Jie Wu

Homotopy Rigidity of the Functor \(\Sigma \Omega\) on co-\(H\)-spaces--Joint with Jelena Grbic

slide: PDF.
video: comming soon.
By means of homotopy rigidity of a functor \(F\) on a subcategory of spaces, we refers to the property that \(F(X)\simeq F(Y)\) implies \(X\simeq Y\). The question that we considered concerns the homotopy rigidity of the functor \(\Sigma \Omega\). This functor is not rigidity for general spaces, for instance, if \(X=\mathbb{C}\mathrm{P}^2\) and \(Y=S^5\times \mathbb{C}\mathrm{P}^\infty\), then \(\Sigma\Omega X\simeq \Sigma \Omega Y\) but \(X\not\simeq Y\). We consider the cases that both \(X\) and \(Y\) are co-\(H\)-spaces. Our statement is as follows:
Suppose that both \(X\) and \(Y\) are simply connected \(p\)-local finite dimensional co-\(H\)-spaces with the property that \(\Sigma \Omega X\simeq \Sigma \Omega Y\). Then \(X\simeq Y\).
If both \(X\) and \(Y\) are atomic, then the canonical composite \(X\to \Sigma\Omega X\simeq \Sigma \Omega Y\to Y\) is a homotopy equivalence. In general cases, the homotopy equivalence \(\Sigma\Omega X\simeq \Sigma\Omega Y\) may move around atomic pieces and so the above composite may not be a homotopy equivalence in general.
Our method for proving the above statement is to produce certain functorial suspension splittings of the self smash products of the loops on co-\(H\)-spaces and then to make a kind of counting on irreducible pieces for concluding that \(X\) and \(Y\) must have the same irreducible pieces with multiplicity.
It seems that the above statement should hold for finite type co-\(H\)-spaces. Our method of the counting game does not seem working well for finite type cases.