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Consider the quotient space of $S^1$ constructed by identifying antipodal points. that is $x\sim y$ if $x=\pm y$ $x,y\in S^1$. Now let $f:S^1\to S^1$ be the antipodal map $f(x)=-x$ and $p:S^1\to S^1/\sim$ be the quotient map. Let $U\subset S^1$ be open. Is it true that $p^{-1}(p(U))=U\cup f(U)? $

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  • $\begingroup$ Both $p^{-1}(p(U))$ and $U \cup f(U)$ can be shown to be the set of points in $S^1$ that are in the equivalence class of some point in $U$. $\endgroup$ – angryavian Jul 31 '17 at 20:59
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Let $U\subseteq S^1$ be open. For $x\in U\cup f(U)$, either $x\in U$, or $-x\in U$. Then, $p(x) = p(-x) = [x]\in p(U)$, and $p^{-1}([x]) = \{x, -x\}$, so $x\in p^{-1}(p(U))$. This implies that $U\cup f(U)\subseteq p^{-1}(p(U))$.

Now, for $x\in p^{-1}(p(U))$, we have that $p(x) = [x]\in p(U)$, which implies that $x\in U$ or $-x\in U$. As $-x\in U$ implies $x\in f(U)$, we have either $x\in U$ or $x\in f(U)$, so $x\in U\cup f(U)$. This implies that $p^{-1}(p(U))\subseteq U\cup f(U)$, so $$p^{-1}(p(U)) = U\cup f(U)$$ Clearly, $f$ is an open map, so for any open $U$, as $p^{-1}(p(U))$ is open in $S^1$, $p(U)$ is open in the quotient space topology on $X/\sim$. Therefore, $p : X\to X/\sim$ is open.

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