Homotopy group of rank 2 of various manifolds I have taken a group theory course (undergrad physics), we did in a very short way the superior homotopy groups (to say, 1 hour of lesson) but it was not clear at all, plus we knew nothing of topology so guess how much we have understood...anyway he left as exercise this in the picture. Can you help me with the ones after $S^9$? (We have made no theorems, the only explicit example was $\pi_2(S^2)$ which we calculated).

 A: This seems like an extremely unreasonable assignment to me. 
Anyway, $\mathbb{R}^3 \setminus \{ a, b \}$ deformation retracts onto a wedge sum of two $2$-spheres $S^2 \vee S^2$. By the Seifert-van Kampen theorem this space is simply connected, so by the Hurewicz theorem its $\pi_2$ is isomorphic to its $H_2$. Its $H_2$ can be calculated e.g. by Mayer-Vietoris to be $\mathbb{Z} \oplus \mathbb{Z}$, one copy for each $2$-sphere. Kostas' answer correctly describes the intuition but proving it rigorously requires real work; I'm not sure what's being asked here. 
$\mathbb{R}^3 \setminus \mathbb{R}$ is homeomorphic to $\mathbb{R} \times (\mathbb{R}^2 \setminus \{ 0 \})$, which deformation retracts onto $\mathbb{R}^2 \setminus \{ 0 \}$, which further deformation retracts onto $S^1$. $\pi_2(S^1)$ can be computed by passing to the universal cover of $S^1$, which is $\mathbb{R}$, hence contractible. The conclusion is that $\pi_2(S^1)$ vanishes. 
A: $\mathbf{R}^3\setminus \{0\}$ deformation retracts onto the 2-sphere, so the answer is the same. For the next one, although it can be deduced by heavy machinery, intuitively it is $\mathbf{Z}^2$ (like how the fundamental group of the torus is that), because you can have a nonhomotopy trivial sphere around either $a$ or $b$, and you can take linear combinations of them. For the last part, what is $A$? 
