The fundamental group of $S^n$

Question

I want to prove that $\pi_1(S^n,x_0)$ is trivial if $2\leq n,$ BUT using universal covering. So let $p:\tilde S^n \rightarrow S^n$ the universal covering. Define $f:D^n\rightarrow S^n$ such that $f(\partial D)=x_0.$ Proof that $f$ rises to $f':D^n\rightarrow \tilde S^n$ and with this function $f'$ define $f'':S^n\rightarrow \tilde S^n$ such that $p\circ f''=id_{S^n}.$ With this applying the fundamental group to $p\circ f''$ get that $\pi_1(S^n,x_0)$ is the trivial group.

I have the definition of $f$ and is easy to prove that $f'$ exist, and is clear that if I find $f''$ with the property $p\circ f''=id_{S^n}$ and using the fundamental group like the hint I get the final of the proof, BUT I don't know how define $f''.$
Thanks.

Answer 1

Hint: Since $S^n \cong D^n/\partial D^n$, a map $f': D^n \to \tilde S^n$ that maps $\partial D^n$ to a point is equivalent to a map $f'':S^n \to \tilde S^n$. And keep in mind that $p \circ f'=f$. (Click below for the full solution.)

We should choose $f: (D^n,\partial D^n) \to (S^n,x_0)$ to be the quotient map $D^n \to D^n/\partial D^n=S^n$ (so it equivalent to the identity $S^n \to S^n$ in the above sense). We get a lift $f': D^n \to \tilde {S^n}$, but we need to check that $\partial D^n$ is mapped to a single point in $\tilde {S^n}$. Since $f': D^n \to \tilde {S^n}$ must satisfy $$p\circ f'(\partial D^n)=f(\partial D^n)=x_0,$$ $f'$ must map $\partial D^n$ to $p^{-1}(x_0)$. Then $f'(\partial D^n)$ is a single point because $p^{-1}(x_0)$ is discrete and $\partial D^n$ is connected. Since $S^n=D^n/\partial D^n$, we get a map $f'':(S^n,x_0)\to (\tilde{ S^n},\tilde x_0)$ such that $p\circ f''=\operatorname{id}_{S^n}$.

Moral of the story: $\,$ In an ideal world, we would directly construct a lift $f'': S^n \to \tilde S^n$ of $\operatorname{id}_{S^n}: S^n \to S^n$, since this implies that $p_*$ is surjective. The problem is that we're not sure that such a lift exists because we don't know $\pi_1(S^n)$, so we have to pass through $D^n$.