After you introduce the homotopy groups and upon restricting to CW complexes, you ask the question: "Do homotopy groups determine a space up to homotopy equivalence?" With the answer being "No, $\mathbb{R}P^2 \times S^3$ and $\mathbb{R}P^3 \times S^2$ have the same homtopy groups, but are not homotopy equivalent since their homology is different.

But there are special cases where it does. If your space has all trivial homotopy groups, then it is contractible. More generally, all Eilenberg-MacLane spaces of the same dimension and same group are homotopy equivalent.

So to me it makes sense to ask the question: "Are spaces with the homotopy groups of a sphere homotopy equivalent to a sphere?"

The answer to this is no because of a kind of funny construction, just take the product of Eilenberg-MacLane spaces for each homotopy group of $S^n$. Or a more restrained example is $S^2$ and $S^3 \times \mathbb{C}P^\infty$ (again, check homology).

However, all of these examples are infinite dimensional, so what about the following "Is a finite dimensional CW complex with the same homotopy groups as $S^n$ homotopy equivalent to $S^n$?"

  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ Apr 24, 2019 at 15:39
  • $\begingroup$ Does this answer my question in a way I am missing or is it just a related concept? $\endgroup$ Apr 24, 2019 at 16:08
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    $\begingroup$ If $X$ is specifically $n$-dimensional, then the answer is yes. I have no clue about the general question, and I would really be interested in the answer. I also think we can say some things if the dimension is not too big, e.g. $\leq 2(n-1)$ (then we are in the range of connectedness of $X$ + connectedness of $X^{k-1}$ where $k=\dim X$) $\endgroup$ May 15, 2019 at 14:45
  • $\begingroup$ @Max In the case $X$ is n-dimensional, are you saying that the generator of $\pi_n (X)$ is a equivalence by constructing an inverse based on the fact that for n-dimensional spaces $S^n$ represents cohomology? $\endgroup$ Sep 19, 2019 at 21:38
  • $\begingroup$ Yes, something like that should do. But being $n$ dimensional is quite restrictive unfortunately $\endgroup$ Sep 20, 2019 at 5:56

1 Answer 1


I have a partial result.

The only compact manifold (of any dimension) which has the homotopy groups of $S^m$ is $S^m$.

(Really this argument shows any space that has at least two copies of $\mathbb{Z}$ somewhere in its reduced homology can't have the homotopy groups of $S^m$.)

In the case $k=2m-1=1$, the result follows from uniqueness of Eilenberg-MacLane spaces. Now suppose $k=2m-1 >1$. Let $M$ be a manifold with the same homotopy groups as $S^k$. Since it is simply connected, the manifold is orientable and must have its top homology equal to $\mathbb{Z}$.

It is also the case that rationally both $M$ and $S^k$ are equivalent to $K(\mathbb{Q},k)$ which is also a Moore space for $\mathbb{Q}$. Since the homology of the rationalization is equal to the rationalization of the homology, we see that there is exactly one copy of $\mathbb{Z}$ in the reduced homology of $M$. Poincare duality then tells us that $M$ has the homology of a sphere. Since our space is simply connected, this implies it is homotopy equivalent to a sphere. By the Poincare conjecture, we are done.

Now let $k=2m$ and $M$ be a manifold with the homotopy groups of $S^k$. The rationalization of $M$ fits into a fibration $K(\mathbb{Q},4k-1) \rightarrow M_\mathbb{Q} \rightarrow K(\mathbb{Q},2k)$. Because the (rational or integral) cohomology groups of $K(\mathbb{Q},2k-1)$ are $H^0 (K(\mathbb{Q},2k-1)) =\mathbb{Q}=H^{2k-1}(K(\mathbb{Q},2k-1))$ and $H^* K(\mathbb{Q},k)=\mathbb{Q}[x]$, the Serre spectral sequence has a very simple form. There is only one possible nontrivial differential, and the fact that our space has bounded cohomology tells us the differential kills off all but $H^k(M _\mathbb{Q})=\mathbb{Q}$.

The same deduction as in the odd case tells us that our space has the homology of $S^k$, so we are done.

  • $\begingroup$ Perhaps this observation about the number of copies of the integers is reflecting the important result that a space with rational cohomology not generated by 1 element has the Betti numbers of its loop space unbounded. Perhaps one could get further with this fact. $\endgroup$ Nov 11, 2020 at 17:37

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