Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?
share|improve this question

1 Answer 1

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 $q$.

share|improve this answer
1  
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.