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What is an example of

  1. a topological embedding that has no continuous left inverse?
  2. a quotient map that has no continuous right inverse?
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1 Answer

up vote 2 down vote accepted

For (1), let $f:[0,1)\to[0,1]:x\mapsto x$.

For (2), let $X=\{0,1\}\times[0,1]$, let $A=\{0,1\}\times\{0\}$, and let $q:X\to X/A$ be the quotient map. In otherwords, $X$ is the disjoint union of two copies of the $[0,1]$, and $X/A$ is obtained by identifying the two copies of $0$. Clearly $X/A$ is homeomorphic to $[0,1]$, so any continuous map from $X/A$ to $X$ must map $X/A$ to a closed interval in either $\{0\}\times[0,1]$ or $\{1\}\times[0,1]$. Thus, a continuous $f:X/A\to X$ that is a right inverse for $q$ must map $X/A$ onto either $\{0\}\times[0,1]$ or $\{1\}\times[0,1]$, taking $q(\langle 0,1\rangle)$ and $q(\langle 1,1\rangle)$, the two endpoints of $X/A$, to the endpoints of that segment. But then $f$ cannot take $q(\langle 0,0\rangle)=q(\langle 1,0\rangle)$ to an endpoint of $\{0\}\times[0,1]$ or $\{1\}\times[0,1]$ therefore can’t be a right inverse for $q$.

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Isn't $[0, 1] \to S^1$ an easier counterexample for (2)? –  Zhen Lin Feb 2 '12 at 20:56
@ZhenLin: I didn’t think about it very hard; I just gave the first one that occurred to me. Basically the same idea either way. –  Brian M. Scott Feb 2 '12 at 21:05
This might be stupid, but how does (1) work? $f$ is clearly an embedding, and it has an inverse $g : [0,1) \subset [0,1] \to [0,1): x \to x$, but I don't see why $g$ cannot be continuous?.. =\ –  Rick Feb 2 '12 at 23:21
And, yes, @BrianM.Scott, I really like your counter-example for (2), even though it is a little more complicated than $[0,1] \to S^1$. –  Rick Feb 2 '12 at 23:22
@Rick: A left inverse for that map is a map $g:[0,1]\to[0,1)$ such that $g\circ f$ is the identity on $[0,1)$: the domain of $g$ has to be $[0,1]$. (If $f:X\to Y$ is an embedding, $f^{-1}:f[X]\to X$ is always continuous.) –  Brian M. Scott Feb 2 '12 at 23:33
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