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 $S^n$?
> 
> 
> 



Converse to Hopf degree theorem
*See the second half of the answer for a complete characterisation of closed orientable manifolds with the Hopf property.*




---


Note that $X$ having the Hopf property is equivalent to the injectivity of $\deg : [M, X] \to \mathbb{Z}$ for every $M$; here $[M, X]$ denotes the free homotopy classes of maps $M \to X$.


If $X$ is simply connected, then for $M = S^n$ we have $[S^n, X] = \pi\_n(X)$ - in general we would obtain a quotient of $\pi\_n(X)$ by an action of $\pi\_1(X)$. With respect to the group operation $\ast$ in $\pi\_n(X)$, we have $\deg([f]\ast [g]) = \deg([f]) + \deg([g])$, so $\deg : \pi\_n(X) \to \mathbb{Z}$ is a group homomorphism which must be injective if $X$ has the Hopf property.



> 
> If $X$ is a simply connected manifold with the Hopf property, then either $\pi\_n(X) = 0$ or $\pi\_n(X) \cong \mathbb{Z}$. Moreover, if $\pi\_n(X) \cong \mathbb{Z}$, then $X$ is a rational homology sphere.
> 
> 
> 


For the final claim, note that $\pi\_n(X) \cong \mathbb{Z}$ means that there exists a map $S^n \to X$ of non-zero degree. It then follows from Poincaré duality that $X$ must be a rational homology sphere, see [here](https://math.stackexchange.com/q/853668/39599).


**Example:** This observation can be used to show that many non-trivial products involving simply connected spheres do not have the Hopf property. For instance, for any simply connected closed manifold $Y$, the product $S^2\times Y$ does not have the Hopf property. To see this, let $n = \dim Y > 1$ and note that $\pi\_{n+2}(S^2\times Y) \cong \pi\_{n+2}(S^2)\oplus\pi\_{n+2}(Y) \cong F\oplus\pi\_{n+2}(Y)$ for some non-trivial finite group $F$.




---


Thanks to Nick L's observation in the comments below, we have a complete characterisation of closed orientable manifolds with the Hopf property.



> 
> $X$ has the Hopf property if and only if $X$ is homeomorphic to $S^n$.
> 
> 
> 


**Proof:** If $X$ is homeomorphic to $S^n$, then it has the Hopf property by the Hopf degree theorem.


Suppose that $X$ is not homeomorphic to $S^n$. Then there is $0 < k < n$ such that $\pi\_k(X) \neq 0$. Choose an essential map $f : S^k \to X$ and define $F : S^k\times S^{n-k} \to X$ by $F(x, y) = f(x)$. If $i : S^k \to S^k\times S^{n-k}$ denotes an inclusion into the first factor, then $F\circ i = f$, so the composition


$$\pi\_k(S^k) \xrightarrow{i\_\*} \pi\_k(S^k\times S^{n-k}) \xrightarrow{F\_\*} \pi\_k(X)$$


is precisely $f\_\* : \pi\_k(S^k) \to \pi\_k(X)$ which is determined by $[\operatorname{id}] \mapsto [f]$. As $f\_\* \neq 0$, we see that $F\_\* \neq 0$ and hence $F : S^k\times S^{n-k} \to X$ is essential. Now note that $F = f\circ\operatorname{pr}\_1$ where $\operatorname{pr}\_1 : S^k\times S^{n-k} \to S^k$ denotes projection onto the first factor. Since $F$ factors through $S^k$ and $k < n$, the map $F$ has degree zero. As $F$ has degree zero but is not nullhomotopic, $X$ does not have the Hopf property. $\square$


