TITLE: What are examples when the equality of some invariants is good enough in algebraic topology?
QUESTION [16 upvotes]: 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?

REPLY [7 votes]: 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 $n>0$.

REPLY [6 votes]: 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.