Is there a (non-Abelian) homology theory that realizes the following:
Let $X,Y$ be manifolds with complexes $C(X),C(Y)$.
Then $X$ and $Y$ are homotopy-equivalent if and only if $C(X)$ and $C(Y)$ are isomorphic.
Or maybe, the following? ("You need the map…")
Let $X,Y$ be manifolds with complexes $C(X),C(Y)$. Let $f:X\to Y$ be a continuous map which induces the "chain map" $f_*:C(X)\to C(Y)$.
Then $f$ is a homotopy equivalence (it admits a $g:Y\to X$ such that $fg$ and $gf$ are homotopic to the identities) if and only if $f_*$ is an isomorphism.
If yes: which one/ones? An introduction to such theory?
If no: not yet, or is it impossible? Why?
No restriction for the objects in the complexes, they can be groups, modules, groupoids, anything else.
Of course, $X,Y$ may be required to be "nice enough" for the above to work. So, please state also the technical requirements. Moreover:
Same questions, but with cohomology instead of homology?
Best Answer
The answer to the title question, for the usual meaning of "homology theory," is no. Homology is always an invariant of the stable homotopy type of a space, and so no homology theory can distinguish two spaces which are stable homotopy equivalent but not homotopy equivalent. For example, no homology theory can distinguish $S^1 \times S^1$ and $S^1 \vee S^1 \vee S^2$, but the former has nontrivial cup products while the latter does not.
On the other hand, we have the following homology version of Whitehead's theorem, which bypasses the above constraint because it does not come from a homology theory in the usual sense.