How to show these spaces are connected?

Question

Let $p:\mathbb{S}^n\to\mathbb{P}^n(\mathbb{R})$ be the natural projection map, which is continuous, closed, open and surjective. I want to show it is a covering map.

Let $x\in \mathbb{S}^n$, and define $U=\{y\in\mathbb{S}^n:(x,y)\ne0\}$ where $(x,y)$ denotes the inner product of $\mathbb{R}^{n+1}$. Then $U$ is open in $\mathbb{S}^n$ and the sets $U_+=\{y\in\mathbb{S}^n:(x,y)>0\}$ and $U_-=\{y\in\mathbb{S}^n:(x,y)<0\}$ are disjoint, open and closed in $U$. I want to show they are the components of $U$, but i don't see how to show they are connected. I was thinking about finding a continous map whose domain is connected and whose image is, say, $U_+$; but I'm not able. Any hints or suggestion? Thank you in advance :)

Answer 1

Define $V_+ = \{ y \in \mathbb{R}^{n+1} \mid (x,y) > 0 \}$ It is easy to verify that this set is convex, hence path connected (and especially connected). Moreover $U_+ \subset V_+ \subset \mathbb{R}^{n+1} \setminus \{ 0 \}$. We have a retraction $r : \mathbb{R}^{n+1} \setminus \{ 0 \} \to S^n, r(y) = y/\lVert y \rVert$. Trivially $U_+ = r(U_+) \subset r(V_+)$. In fact, we have $r(V_+) = U_+$: If $(x,y) > 0$, then $(x,r(y)) = (x,y/\lVert y \rVert) = (x,y)/\lVert y \rVert > 0$. This shows that $U_+$ is connected.

The connectedness of $U_-$ is shown analogously.

Remark:

You mentioned in your comments that the definition of covering map you are using requires evenly covered sets to be connected. In my opinion this is unusual. However, if the base space $X$ of $p : Y \to X$ is locally connected, then this agrees with the usual definition. To see this, take any open neigborhood $U$ of $x$ which is evenly covered, i.e. $p^{-1}(U)$ is the disjoint union of open $V_\alpha$ which are mapped by $p$ homeomorphically onto $U$. By local connectedness there is a connected open neighborhood $U'$ of $x$ such that $U' \subset U$. Let $V'_\alpha = p^{-1}(U') \cap V_\alpha$. Then $U'$ is evenly covered with connected sheets $V'_\alpha$.

What might be the reason to use your definition? If you require the sheets to be connected, then they are the components of $p^{-1}(U)$. In particular, the set of sheets is uniquely determined. This is no longer true for non-connected sheets: You will find various decompositions of $p^{-1}(U)$ into disjoint open sets which are mapped by $p$ homeomorphically onto $U$.

In your question $S^n$ is locally connected and $p$ is easily seen to be an open map, hence also $\mathbb{P}^n(\mathbb{R})$ is locally connected.

Answer 2

Let $E:\mathbb{R}^n\to x^\perp\subset\mathbb{R}^{n+1}$ be a linear isometry ($x^\perp$ is $n$-dimensional). Then the map $f:\mathbb{R}^n\to\mathbb{S}^n$ defined by $$f(y):=\sqrt{1-\|y\|^2}x/\|x\|+Ey$$ is continuous and maps the open unit disk onto $U_+$. Note that $$\|f(y)\|^2=(1-\|y\|^2)+\|y\|^2=1$$ since $Ey\perp x$. Also for any $p\in U_+$, take $\|x\|=1$ without loss of generality, and let $y:=E^{-1}(p-\langle p,x\rangle x)$, then $$\|y\|^2=\|p-\langle p,x\rangle x\|^2=1-\langle p,x\rangle^2<1$$ and $f(y)=\langle p,x\rangle x+p-\langle p,x\rangle x=p$.