TITLE: Converse to Hopf degree theorem
QUESTION [15 upvotes]: Below, I mean smooth oriented closed connected manifolds and smooth maps (but am happy to hear about the topological category, or unoriented manifolds, etc instead).
Say that $X^n$ has the Hopf property if two maps $f_0,f_1 : M^n\to X^n$ are homotopic if and only if they have the same degree.
Say that $X$ has the self-Hopf property if the Hopf property holds for $M=X$.
The Hopf degree theorem says that $S^n$ has the Hopf property. It's easy to see that $T^n$ doesn't have the (self-)Hopf property. My question is:

If $X$ has the (self-)Hopf property is it homeomorphic to $S^n$?

REPLY [7 votes]: A manifold $M$ of dimension $>1$ with $H^1(M) \neq 0$ does not have the self-Hopf property. For if so, then there is a map $f: M \to S^1$ that induces a non-trivial homomorphism $f_*: H_1(M) \to \mathbb{Z}$. Compose this with a map $g: S^1 \to M$ carrying a non-torsion homology class to get a map $M \to M$ that induces a non-trivial map in homology and hence is not null-homotopic. But it factors through $S^1$ and so has degree $0$.
This was a counterexample to an attempted proof that a hyperbolic manifold with trivial isometry group has the self-Hopf property. (Hyperbolic manifolds with trivial isometry group exist, at least in dimension $3$.) The argument is that a map of non-zero degree must in fact have degree one (using Gromov's norm) and hence are homotopic to an isometry. (This uses Gromov's proof of Mostow rigidity; see Haagerup, Uffe; Munkholm, Hans J.
Simplices of maximal volume in hyperbolic n-space. Acta Math. 147 (1981), no. 1-2, 1–11.) So by the hypothesis on the isometry group must be homotopic to the identity.  But the previous paragraph says that this breaks down for degree 0.