What are examples when the equality of some invariants is good enough in algebraic topology?

Question

As far as my understanding goes, most of the tools of algebraic topology (homotopy groups, homology groups, cup product, cohomology operations, Hopf invariant, signature, characteristic classes, knot invariants...) are mostly designed with "discrimination" in mind: if two spaces/maps/bundles/... do not have the same invariant, then they are not "equivalent" for the correct notion of equivalence. And of course two things having the same invariants in some arbitrary big class is not enough for the things to be equivalent (e.g. there are spaces that have the same homotopy and homology groups that are not homotopy equivalent).

But sometimes computing only a few of those invariants is enough:

• If a knot has the same crossing number as the unknot (zero), then it's the unknot. (Maybe a silly example.)
• If a space has the same homotopy groups as an Eilenberg–MacLane space (i.e. they all vanish except one), then it has the same homotopy type as an Eilenberg–MacLane space.
• If a closed (connected) 3-manifold has the same fundamental group as the 3-sphere, then it's actually homeomorphic to the 3-sphere (the famous Poincaré conjecture / Perelman theorem).
• The result that made me think of this question: under some technical conditions, if a simplicial operad has the same rational homology as the little $n$-disks operad (as Hopf $\Lambda$-operads), then it is rationally equivalent to the little $n$-disks operad (Fresse–Willwacher).

The above results are the ones I could think of (and are probably very skewed towards my interests).

What are some other examples of situations where an algebro-topological invariant, that in general is not enough to characterize an object, is enough to actually completely characterize the object?

Answer 1

Classification of simply-connected 4-5 and 6-manifolds.

• The homotopy type of a simply connected compact 4-manifold only depends on the intersection form on the second homology group. Moreover a famous theorem of M. Freedman implies that the homeomorphism type of the manifold only depends on this intersection form, and on a $\mathbb{Z}/2\mathbb{Z}$-invariant, the Kirby–Siebenmann invariant
• Barden in 1965 has given the following classification result for simply-connected compact smooth manifolds. Let $M$ and $N$ be simply-connected, closed, smooth 5-manifolds and let $\phi:H_2(M)\cong H_2(N)$ be an isomorphism preserving the linking form and the second Stiefel-Whitney class. Then $\phi$ is realised by a diffeomorphism.
• A system of invariants is also known for simply-connected 6-manifolds, it is described in the paper "CUBIC FORMS AND COMPLEX 3-FOLDS" by Okonek and Van de Ven (L'enseignement mathématique, 1995).

Answer 2

• Stiefel-Whitney numbers detect (unoriented) bordism classes and together with Pontryagin numbers, they determine oriented bordism classes.
• Ranicki's total surgery obstruction of a finite $n$-dimensional Poincaré complex vanishes if and only the space is homotopy equivalent to a closed topological manifold of the same dimension.

Answer 3

Maybe this is a little too on the nose, but: obstruction theory to extension and lifting problems in the sense of, e.g., Whitehead Chapter 6 or Davis-Kirk Chapter 7. Here vanishing of the obstruction classes (which can sometimes be guaranteed by properties of the spaces) suffices to demonstrate the existence of extensions of maps or lifts of maps from the base space to the total space of a fibration.

Answer 4

Here are three results where the conclusion is not exactly an equality in the topological category, but is something quite close to that. All three are of fundamental importance.

• If $f\colon X\to Y$ is a map of simply connected CW complexes or of connective CW spectra, and $H_*(f)$ is an isomorphism, then $f$ is a homotopy equivalence.
• If $f\colon X\to Y$ is a map of finite spectra, and $H_*(f;\mathbb{Q})=0$, then $nf=0\colon X\to X$ for some $n>0$.
• If $f\colon\Sigma^d X\to X$ is a map of finite spectra, and $MU_*(f)=0$, then $f^n=0\colon\Sigma^{nd}X\to X$ for some $n>0$.

Answer 5

Here is the easiest generalization of the fact you cite about Eilenberg-MacLane spaces. Spaces $X$ with exactly two nontrivial homotopy groups $\pi_n(X), \pi_m(X), 2 \le n < m$ are classified (up to (weak) homotopy equivalence) by these two homotopy groups together with one additional Postnikov invariant, which is a cohomology class in $H^{m+1}(B^n \pi_n(X), \pi_m(X))$, where $B^n A$ denotes the $n$-fold delooping $K(A, n)$ of a discrete abelian group $A$. This class classifies the fibration

$$B^m \pi_m(X) \to X \to B^n \pi_n(X).$$

If $n = 1$ then we need the additional data of the action of $\pi_1(X)$ on $\pi_m(X)$, and then the cohomology above is group cohomology with nontrivial / local coefficients.

Example. The $3$-truncation of $S^2$ has two nontrivial homotopy groups $\pi_2(X) \cong \pi_3(X) \cong \mathbb{Z}$, and all other homotopy groups vanish. The Postnikov invariant is a class in $H^4(B^2 \mathbb{Z}, \mathbb{Z}) \cong H^4(\mathbb{CP}^{\infty}, \mathbb{Z}) \cong \mathbb{Z}$, and I believe it turns out to be a generator.

The natural generalization of this fact is the theory of Postnikov towers, although there the relevant Postnikov invariants are defined in terms of spaces defined in terms of other Postnikov invariants, so it's trickier to tell whether the're different or the same.