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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?

1 Answers 1

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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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    Oh! So it has it be continuous as a map from [0,1**]** to $[0,1)$, not [0,1**)** to $[0,1)$! I see, it's all clear to me now! Thank you very much!!2012-02-02