Degree of maps and coverings Following a recent question I had concerning degree $1$ maps from spheres, I came up with an assumption, which might either be very easily proven false, or, if not, still hasn't been answered. It goes as follows:
Let $M$ be a closed, orientable $n$-manifold and $f: S^n \to M$ a map of degree $m$. Then $S^n$ is a $k$-sheeted covering space of $M$, where $k$ is some divisior of $m$.
Some remarks:

*

*Since degree $1$ maps from the sphere are precisely homotopy equivalences (see here), the generalized Poincaré-conjecture in the TOP-category follows as a special case for $m=1$.


*This assumtion can very easily be proven false if $S^n$ is replaced by a general $n$-manifold. For $n=2$, there is a simple map of degree $1$ from $S_g$, the surface of genus $g \geq 1$, to $S^2$ as a result from collapsing the 1-skeleton (in its elementary CW-structure) to a point.


*It is also false to assume that $f$ has the homotopy-type of an $m$-sheeted cover. I believe I have read that already for $m=1$, there are some highly non-trivial counterexamples in dimension 6 or higher.
Can somebody enlighten me on this ?!
Edit: In my original question, I assumed that the degree of the cover would have to equal the degree of the Initial map. This is easily proven false by considering maps from $S^n$ to itself that are suspensions of the map $z \to z^m$ on $S^1$. This has degree $m$.
Edit 2: As suggested by Eric Wofsey, an equivalent assumption would be the following:
Let $M$ be a closed, simply-connected $n$-manifold with $M \neq S^n$ and let $f: S^n \to M$ be a map. Then $deg(f)=0$.
Now it becomes obvious that one has to go to dimension at least $4$, in order to find counterexamples.
 A: As mentioned in the comments, such a thing needs to be a rational homology sphere, and in particular a $\Bbb Z[1/d]$-homology sphere, where $d$ is the degree. It also needs to have fundamental group of order coprime to $d$. As in the comments, we may as well assume the manifold is simply connected by passing to the universal cover.
First, there are no simply connected rational homology spheres in dimensions 3 or 4 other than $S^4$, up to homeomorphism - Poincare duality forces the integral homology to be $H^*(S^n)$, and you can pick a degree 1 map $M \to S^n$, which is hence a homotopy equivalence; now invoke the Poincare conjecture.
The first place this is possible is dimension 5, and there are examples in all dimensions at least 5, as produced by Danny Ruberman in this paper. Indeed, he points out that as long as $M$ is a simply connected $\Bbb Z[1/d]$-homology sphere, there is a map $S^n \to M$ of degree $d^r$ for some $r$. A very explicit example of such a manifold is $SU(3)/SO(3)$, a simply connected 5-manifold whose homology is $H_2 = \Bbb Z/2$, $H_0 = H_5 = \Bbb Z$, and all others zero (which I got from here).
Here is a precise statement of the "Hurewicz mod $\mathcal C$" theorem he invokes:
Let $\mathcal C$ be the set of finite abelian groups, or abelian $d$-groups (meaning groups whose orders have the same prime divisors as $d$). (There are other choices of $\mathcal C$ that work, but I don't remember the precise requirements.) Then if $\pi_k(X) \in \mathcal C$ for all $1 \leq k \leq n-1$, then $H_k(X) \in \mathcal C$ as well for $i \leq k \leq n-1$, and the Hurewicz map $h: \pi_n(X) \to H_n(X)$ is an isomorphism mod $\mathcal C$, which means that the kernel and cokernel of $h$ are both in $\mathcal C$.
In our specific case, by assuming that $M$ was a $\Bbb Z[1/d]$ homology sphere means we can induct upwards to see that $\pi_k(X)$ must also be in $\mathcal C$,  and finally conclude that the map $\pi_n(X) \to H_n(X) = \Bbb Z$ has finite abelian cokernel - and in particular, the map has nontrivial image. We get the precise statement about a map of degree $d^r$ by picking $\mathcal C$ the be the set of $d$-groups.
