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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?

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Homology with local coefficient, see Whitehead's theorem. –  Fernando Muro Apr 21 at 7:11
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Homology is algorithmic while homotopy is not. At least that's the idea. –  Wlodzimierz Holsztynski Apr 21 at 7:17
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What sort of "nonabelian homology" do you have in mind exactly? Most generalizations of homology that I know of do not come from any sort of complexes. –  Eric Wofsey Apr 21 at 13:37
    
@WlodzimierzHolsztynski Homotopy groups of CW-complexes with no 1-cells can be computed via an algorithm. –  Fernando Muro Apr 22 at 20:47

3 Answers 3

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.

Theorem: Suppose $f : X \to Y$ is a map of path-connected spaces inducing an isomorphism on $\pi_1$ and an isomorphism on homology with all local coefficients. Then $f$ is a weak equivalence. (In particular, if $X, Y$ are simply connected, we only need $f$ to induce an isomorphism on homology.)

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If $X$ and $Y$ are finite type nilpotent spaces, then $X$ and $Y$ are weakly equivalent if and only if their cochain complexes are quasi-isomorphic as $E_\infty$-algebras. Moreover, assuming $X$ and $Y$ are also CW-complexes, two maps $f,g:X\to Y$ are homotopic if and only if they induce homotopic maps of $E_\infty$-algebras at the level of cochain complexes. In particular, if a map between such spaces $f:X\to Y$ induces a quasi-isomorphism in singular cohomology, then it is an homotopy equivalence. All these results are consequences of more precise theorems from this paper:

M. Mandell, Cochains and homotopy type, Publ. Math. IHES 103 (2006), 213-246.

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What you could be looking for is Theorem 6 of J.H.C. Whitehead's "Combinatorial Homotopy II" Bull. Amer. Math. Soc. 55 (1949) 453-496. For a CW-complex K he defines what he calls a "homotopy system" $\rho (K)$ and the theorem says that for a map $f:K \to L$ of CW-complexes, $f$ is an equivalence if and only if $\rho(f)$ is an equivalence. However to get "only if $\rho(f)$ is an isomorphism" is unlikely for this functor.

We have come to call $\rho(K)$ the fundamental crossed complex $\Pi(K_*)$ of the filtered space $K_*$, the filtration being by skeleta. Many aspects of Whitehead's work are dealt with in the book Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids, EMS Tract Vol 15, 2011.

To add to the above, a recent preprint available from Graham Ellis' preprint page at Galway

G. Ellis and Le Van Luyen, "Homotopy 2-types of low order", preprint.

lists isomorphism classes and quasi-isomorphism classes of some crossed modules of low order.

This answer is related to my answer to this mathoverflow question.

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This is incredibly interesting! –  geodude Apr 24 at 11:07

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