Homotopy groups of compact surfaces I want to calculate the higher homotopy groups of $\Sigma_g$ and $\mathbb{R}P^2\# \mathbb{R}P^2\#\cdots\# \mathbb{R}P^2$. But I haven't found the methods to calculate the homotopy groups of connected sum. So how to solve this question?
 A: If $X$ is a path-connected topological space nice enough so that universal cover $\tilde{X}$ exists, then the covering map $p : \tilde{X} \to X$ induces an isomorphism $p_* : \pi_n(\tilde{X}, \tilde{x_0}) \to \pi_n(X, x_0)$ for all $n > 1$. This is a simple application of the functoriality of $\pi_n$ coupled with the lifting lemma (key point is $S^n$ for $n > 1$ is simply connected). You should work it out if you don't know this fact already.
Universal cover of $M_1 = T^2$ is just $\Bbb R^2$. For $g > 1$, it can be seen that it is the upper half plane $\Bbb H^2$ or the open unit disk: there is a triangulation of $\Bbb H^2$ by geodesic $4g$ sided polygons, and an appropriate isometry group of $\Bbb H^2$ acts so that the quotient identifies the sides of the fundamental polygon appropriately. Alternatively apply classification of $2$-manifolds: there are not too many simply connected noncompact $2$-manifolds out there (noncompact because $\pi_1(M_g)$ is infinite, so the cover has to be infinite sheeted).
In both cases, the universal cover is contractible. Thus, $\pi_n(M_g) \cong \pi_n(D^2) = 0$ for $n > 1$ whenever $g \neq 0$. The nonorientable surface with $n$ cross caps has the same universal cover as the orientable surface of genus $n - 1$ (why?), so from that you can similarly conclude. Note that for $g = 0$ we would end up computing $\pi_n(S^2)$ which is not quite entirely trivial so I'm not including it. 
