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Take the closed interval [0,1] with the equivalence relation $t\sim t$, and $0\sim 1$. Let $\pi:[0,1]\rightarrow[0,1]/\sim$ be the quotient map. The set $U=[0,\frac{1}{2})$ is open in [0,1]; show that $\pi(U)$ is not open in $[0,1]/\sim$.

My confusion is that $V\subseteq X/\sim$ is open if and only if $\pi^{-1}(V)$ is open in $X$, so if we let $X=[0,1]$ and $V=\pi(U)$, shouldn't $\pi(U)$ be open?

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    $\begingroup$ Actually, $\pi$ is a homeomorphism. Are you sure that only $t\sim t$ and not perhaps also $0\sim 1$? $\endgroup$ – Hagen von Eitzen Dec 6 '17 at 20:39
  • $\begingroup$ Rereading the question, I actually think that is what was meant. I will edit. $\endgroup$ – mrose Dec 6 '17 at 20:57
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$\pi(U)$ isn't open because the equivalence class $[0]$ of an endpoint isn't an interior point of $\pi(U)$. (This is quite clear if you think of $[0,1]/{\sim}$ as a circle.) Indeed, any neighborhood $N$ of $[0]$ is pulled back by $\pi$ to a neighborhood of $\{0,1\}$ in $[0,1]$, which contains a subset of the form $[0,\epsilon) \cup (\epsilon,1]$ for some small enough $0 < \epsilon < 1$. Thus $N = \pi(\pi^{-1}(N))$ contains $[1 - \frac{\epsilon}{2}]$, whereas $[1 - \frac{\epsilon}{2}] \notin \pi(U)$.

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