopf invariant will even have primitive homology. Pick any of those which is not divisible by p, and you have a counterexample, if I understand your question correctly. – Achim Krause Jun 10 '14 at 6:45
• Duplicating what Achim Krause said somewhat: Any suspension has all homology primitive, but you can attach an $(n+4)$-cell to $S^n$ for large $n$ by a map that generates a summand of order three in the homotopy group. – Tom Goodwillie Jun 10 '14 at 11:46