I am looking for a proof of the following fact:

Every compact topological $n$-manifold $M$ has a continuous and not nullhomotopic map $f: S^k \rightarrow M$ for some sphere $S^k$ with $1 \leq k \leq n$.

I came into this while reading a proof of Lyusternik-Fet theorem that every compact Riemannian manifold has a closed non trivial geodesic, but can't prove it, nor find any reference. The statement is equivalent to saying that a compact topological manifold has at least a non trivial homotopy group in the range $1,...,n=dim(M)$. The statement is obvious if $M$ is not simply connected, so let's see what can be said when $M$ is simply connected.

If $M$ is closed, [connected] and orientable, then Poincaré Duality applies; so we have $H_n (M) = \mathbb{Z}$ (singular homology with integer coefficients). Now, even if all $\pi _k$ are trivial for $1 \leq k \leq n-1$, we can still conclude by Hurewicz Isomorphism theorem that $\pi_n(M)=H_n(M)=\mathbb{Z}$, and so we are done.

However I need the more general case of $M$ compact, regardless of orientability, and I don't understand how/why simply connectedness + compactness together imply the existence of a non trivial map from some higher sphere.

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    $\begingroup$ You probably always mean closed (i.e. without boundary) manifolds since the statement is not true for a disk. Could you then not pass to the orientable double cover and apply your reasoning? $\endgroup$ May 21, 2012 at 13:50
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    $\begingroup$ In other words, simply connected implies orientable. $\endgroup$ May 21, 2012 at 14:03
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    $\begingroup$ @Michael : you're right, I always mean closed. Thank you both, now I see how things work; I was not aware of the orientable double cover construction, as well as the implication "simply connected => orientable". $\endgroup$
    – Lor
    May 21, 2012 at 16:15
  • $\begingroup$ Do you happen to know a reference to your theorem? $\endgroup$
    – ABIM
    Dec 27, 2020 at 17:11

1 Answer 1


Any connected covering space of $M$ corresponds to a conjugacy class of subgroups of $\pi_1(M)$, and so if $M$ is simply connected then the only connected covering space of $M$ is $1\colon M\to M$. But if $M$ is non-orientable, then we would have a connected double cover since $M$ is closed, a contradiction, and so $M$ must be orientable.


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