I'd like some verification for the following two claims here on page four (b) and (c). That is, if we first suppose $S^{2n}$ covers $X$, then if $d:S^{2n} \rightarrow S^{2n}$ is a nontrivial deck transformation we know that $deg(d) = (-1)^{2n+1}$.
Similarly if $S^{2n+1}$ covers $X$ then nontrivial deck transformations have $deg(d) = (-1)^{2n+2}$.
Here's what I've tried thinking about. In my book (Bredon), we define the degree of a map $d:S^n \rightarrow S^n$ to be the number $a$ such that $d_*(\gamma) = a \gamma$ for all $\gamma \in \tilde{H}_n(S^n) \simeq \mathbb{Z}$. This question here showed me that only $\mathbb{Z}_2$ acts freely on $S^{2n}$. I don't know why that is, but assuming it is true this means that the action is like the antipodal identification giving us projective space, which I have shown before has degree $-1$ when doing problem related to the Hairy Ball Theorem.
Deck transformations of connected covers are determined by their value on one point which means nonidentity transformations have no fixed points. Any map $S^n \rightarrow S^n$ with no fixed points is homotopic to the antipodal map which makes its degree $-1^{n+1}$.