As we know, $S^0\to S^n \to \mathbb RP^n$ is a fiber bundle with the usual covering space projection. I wonder if we can construct a projection $S^n \to S^n$ such that $S^0 \to S^n \to S^n$ becomes a fiber bundle.
I need this result to complete the proof in here.
The long exact sequence in homotopy gives
$$\dots \to \pi_1(S^n) \to \pi_0(S^0) \to \pi_0(S^n) \to \pi_0(S^n) \to 0.$$
If $n = 0$, then we get an exact sequence $0 \to \pi_0(S^0) \to \pi_0(S^0) \to \pi_0(S^0) \to 0$, but this is impossible so $n > 0$. Then $\pi_0(S^n) = \pi_0(S^n) = 0$, so $\pi_1(S^n)$ surjects onto $\pi_0(S^0) \cong \mathbb{Z}_2$. As $\pi_1(S^n) = 0$ for $n \neq 1$, we see that if an example exists, then we must have $n = 1$.
Such an example exists. The map $S^1 \to S^1$, $z \mapsto z^2$ is a double covering, i.e. a fibre bundle $S^0 \to S^1 \to S^1$.