Vrftsjtryk

I'm self-studying algebraic topology and need help with the following problem (I'm only at part a.)

The relevant definitions are as follows.

Definition: Let F be a discrete space and X be any space. Then X $times$ F is a disjoint union of copies of X, indexed by F. The projection $pi:$ X $times$ F $to$ X is called a trivial covering.

Definition: A map $p:tilde{X} to X$ is a covering map if it is locally a trivial covering. That is, if X has an open cover ${N_alpha: alpha in A}$ by trivializing neighborhoods for p, i.e. there exists discrete $F_alpha$, and a homeomorphism $varphi_alpha: p^{-1}(N_alpha) to N_alpha times F_alpha$ such that $p = pi circ varphi_alpha$ on $p^{-1}(N_alpha)$.

In terms of diagram, my interpretation of the problem is as follows.

To show that $p$ is a covering map, I guess I'd have to find the discrete space $F$ and the homeomorphism $varphi_alpha$, but I don't know how to proceed. Also, I understand that $mathbb{R}P^n$ is the space obtained by identifying antipodal points of $S^n$, but I can't figure out what its open sets look like.

Your picture is misleading because it assumes that $p$ is a trivial covering. This is not true, it is only a locally trivial covering. To see that, define
$U_i^pm = { (x_1,dots,x_{n+1}) in S^n mid (-1)^{pm 1} x_i > 0}$. These set are the intersections of $S^n$ with open half-spaces in $mathbb{R}^{n+1}$, thus open subsets of $S^n$. Note that the $U_i^pm$ cover $S^n$.

We have $x= (x_1,dots,x_{n+1}) in U_i^+$ if and only if $- x = (-x_1,dots,-x_{n+1}) in U_i^-$. Hence $p(U_i^+) = p(U_i^-)$, and we denote this subset of $mathbb{R}P^n$ by $V_i$.

$V_i$ is open in $mathbb{R}P^n$ because $p^{-1}(V_i) = U_i^+ cup U_i^-$. Clearly $p_i^pm : U_i^pm stackrel{p}{rightarrow} V_i$ is a bijection. It is even a homeomorphism because it maps open sets to open sets.

Now let $F = {+1,-1 }$. Then we get a homeomorphism
$$phi_i : p^{-1}(V_i) to V_i times F, phi_(x) =
begin{cases}
(p_i^+(x),+1) & x in U_i^+ \
(p_i^-(x),-1) & x in U_i^-
end{cases}
$$