Jie Wu

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

slide: PDF.
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:
 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.