Let $\mathcal{Man}$ be the set of all connected closed orientable manifolds up to homeomorphism. It is an abelian semigroup with respect to the cartesian product $\times$ (multiplication) and unit $\ast$ (zero dimensional manifold ).
In the same way lets define $\mathcal{Man}^{h}$ as the set of all connected closed orientable manifolds up to homotopy equivalence. It is an abelian semigroup with respect to the cartesian product $\times$ (multiplication) and unit $\ast$ (zero dimensional manifold ).
There is an evident (surjective) homomorphism of abelian semigroups :
$$i:\mathcal{M}\rightarrow \mathcal{M}^{h} $$ $$M\mapsto M $$
which induces a homomorphism of abelian groups after group completion (Grothendieck construction). $$K(i):K(\mathcal{Man})\rightarrow K(\mathcal{Man}^{h}) $$
I do believe that my questions are naive, but let me try to ask them:
- Is there some (partial) understanding of the groups $K(\mathcal{Man})$ and $ K(\mathcal{Man}^{h})$ ? (e.g. Do they have torsion elements? what are other relations...?)
- Is there some (partial) understanding of the Kernel of the map $K(i)$ ? (torsion elements, generators, other relations,...)
First Edit: let me ask a more direct question: suppose that $X$ and $Y$ are connected closed orientable manifolds such that $X\times X$ is homeomorphic to $Y\times Y$, is $X$ homeomorphic to $Y$, or is there a counterexample?
Second Edit (Motivation): I'm not a specialist of this field, but reading some articles and some questions/answers in mathoverflow, it seems to me that lots of conjecture and theorems can be reformulated in the setting of the abelian semigroups $\mathcal{Man}$ and $\mathcal{Man}^{h}$.
- The semigroups $\mathcal{Man}^{h}$ and $\mathcal{Man}$ do not have cancellation property: the equation $a+b=a+c$ does not imply that $b=c$ in general.
- the semigroup $\mathcal{Man}$ does have torsion elements (Lens spaces $L(7,1)$ and $L(7,2)$), I do believe that $\mathcal{Man}^{h}$ does have also torsion elements.
- Let $T^{1}$ be the torus of dimension 1 (circle). Let $\mathcal{T} \subset \mathcal{Man}$ be the smallest subsemigroup containing $T^1$ and all $M$ such that $M$ is homotopy equivalent to $T^{n}$ for some natural number $n$. If we can prove that $(\mathcal{T},\times,\ast) $ is isomorphic to $(\mathbb{N},+,0)$ then the Poincare conjecture follows in dimension 3 (because it will imply that if $T^3$ is homotopy equivalent to $M$ then $T^3$ is homeomorphic to $M$).
- The Borel conjecture can be formulated as follows: Let $\mathcal{Man}_{aspherical}$ be the subsemigroup generated by aspherical closed connected orientable manifolds and let Let $\mathcal{Man}_{aspherical}^{h}$ be the subsemigroup generated by aspherical closed connected orientable manifolds (up to homotopy equivalence) then the Borel conjecture is true if the natural homomorphism $\mathcal{Man}_{aspherical}\rightarrow \mathcal{Man}_{aspherical}^{h} $ is an isomorphism.
- In general it is very hard to work with semigroups even if they are abelian. For all these reasons, it seems that is reasonable to pass to the Grothendieck construction and study the associated abelian groupes, of course some important information will be lost (the failure of the cancelation property).
