# If $M$ is a rational homology sphere, is there a map $S^n \to M$ of non-zero degree?

Let $M$ be a closed orientable $n$-dimensional manifold.

It is not hard to show that $M$ admits a degree one map $M \to S^n$ (see here for example). In fact, because $S^n$ admits maps $S^n \to S^n$ of all degrees, $M$ admits a map $M \to S^n$ of any degree.

On the other hand, not every manifold $M$ admits a map $S^n \to M$ of non-zero degree (there are always degree zero maps, constant maps for example). As is shown in this question, if $f : S^n \to M$ is a map of degree $k$, then $ku = 0$ for all $u \in H_i(M; \mathbb{Z})$ where $0 < i < n$. So if $k = \pm 1$ (the sign depends on the choice of orientations), then $M$ is an integral homology sphere, and if $k \neq -1, 0, 1$, then $M$ is a rational homology sphere. What about the converse?

Let $M$ be a rational homology sphere. Does there exist a map $f : S^n \to M$ of non-zero degree?

If the homology of $M$ has torsion of degrees $d_1, \dots, d_m$, then the degree of such a map must be a multiple of $\operatorname{lcm}(d_1, \dots, d_m)$.

In the special case of an integral homology sphere, we can ask a more precise question.

Let $M$ be an integral homology sphere. Does there exist a map $f : S^n \to M$ of degree one?

Update: Thanks to Mike's answer, we can say the following:

There is a map $f : S^n \to M$ of non-zero degree if and only if $M$ and its universal cover are rational homology spheres; moreover, if $M$ is not simply connected, the degree is a multiple of $|\pi_1(M)|$ (which is finite because the universal cover of $M$ is compact). If $n$ is even, then $M$ must be a simply connected rational homology sphere (see my comments on Mike's answer). Interestingly, when $n = 4$, it follows that $M$ must be $S^4$.

As for the second question, the only integral homology sphere which admits a degree one map $f : S^n \to M$ is $M = S^n$ (see my comment below).

• Just as Mike's answer appeared, I remembered that a degree one map induces a surjection of fundamental groups, so the answer to the second question is no unless $M = S^n$. – Michael Albanese May 8 '17 at 14:55
• I am not sure if it is relevant, but it is related: Manifolds that have degree 1 maps from spheres by Kevin Iga. He proves that such a map exists iff $M$ is a homotopy sphere. – Chris Gerig May 12 '17 at 23:59
• @ChrisGerig: Thanks, it is certainly relevant. It is not hard to show. You only need to know that degree one maps induce surjections on fundamental groups and that $M$ must be a homology sphere (the argument I linked to seems to be simpler than the one used by Kevin Iga, but essentially the same). Once you know that, $M$ must be homeomorphic to $S^n$ (Whitehead + Poincaré conjecture). In the category of smooth manifolds, that means $M$ is a homotopy sphere. The map $\operatorname{id} : S^n \to M$ is a degree one map (it can also be homotoped to a smooth map and it will still have degree one). – Michael Albanese May 13 '17 at 1:45

When $\Sigma$ is simply connected, and its homology is $k$-torsion, then there is a degree $d$ map $f: S^n \to \Sigma$ for some $d$ with $\gcd(d,k) = k$. Note that $H_*(\Sigma;\Bbb Z)$ is $k$-torsion for all $* \leq n-1$; so we should apply Serre's "Hurewicz mod C" theorem to the smallest Serre class that includes $k$-torsion groups. This is the class of all groups whose order has the same prime divisors as $k$. Applying Hurewicz mod C we see the map $\pi_n(\Sigma) \to H_n(\Sigma) = \Bbb Z$ is an isomorphism mod C; this means that $\pi_n(\Sigma)$ is $\Bbb Z$ plus a $k^r$-torsion group, and the map $\Bbb Z \to \Bbb Z$ has $k^r$-torsion cokernel. In particular, there is a map $f: S^n \to \Sigma$ of degree $d>0$ where $d \mid k^r$ for some $r$. (This simplifies drastically when $k$ is prime, and to the classical Hurewicz theorem for integer homology spheres.) I learned this argument from this paper of Danny Ruberman's.
When $\Sigma$ is not simply connected, any map of nonzero degree from a sphere factors through the universal cover. So for such a map to exist, $\widetilde \Sigma$ must be a rational homology sphere. In this case, the above argument produces a map of nonzero degree. Any map $S^n \to \Sigma$ has degree divisible by $|\pi_1(\Sigma)|$, so even if $\Sigma$ is an integer homology sphere (eg the Poincare homology sphere), you still cannot get maps of degree 1. (In fact, note that a simply connected integer homology sphere is homotopy equivalent and hence homeomorphic to $S^n$.)
• Thanks as always. I should have remembered that a degree one map induces a surjection on fundamental groups. If $n$ is even and $\Sigma$ is not simply connected, $\chi(\tilde{\Sigma}) = |\pi_1(\Sigma)|\chi(\Sigma) = 2|\pi_1(\Sigma)|$ so $\tilde{\Sigma}$ will not be a rational homology sphere. In odd dimensions, I guess Lens spaces are examples of rational homology spheres with universal cover a rational homology sphere. – Michael Albanese May 8 '17 at 15:02
• There are also examples of simply connected rational homology spheres like $SU(3)/SO(3)$ which can have potentially interesting quotients. – user98602 May 8 '17 at 15:04
• I should have been more careful in my first comment. If $\Sigma$ is an even-dimensional rational homology sphere which is not simply connected, then $\tilde{\Sigma}$ is not a rational homology sphere. To see this, note that either $\pi_1(\Sigma)$ is finite, in which case the argument I gave in the first comment holds, or $\pi_1(\Sigma)$ is infinite in which case $\tilde{\Sigma}$ is non-compact and therefore not a rational homology sphere. – Michael Albanese May 10 '17 at 14:50