Higher homotopy groups of surface of genus 2, without using universal cover In a course of algebraic topology I am self-studying I encountered the following problem: calculate the homotopy groups $\pi_n(S_2)$, where $S_2=T\#T$. Now I found a solution online involving taking its universal covering, which is the hyperbolic half-plane (I think?). But I have no knowledge of hyperbolic spaces, and it this topic is well beyond the content of this course. Now that I am looking for is some other, more basic, but possibly less general (i.e. not generalizable to surfaces of genus $g\geq 2$) solution.
Interestingly, I later found that this problem seems to have been removed from the problem list, so I am not sure if this simpler solution that I am looking for even exists. In fact, this may well be the main reason why I am interested in it?
TL;DR: What ways of computing $\pi_n(S_2)$, other than using its universal cover, do you know?
 A: There are a few tools for computing homotopy groups of connected sums, but they're narrow in scope and involve some complicated machinery. (That having been said, the case of $\pi_1$ can be handled easily with the Seifert-van Kampen theorem. You might be able to get the low $\pi_n$ for free in some cases with the Hurewicz theorem, but the manifolds here aren't simply connected.) Homotopy groups are pretty complicated to compute in general; the easiest techniques if you don't have a convenient covering space are using the sequence from a fibration, somehow reducing the problem to homology or cohomology, computing a minimal model (for rational homotopy, at least), or using other properties of the manifold in question (e.g., compact Lie algebras have $\pi_2 = 0$). At the very least, I would expect any result on the homotopy of the connected sum of $n$-manifolds to depend on that of $S^n$ (as in excision or Seifert-van Kampen), which is already a ridiculously complicated thing.
Instead, use the fact that for a universal cover $\tilde X \to X$ with $X$ connected, the induced maps $\pi_n(\tilde X) \to \pi_n(X)$ are isomorphisms for $n\geq 2$. (I should probably mention that we're working in something like the CW-category, but the specific problem here deals with smooth closed surfaces.) This follows immediately from the long exact sequence for a fibration; Hatcher, for example, might have a more direct proof. Closed surfaces of genus $g> 0$ have contractible universal cover, which follows from their construction from the identification of a $4g$-polygon in the plane. (The setup isn't immediate, but it can be found in, say, Hatcher.) The fact that the cover is actually hyperbolic space isn't immediately relevant, since we just care that it's contractible, and being hyperbolic is a geometric property.
(Of course, geometric and topological properties are closely connected, especially in low dimensions (e.g., Gauss-Bonnet) and especially for hyperbolic manifolds (e.g., Mostow rigidity). Oriented surfaces of genus $g > 1$, like the one in this problem, admit a hyperbolic metric. Thus it's covered by $\mathbb{H}^2$ by the Cartan-Hadamard theorem. (In short, the idea behind the theorem is that the exponential map on the tangent space turns out to be a covering map, with completeness ensuring that it's defined everywhere and the negative curvature showing that it's a local diffeomorphism. (The latter in particular requires a bit of machinery but is straightforward.))
A: There is a way to do this without knowing apriori any of the (geometric) structure for $\mathbb H^2$.
Start with the usual octogon, denoted  $D_1$, and take the quotient map to $\Sigma_2$. Then what you want to do is add an octogon to each edge of $D_1$ an octogon (it definitely cannot be a regular octogon) and continue in this way to get all of $\mathbb R^2$ which is contractible and implies the result by functoriality and some lifting.
A different way to see it is by taking the long exact sequence
$$\pi_n(F) \to \pi_n(\mathbb R^2) \to \pi_n(\Sigma_2) \to \pi_{n-1}(F)$$
but since $F$ is discrete, $\pi_n$ vanish for $n \geq2$, and since $\pi_n(\mathbb R^2)=0$, you get that $\pi_n(\Sigma_2)=0$.
