I read somewhere the following sentence: the homotopy type of an aspherical manifold is determined by its fundamental group. Recall that $M$ is called aspherical if $\pi_n(M)=0$ for $n>1$. By Whitehead's theorem we know if $f\colon X \to Y$ is a mapping between spaces having a homotopy type of a CW-complex and $f$ induces isomorphisms on homotopy groups, then $f$ is a homotopy equivalence. Since every smooth manifold has a homotopy type of a CW-complex we can use this theorem: however if $N$ is another CW-complex (up to homotopy) and $N$ has the same homotopy groups as $M$ (where $M$ is aspherical) why do we know that there is a single mapping $f\colon M \to N$ inducing an isomorphism on homotopy groups?


The thing you're stating is not really about manifolds. An Eilenberg MacLane space $K(G,n)$ is a space $X$ with $\pi_n X = G$ and all other homotopy groups equal to zero. These are unique up to weak homotopy equivalence (hence, by Whitehead, CW-complex Eilenberg MacLane spaces are unique up to homotopy equivalence). In fact, there exists a unique continuous map $K(G,n) \to K(H,n)$ up to homotopy that induces a given homomorphism on the level of $\pi_n$. The proof is, essentially, to just do what you want on the 1-skeleton and check that the 2-skeleton doesn't get mad when you do so. An actual proof is given in Hatcher, 1B.9, in the case $n=1$, and 4.30 for the general case.

The thing you might be thinking about in the case of manifolds is even cooler: the Borel conjecture says that if $M, N$ are closed aspherical manifolds with isomorphic fundamental groups, there is a homeomorphism $M \to N$ inducing any given isomorphism on the fundamental groups. This seems likely to be true, and it's known for a wide class of manifolds, including manifolds of dimension $n \leq 3$ (by results of Waldhausen and geometrization) and all hyperbolic manifolds (by the even stronger Mostow rigidity).

  • $\begingroup$ You meant "these are well defined up to weak homotopy equivalence" or maybe "these are unique up to weak homotopy equivalence"? $\endgroup$ – truebaran Oct 2 '15 at 21:58
  • $\begingroup$ Sure, unique. If you're trying to define a space up to weak homotopy equivalence that's what it would mean to be well-defined :-) $\endgroup$ – user98602 Oct 2 '15 at 22:18

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