Proof that if $S^{k}\rightarrow S^{m}\rightarrow S^{n}$ is a fiber bundle, then $k=n-1$ and $m=2n-1$. I was reading a proof  online (7a) here: http://www.math.wisc.edu/~dummit/sets/752-fs.pdf
that if $S^{k}\rightarrow S^{m}\rightarrow S^{n}$ is a fiber bundle, then $k=n-1$ and $m=2n-1$ but I'm somewhat confused on a number of things....
1. What is the exactness be is refering to? We don't have a sequence of groups
2. What is the relevance of  $\mathbb{RP}^{n}\not\cong S^{n} \: \forall n>1$?
3. Why $n=m=1$ not just $n=m$?
4. Should it not be the case where $1\leq k\leq n-2$ that should be considered instead of  $1\leq j\leq n-2$?
 A: *

*The takeaway point from what he calls "exactness" is that the dimension of the total space
($S^m$) must be the sum of the dimension of the fiber ($S^k$) and the dimension of the base ($S^n$). (You can sort of think of a short exact sequence of maps $0 \to S^k \to S^m \to S^n \to 0$, but this is a little loose.)

*If $k = 0$, then the fiber is $S^0$, which consists of two points (the two points of distance 1 from the origin in $\mathbb{R}$). In that case, $n = m$ by the above remark, and the map $S^m \to S^m$ must be a double cover. I think this means that the base $S^m$ must be homeomorphic to $RP^m$, which only occurs when $m = 1$.
(Edit: to clarify the last point, if $m > 1$, then $S^m$ is simply connected ($\pi_1(S^m) = 0$), so there cannot exist a double cover $S^m \to S^m$, since such a double cover would imply that the fundamental group of the base $S^m$ is $\mathbb{Z}/2\mathbb{Z}$, which is a contradiction.)

*This is answered by my answer to question 2.

*$1 \leq j \leq n-2$ is correct. He is thinking of $k$ and $n$ being fixed (giving $m = n + k$), and he is making a statement about $\pi_j(S^k)$ for $1 \leq j \leq n-2$.
