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I have a metric space $X$ and an isometric involution defined on it $i:X\rightarrow X$. My intuiton tells me that I can find a (continous) section $s:X/i \rightarrow X$. Is this true? Any references where I might read about similar situations? (For example studying when do covering spaces have sections could be similar)

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I haven't seen section used in that sense before, what would this mean with say $S^1$ and rotation by $\pi$? – muzzlator Feb 19 '13 at 20:12
Maybe I'm misusing the word. I want a continuous function $s:X/i \rightarrow X$ such that $\pi \circ s = id$, where $\pi$ is the canonical projection. – Sak Feb 19 '13 at 20:14
Sorry ignore my last comment, I think you may have to look at higher order spheres like $S^2$ and $\mathbb{R}P^2$. – muzzlator Feb 19 '13 at 20:17
Actually no, there is no section which will work for $S^1$ and $S^1 / \{-1,1\}$. If you continuously map $\{x,-x\}$ to $x$ or something, eventually you'll have two points not connected at the ends. – muzzlator Feb 19 '13 at 20:24
@muzzlator Why not post this as an answer? – user53153 Feb 20 '13 at 1:57
up vote 1 down vote accepted

Consider the mapping of $S^1$ into $S^1 / \{-x, x\}$. If we are to find a continuous section $s : S^1 / i \rightarrow S^1$, then because $S^1 / i$ is connected, the image must be too. This means that you will get the image of $s$ being a semi-circle. Let $x_0$ be a point where the semicircle breaks. Then $s$ isn't continuous at $\pi(x_0)$.

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