Exercise 60.2

Question

Let $X$ be the quotient space obtained from $B^2$ by identifying each point $x$ of $S^1$ with its antipode $-x$. Show that $X$ is homeomorphic to the projective plane $P^2$.

Amit Awasthi · Accepted Answer

\begin{proof}
  Consider $X$ as constructed from $B^2$ identified with the closed upper
  hemisphere of $S^2$. Let $p\colon S^2 	o P^2, q\colon B^2 	o X$ be the
  quotient maps, and $\pi\colon S^2 	o B^2$ the map sending $x$ to either itself or $-x$ if $x 
otin B^2$; note this is a quotient map since $U \subseteq S^2$ open implies $\pi(U)$ open, and $V \subseteq B^2$ open implies $\pi^{-1}(V) = V \cup -V$ open. We then have the commutative diagram
  \begin{center}
    \begin{tikzcd}
      S^2 \arrow[swap]{d}{p}\arrow{r}{\pi} & B^2 \arrow{d}{q}\
      P^2 \arrow[dashed]{r}{f} & X
    \end{tikzcd}
  \end{center}
  and since $q \circ \pi$ is a quotient map by p.~141, we see $f$ is a quotient map by Theorem $22.2$. But since for $x \in X$, $(q \circ \pi)^{-1}(x) = \{x,-x\} \in P^2$, and moreover any equivalence class $\{x,-x\}$ can be realized as the inverse image of this type, $f$ is a bijection and $P^2 = \{(q \circ \pi)^{-1}(x) \mid x \in X\}$, and so $f$ is a homeomorphism by Corollary $22.3(a)$.
\end{proof}

Exercise 60.2

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Exercise 60.2

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