---- > [!definition] Definition. ([[projective plane]]) > The **projective plane** $\mathbb{R}P^{2}$ is the [[quotient space]] (a [[surface]]) obtained from $\mathbb{S}^{2}$ by identifying each point $x$ of $\mathbb{S}^{2}$ with its antipodal point $-x$. > [!basicproperties] > - $\mathbb{R}P^{2}$ is [[compact]] and the [[topological quotient map|quotient map]] $p:\mathbb{S}^{2} \to \mathbb{R} P^{2}$ is is a [[covering space|covering map]]. First we show $p$ is an [[open map]]. A set $V \subset \mathbb{R}P^{2}$ is open iff $p ^{-1}(V)=\bigsqcup_{v \in V}[v]$ is open in $\mathbb{S}^{2}$. Now let $U \subset \mathbb{S}^{2}$ be open in $\mathbb{S}^{2}$. Let $a$ be the antipodal map $x \xmapsto{a} -x$; it is a [[homeomorphism]] and therefore an [[open map]]. Hence $\begin{align} p ^{-1}p(U)= & \{ s \in \mathbb{S}^{2} : p(s) \in p(U) \} \\ = & \{ s \in \mathbb{S}^{2} : [s] = [u] \text{ for some } u \in U \} \\ = & \bigcup_{u \in U}^{}[u] \\ = & U \cup a(U) \end{align}$ is open. A similar proof shows that $p$ is a closed map. Now we show $p$ is a [[covering space|covering map]]. Let $y \in \mathbb{R}P^{2}$; WTS $y$ is [[evenly covered]] by $p$. Choose $x$ in $p ^{-1}(y)=\{ y,-y \}$. Wrap an $\varepsilon$-ball $U$ around $x$ such that $U$ contains no pair $\{ z, a(z) \}$ of antipodal points. Then $p |_{U}$ is [[bijection|bijective]]; being [[continuous]] and [[open map|open]] it is a [[homeomorphism]]. Similarly for $p |_{a(U)}$. Thus $p ^{-1} p(U)$may be partitioned into two slices and we conclude $p(U)$ is a neighborhood witnessing even covering of $y$. - The [[fundamental group]] of $\mathbb{R}P^{2}$ is [[group isomorphism|isomorphic]] to $\mathbb{Z} / 2\mathbb{Z}$. We will use [[path-connected covering space yields surjective lifting correspondence|this result]] regarding [[lifting correspondence derived from a covering map|the lifting correspondence]]... since $\mathbb{S}^{2}$ is [[simply connected]] ([[past one dimension, the sphere is simply connected|see here]]), $\pi_{1}(\mathbb{R}P^{2}, y_{0})$ is in [[bijection|bijective correspondence]] with $p ^{-1}(y_{0})$. The latter set has two elements, so $\pi_{1}(\mathbb{R}P^{2}, y_{0})$ must have two elements also. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` #reformatrevisebatch04