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.