Homotopy Rigidity of the Functor \(\Sigma \Omega\) on
co\(H\)spacesJoint with Jelena Grbic
video: comming soon.
abstract:
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:
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.
