Homotopy classification of selfmaps of product of spheres? Self-maps of n-torus $T^n=S^1\times ...\times S^1$ are classified by the induced homorphism of fundamental group $\pi_1 T^n=Z^n$.
Is a similar result true form self-maps of $S^k\times ...\times S^k$ (n times)? Are they  classified by homomorphisms of
 $\pi_k(S^k\times ...\times S^k)=Z^n$?
 A: No chance.  For example, take the self maps of $S^3 \times S^3$. Then based maps gives
$$
\text{maps}_\ast(S^3\times S^3,S^3 \times S^3) = \text{maps}_\ast(S^3\times S^3,S^3) \times \text{maps}_\ast(S^3\times S^3,S^3)\, .
$$
On the other hand $S^3$ has the homotopy type of $\Omega X$ where $X= BS^3$
is the classifying space for the topological group $S^3 = SU(2)$.
Inserting this in and taking adjoints we get
$$
\text{maps}_\ast(S^3\times S^3,S^3 \times S^3) \simeq 
\text{maps}_\ast(\Sigma(S^3\times S^3),X) \times \text{maps}_\ast(\Sigma(S^3\times S^3),X)\, .
$$
Finally, for any based CW complexes $A,B$ there's a equivalence $\Sigma (A\times B) \simeq \Sigma A \vee \Sigma B \vee \Sigma (A\wedge B)$, so inserting and adjointing back, we get
$$
\text{maps}_\ast(S^3\times S^3,S^3 \times S^3) \simeq
(\Omega^3S^3)^{\times 4} \times (\Omega^6 S^3)^{\times 2}\, .
$$
Taking $\pi_0$ we obtain
$$
[S^3\times S^3,S^3 \times S^3] \cong  \Bbb Z^{\oplus 4} \oplus  
 (\Bbb Z_{12})^{\oplus 2} \, .
 $$
The torsion term on the right shows this is a counterexample.
Note Added, December 31: The equivalence $\Sigma (A\times B) \simeq \Sigma A \vee \Sigma B \vee \Sigma (A\wedge B)$ is not a map of co-H spaces, so the  displayed isomorphism is a priori just one of sets, not groups. However, there's definitely a short exact sequence
$$
0 \to (\Bbb Z_{12})^{\oplus 2} \to [S^3\times S^3,S^3 \times S^3] \to \Bbb Z^{\oplus 4} \to 0 
$$
induced from the cofiber sequence $S^3 \vee S^3 \to S^3 \times S^3 \to S^6$. This sequence obviously splits, so the isomorphism we displayed is one of groups if we use the group structure on the middle term given from the topological group structure of $S^3 \times S^3$.
Note Added, January 2: Gustavo Granja (thanks!) pointed out that the seqeunce is not split. So what I claimed above is not correct at stated. What's remains true is firstly,
$$
\text{maps}_\ast(S^3\times S^3,S^3 \times S^3) \simeq
(\Omega^3S^3)^{\times 4} \times (\Omega^6 S^3)^{\times 2}\, 
$$
as spaces. Secondly, we have a non-split short exact sequence of groups
$$
0 \to (\Bbb Z_{12})^{\oplus 2} \to [S^3\times S^3,S^3 \times S^3] \to \Bbb Z^{\oplus 4} \to 0 
$$
So, what I give above is a still counterexample to the question.
